On the Efficiency of Random Permutation for ADMM and Coordinate Descent

Ruoyu Sun, Zhi-Quan Luo, Yinyu Ye

Introduction

A simple yet powerful idea for solving large-scale computational problems is to iteratively solve smaller subproblems. The applications of this idea include coordinate descent (CD), POCS (Projection onto Convex Sets), SGD (Stochastic Gradient Descent). They are well suited for large-scale unconstrained optimization problem (see, e.g. Wright , for a recent survey of CD) since it decomposes a large problem into small subproblems. The decomposition idea is crucial for huge problems due to both the cheap per-iteration cost and small memory requirement. Moreover, this idea is “orthogonal” to other large-scale optimization ideas such as first-order methods (using only gradient information) and random projection, and thus can be easily combined with other ideas.

This paper is motivated by a natural question: how should we extend the decomposition idea to solve problems with constraints? We consider a constrained minimization problem with a convex objective function and linear constraints (this is for motivation; our analysis is for a much simpler version):

To apply the decomposition idea to a constrained problem, one possible way is to form the augmented Lagrangian function and perform coordinate descent for the primal problem and a gradient step for the dual problem, i.e. combining BCD with augmented Lagrangian method, to obtain the so-called alternating direction method of multipliers (ADMM). ADMM was originally proposed in Glowinski and Marroco (see also Chan and Glowinski , Gabay and Mercier ) to solve problem (1) when there are only two blocks (i.e. n=2n=2) and the objective function is separable. It is natural and computationally beneficial to extend the original ADMM directly to solve the general nn-block problem (1) via the following procedure:

The convergence of the direct extension of ADMM to multi-block case had been an open question, until a counter-example was recently given in Chen et al. . More specifically, Chen et al. showed that even for the simplest scenario where the objective function is and the number of blocks is 33, ADMM can be divergent for a certain choice of A=[A1,A2,A3]A=[A_{1},A_{2},A_{3}]. There are several proposals to overcome the drawback (see, e.g., ), but they either need to restrict the range of original problems being solved, add additional cost in each step of computation, or limit the stepsize in updating the Lagrange multipliers. These solutions typically slow down the performance of ADMM for solving most practical problems. Moreover, it is not clear how to compare the convergence speed of these algorithms as they typically contain different parameters. One may ask whether a “minimal” modification of cyclic multi-block ADMM (2) can lead to convergence, and whether we can provide some convergence speed analysis that is easy to interpret.

One of the simplest modifications of (2) is to add randomness to the update order. Randomness has been very useful in the analysis of block coordinate descent (BCD) methods and stochastic gradient descent (SGD) methods. In particular, a recent work Sun and Ye showed that randomized CD (R-CD) can be up to O(n2)O(n^{2}) times faster than cyclic CD (C-CD) for quadratic minimization in the worst case, where nn is the number of variables Rigorously speaking, these two bounds are not directly comparable since the result for the randomized version only holds with high probability, while the result for the cyclic version always holds; anyhow, this O(n2)O(n^{2}) gap is still meaningful if ignoring this difference between deterministic and randomized algorithm.. Another example is the comparison of IAG (Incremental Aggregated Gradient) in Blatt et al. and its randomized version SAG (Stochastic Average Gradient) : it turns out that the introduction of randomness leads to better iteration complexity bounds. There is also some study on randomly permuted version of pure SGD . These examples show that randomization may improve the algorithm in theory and in practice.

It is important to note that the iteration complexity bounds for randomized algorithms are usually established for independent randomization (sampling with replacement), while in practice, random permutation (sampling without replacement) has been reported to exhibit faster convergence (e.g. Shalev et al. , Recht and Re , Sun ). Interestingly, our simulation shows that for solving linear system of equations, randomly permuted ADMM (RP-ADMM) always converges, but independently randomized versions of ADMM can be divergent even for Gaussian data. Therefore, we focus on the analysis of RP-ADMM in this paper.

Random permutation is known to be notoriously difficult to analyze. Even for unconstrained quadratic minimization, the convergence rate of RP-BCD is poorly understood. Many existing works treated cyclic BCD and RP-BCD together , and thus the best known convergence rate of RP-BCD for general convex problems are in fact the same as that of C-BCD . However, in light of a recent study which established an up to O(n2)O(n^{2}) gap between cyclic CD and R-CD , it is unlikely that RP-CD has the same convergence rate as C-CD since that would imply RP-CD could be O(n2)O(n^{2})-times slower than R-CD. For the special example that demonstrates the gap between C-CD and R-CD, it was shown recently that RP-CD is faster than R-CD This paper appeared after the first version of the current paper. . However, the general quadratic case seems to be quite difficult, probably due to its close connection to a matrix AM-GM (algebraic mean-geometric mean) inequality , the difficulty of which is essentially to prove an inequality in non-commutative algebra.

We consider two extremes of a general RP-ADMM: i) the objective is zero, i.e., RP-ADMM for solving a linear system; ii) the constraint is zero and the objective is a quadratic function, i.e., RP-BCD for solving quadratic minimization. Due to the lack of understanding of random permutation for quadratic minimization as discussed previously, we restrict to the two cases in this paper.

The first result of this paper is the expected convergence of RP-ADMM for solving linear systems. More specifically, when the objective function is zero and the constraint is a non-singular square linear system of equations, the expected output of randomly permuted ADMM converges to the unique primal-dual optimal solution. A major technical result in this proof is that the eigenvalues of the expected iteration matrix of RP-BCD for quadratic problems lie in (−1/3,1)(-1/3,1), instead of the typical range (−1,1)(-1,1).

The second result is about the expected convergence rate of RP-ADMM for solving linear systems and RP-BCD for solving quadratic problems. We show that RP-BCD for a convex quadratic minimization problem with equal diagonal entries has expected iteration complexity O(nλavgλmin⁡log⁡(1/ϵ))O(n\frac{\lambda_{\text{avg}}}{\lambda_{\min}}\log(1/\epsilon)), where λavg\lambda_{\text{avg}} and λmin⁡\lambda_{\min} are the average eigenvalue and the minimum eigenvalue of the coefficient matrix, and one “iteration” here means a cycle of updating all blocks. This improves an existing bound of O(n2λavgλmin⁡log⁡(1/ϵ))O(n^{2}\frac{\lambda_{\text{avg}}}{\lambda_{\min}}\log(1/\epsilon)) for RP-BCD by a factor of nn. Built on this result, we further show that RP-ADMM for solving linear systems achieves the same expected iteration complexity bound O(nλavgλmin⁡log⁡(1/ϵ))O(n\frac{\lambda_{\text{avg}}}{\lambda_{\min}}\log(1/\epsilon)).

Technically, we provide a simple and clean proof of the expected convergence, by applying a classical result on the eigenvalues of Jordan product. For proving the expected convergence rate, we propose a new variant of the matrix AM-GM inequality conjecture, and prove a weaker version of this conjecture.

Our result shows that random permutation may be a good answer to the question “how to apply the decomposition idea to solve constrained problems”. As multi-block BCD is widely used for large-scale unconstrained problems, we expect multi-block RP-ADMM to be a good candidate for large-scale linearly constrained problems. Our result provides one of the few direct analyzes of random permutation in optimization algorithms, and offers an explanation of the mysterious gap between RP-ADMM and cyclic ADMM. As reflected by the proof, the intuition is that random permutation provides “3-level symmetrization” that adjusts the spectrum of the update matrix. Based on the analysis for RP-ADMM, we are able to improve the best known complexity of RP-BCD for equally-diagonal quadratic problems by a factor of nn, when expressing the complexity only in terms of the quantity λavgλmin⁡\frac{\lambda_{\text{avg}}}{\lambda_{\min}}.

2 Related Works

This paper is a stronger version of a previous technical report Sun et al. which was not published. Another related work is the paper Chen et al. , which modifies the proof of to make it work with a quadratic objective function.

We highlight a few novel contributions of the current paper (neither in the original technical report nor in the paper ).

(i) The current paper provides a much simpler proof for the result of expected convergence.

(ii) The current paper provides the first convergence rate analysis of RP-ADMM. See Theorem 4 and the proof in Section 4.5, Section 7.2 and Section 7.1.

(iii) The current paper provides an improved convergence rate analysis of RP-BCD, See Theorem 3 and the proof in Section 4.3 and Section 7.3.

(iv) The current paper introduces a theory-motivated algorithm Bernoulli-ADMM, which reduces the sampling time yet still achieves the expected convergence. This update order has not appeared before even in other algorithm setups to our knowledge. See Section 2.5 and Proposition 1.

Besides the technical contributions, we want to emphasize that the current paper is not just adding new result to our previous technical report , but actually completes a missing step of the story. From a mathematical point of view, the most striking consequence of our original proof is that the spectral radius of RP-BCD lies in a smaller region (−1/3,1)(-1/3,1). It is natural to think that this fundamental fact should have an impact on the analysis of original RP-BCD. Our current paper fills this gap by showing that this result can help build an O(n)O(n) gap between the (expected) onvergence rate of RP-BCD and cyclic BCD. A general message is that on one hand, to understand constrained optimization we have to understand unconstrained optimization (analyzing ADMM reduces to analyzing BCD); on the other hand, analyzing constrained optimization helps improve the understanding of unconstrained optimization (the analysis of ADMM leads to progress in BCD). We find this interaction between unconstrained optimization (BCD) and constrained optimization (ADMM) fascinating. The whole story is only revealed in the current paper, but not in the previous technical report or Chen et al. .

Besides the above unique aspects, the current paper inherits some interesting numerical findings from the technical report Sun et al. which do not appear in Chen et al. . We find that cyclic ADMM diverges with probability 1 for many random distributions of data, thus showing that the seemingly surprising divergence behavior reported in is quite common. However, it is easy to miss this finding if one uses the Gaussian distribution to generate data. Another interesting finding is that the independently randomized version of ADMM diverges with probability 1 for Gaussian data but not for the counter-example in , preventing us from analyzing the independently randomized version. Without these findings, the motivation of studying RP-ADMM would be less clear. See Section 2.4 and Section 8.

3 Notation and Organization

Organization. In Section 2, we present three versions of randomized ADMM, with an emphasis on RP-ADMM. In Section 3, we present our main results Theorem 1, Theorem 2 and their proofs. The subsequent sections are devoted to the proofs of the two technical results Lemma 1 and Lemma 2, which are used in the proof of Theorem 2. In particular, the proof of Lemma 1 is given in Section 5, and the proof of Lemma 2 is given in Section 6.

Algorithms

In this section, we will present both randomly permuted and independently randomized versions of ADMM for solving (1), and specialize RP-ADMM for solving a square system of equations. We also present a rather novel algorithm Bernoulli-randomized ADMM (motivated by our proof).

In this subsection, we first propose RP-ADMM for solving the general optimization problem (1), then we present the update equation of RP-ADMM for solving a linear system of equations.

At each round, we draw a permutation σ\sigma of {1,…,n}\{1,\dots,n\} uniformly at random from Γ\Gamma, and update the primal variables in the order of the permutation, followed by updating the dual variables in a usual way. Obviously, all primal and dual variables are updated exactly once at each round. See Algorithm 1 for the details of RP-ADMM. Note that with a little abuse of notation, the function L(xσ(1),xσ(2),…,xσ(n);μ)\mathcal{L}(x_{\sigma(1)},x_{\sigma(2)},\dots,x_{\sigma(n)};\mu) in this algorithm should be understood as L(x1,x2,…,xn;μ)\mathcal{L}(x_{1},x_{2},\dots,x_{n};\mu). For example, when n=3n=3 and σ=(231)\sigma=(231), L(xσ(1),xσ(2),xσ(3);μ)=L(x2,x3,x1;μ)\mathcal{L}(x_{\sigma(1)},x_{\sigma(2)},x_{\sigma(3)};\mu)=\mathcal{L}(x_{2},x_{3},x_{1};\mu) should be understood as L(x1,x2,x3;μ)\mathcal{L}(x_{1},x_{2},x_{3};\mu).

Throughout this paper, we assume AA is non-singular. Then the unique solution to (8) is x=A−1bx=A^{-1}b, and problem (7) has a unique primal-dual optimal solution (x,μ)=(A−1b,0)(x,\mu)=(A^{-1}b,0). The augmented Lagrangian function (3) for the optimization problem (7) becomes

Throughout this paper, we assume β=1\beta=1; note that our algorithms and results can be extended to any β>0\beta>0 by simply scaling μ\mu.

1.2 Example of 333-block ADMM

then the update equation (10) becomes Lˉyk+1=Rˉyk+bˉ\bar{L}y^{k+1}=\bar{R}y^{k}+\bar{b}, i.e.

1.3 General Update Equation of RP-ADMM

In general, for the optimization problem (7), the primal update (5) becomes

Replacing σ(i),σ(j),σ(l)\sigma(i),\sigma(j),\sigma(l) by i,j,li,j,l, we can rewrite the above equation as

where σ−1\sigma^{-1} denotes the inverse mapping of a permutation σ\sigma, i.e. σ(i)=t⇔i=σ−1(t)\sigma(i)=t\Leftrightarrow i=\sigma^{-1}(t). Denote the output of Algorithm 1 after round (k−1)(k-1) as

The update equations of Algorithm 1 for solving (7), i.e. (15) and (6), can be written in the matrix form as (when the permutation is σ\sigma and β=1\beta=1)

where Lˉσ,Rˉσ,Lσ,Rσ,bˉ\bar{L}_{\sigma},\bar{R}_{\sigma},L_{\sigma},R_{\sigma},\bar{b} are defined by

Another expression of LσL_{\sigma}, equivalent to (19), is the following:

A user-friendly rule for writing LσL_{\sigma} is described as follows (use σ=(231)\sigma=(231) as an example). Start from a zero matrix. First, find all reverse pairs of σ\sigma; here, we say (i,j)(i,j) is a reverse pair if ii appears after jj in σ\sigma. For the permutation (231)(231), all the reverse pairs are (1,3),(3,2)(1,3),(3,2) and (1,2)(1,2). Second, in the positions corresponding to the reverse pairs, write down the corresponding entries of ATAA^{T}A, i.e. a1Ta3,a3Ta2a_{1}^{T}a_{3},a_{3}^{T}a_{2} and a1Ta2a_{1}^{T}a_{2}, respectively. At last, write aiTaia_{i}^{T}a_{i} in the diagonal positions. Using this rule, we can write down the expression of L(231)L_{(231)} as

A user-friendly rule to quickly check the correctness of an expression of LσL_{\sigma} is the following (still take σ=(231)\sigma=(231) as an example). According to the order of the permutation (231)(231), the 22nd row, the 33rd row and the 11st row should have a strictly decreasing number of zeros (22 zeros, 11 zero and no zero). In contrast, the 22nd column, the 33rd column and the 11st column should have a strictly increasing number of zeros.

For the general case that di≥1,∀id_{i}\geq 1,\forall i, we can write down the block partitioned LσL_{\sigma} in a similar way. For example, when n=3n=3 and σ=(231)\sigma=(231), we have

2 Randomly Permuted BCD

RP-ADMM is a generalization of RP-BCD. In fact, when the constraint does not exist, RP-ADMM reduces to RP-BCD. In this subsection, we present RP-BCD for solving convex quadratic problems. Note that RP-ADMM for solving linear systems and RP-BCD fo solving quadratic problems are two extremes of general RP-ADMM: in the former case the objective function is zero, and in the latter case the constraint is zero. Interestingly, the two extreme cases are related as the expected iteration matrix of RP-BCD appears as a component of the expected iteration matrix of RP-ADMM. We will show later that their eigenvalues are closely related.

In the augmented Lagrangian function given in (9), if we delete the first term which depends on the dual variable μ\mu, we obtain the quadratic function 12∥Ax−b∥2\frac{1}{2}\|Ax-b\|^{2}. Thus if we eliminate the dual variable μ\mu in the update equations of RP-ADMM, we will obtain the update equations for RP-BCD. Suppose xkx^{k} is the iterate after the k-th epoch (i.e. go through all coordinates once), and σ\sigma is the order used in the kk-th iteration, then, as a simpler version of (17), we have

where LσL_{\sigma} and RσR_{\sigma} are defined as in (21) and (20), and σ\sigma is a random permutation.

3 Residual Trick for Efficient Implementation of ADMM and BCD

We note here that when di=1,∀id_{i}=1,\forall i (in this case BCD becomes CD), per-epoch computation time of ADMM and CD (no matter what order) is O(n2)O(n^{2}); or in other words, per-coordinate-update time is O(n)O(n). For instance, updating xk+1x^{k+1} by (24) in RP-BCD or updating yk+1y^{k+1} by (17) in RP-ADMM only takes time O(n2)O(n^{2}). As mentioned in Section 3.1 of , the trick is to keep track of the residual. For both efficient practical implementation and calculation of computation complexity, one should use this residual trick, but for the ease of theoretical analysis we use the matrix update forms (17) and (24) in this paper; there is no contradiction as our theory only depends on the value of xkx^{k} but not the specific procedure to compute xkx^{k}.

For completeness, we briefly explain how this trick works in our settings. Suppose di=1,∀id_{i}=1,\forall i, and we use CD methods to solve (23) with a certain update order (could be any order, such as cyclic, randomized or randomly permuted). Suppose the coordinate ii is picked, then xix_{i} is updated by by

where A−iA_{-i} contains all columns of AA except AiA_{i}, x−ix_{-i} contains all elements of xx except xix_{i} and represents the current values, and xi+x_{i}^{+} represents the new value. A straightforward implementation of (25) requires multiplying x−ix_{-i} by A−iA_{-i} which takes O(n2)O(n^{2}) operations. With the residual trick (e.g. ), we introduce the residual r=Ax−br=Ax-b, and replace (25) by

Now the calculation of xiTx_{i}^{T} and r+r^{+} takes time O(n)O(n), and thus one epoch of BCD takes time O(n2)O(n^{2}). The same trick can be applied to the primal update of ADMM; with this trick, the dual update (6) can be rewritten as μ+=μ−βr\mu^{+}=\mu-\beta r which takes time O(n)O(n), and thus one epoch of ADMM takes time O(n2)O(n^{2}).

Finally, when di>1d_{i}>1, similar update equations can still be used except a minor difference that 1AiTAi\frac{1}{A_{i}^{T}A_{i}} should be replaced by (AiTAi)−1(A_{i}^{T}A_{i})^{-1}. In a special case that di=d,∀id_{i}=d,\forall i and N=dnN=dn, each iteration of BCD takes time O(Nd+d3)O(Nd+d^{3}) and each epoch takes time O(N2+Nd2)O(N^{2}+Nd^{2}). This cost can be reduced if we use BCGD (i.e. not solving the subproblem exactly but updating each block of variables by a gradient step). In order not to make the paper more complicated, we will not discuss the inexact versions of BCD and ADMM in this paper.

4 Two Versions of Independently Randomized ADMM

In this subsection, we present two other versions of randomized ADMM which can be divergent according to simulations. The failure of these versions makes us focus on analyzing RP-ADMM in this paper. These versions can be viewed as natural extensions of R-BCD (randomized BCD) and .

In the first algorithm, called primal-dual randomized ADMM (PD-RADMM), the whole dual variable is viewed as the (n+1)(n+1)-th block. In particular, at each iteration, the algorithm draws one index ii from {1,…,n,n+1}\{1,\dots,n,n+1\}, then performs the following update: if i≤ni\leq n, update the ii-th block of the primal variable; if i=n+1i=n+1, update the whole dual variable. The details are given in Algorithm 2. We have tested PD-RADMM for the counter-example given in Chen et al. , and found that PD-RADMM always diverges (for random initial points).

A variant of PD-RADMM has been proposed in Hong et al. with two differences: first, instead of minimizing the augmented Lagrangian L\mathcal{L}, that algorithm minimizes a strongly convex upper bound of L\mathcal{L}; second, that algorithm uses a diminishing dual stepsize. With these two modifications, shows that each limit point of the sequence generated by their algorithm is a primal-dual optimum with probability 1. Note that also proves the same convergence result for the cyclic version of multi-block ADMM with these two modifications, thus it does not show the benefit of randomization.

In the second algorithm, called primal randomized ADMM (P-RADMM), we only perform randomization for the primal variables. In particular, at each round, we first draw nn independent random variables j1,…,jnj_{1},\dots,j_{n} from the uniform distribution of {1,…,n}\{1,\dots,n\} and update xj1,…,xjnx_{j_{1}},\dots,x_{j_{n}} sequentially, then update the dual variable in the usual way. The details are given in Algorithm 3. This algorithm looks quite similar to RP-ADMM as they both update nn primal blocks at each round; the difference is that RP-ADMM samples without replacement while this algorithm P-RADMM samples with replacement. In other words, RP-ADMM updates each block exactly once at each round, while P-RADMM may update one block more than one times or does not update one block at each round.

We have tested P-RADMM in various settings. For the counter-example given in Chen et al. , we found that P-RADMM does converge. However, if n≥30n\geq 30 and AA is a Gaussian random matrix (each entry is drawn i.i.d. from N(0,1)\mathcal{N}(0,1)), then P-RADMM diverges in almost all cases we have tested. This phenomenon is rather strange since for random Gaussian matrices AA the cyclic ADMM actually converges (according to simulations). An implication is that randomized versions do not always outperform their deterministic counterparts in terms of convergence.

Since both Algorithm 2 and Algorithm 3 can diverge in certain cases, we will not further study them in this paper. In the rest of the paper, we will focus on RP-ADMM (i.e. Algorithm 1).

5 Bernoulli-Randomized ADMM

To implement randomly permuted ADMM, one needs to sample from all blocks without replacement. To save the sampling time, we propose another algorithm which we call Bernoulli-randomized ADMM. This algorithm is motivated by the proof of Theorem 1. This updating scheme can be applied to other algorithms such as SGD and coordinate descent methods.

The new update order combines the well-known double-sweep order and Bernoulli-randomization. The original double-sweep order is (1,2,...,n−1,n,n−1,n−2,...,1)(1,2,...,n-1,n,n-1,n-2,...,1), meaning that x1,x2,…,xn−1,xn,xn−1,xn−2,…,x1x_{1},x_{2},\dots,x_{n-1},x_{n},x_{n-1},x_{n-2},\dots,x_{1} are updated sequentially in each “cycle”. It combines the normal cyclic order (1,2,…,n)(1,2,\dots,n) and a reverse order (n,n−1,…,1)(n,n-1,\dots,1). We propose the following updating scheme: add a check box to each block, and in each cycle we perform the following operations.

Phase I: go through the blocks x1,x2,…,xnx_{1},x_{2},\dots,x_{n} one by one sequentially as follows: for each block xix_{i}, flip a fair coin and:

if the outcome is “head”, update the block xix_{i} and check the check box;

if the outcome is “tail”, do nothing about xix_{i} and uncheck the check box.

Phase II: go through the blocks xn,xn1,…,x1x_{n},x_{n_{1}},\dots,x_{1} in the reverse order, and update xix_{i} if the box is unchecked.

Note that in each cycle we go through each block twice but update each block exactly once so that the number of totally updated blocks remains nn. For example, when n=5n=5, (35421)(35421) is a possible update order, as shown in the following diagram.

Similarly, (13542)(13542) is also a possible update order. But (13524)(13524) and (35412)(35412) are not possible. The set of all possible update orders is given by

where Γ\Gamma is the set of permutations of {1,2,…,n}\{1,2,\dots,n\} as defined in (4). In other words, a sequence from ΓBR\Gamma_{\text{BR}} is a concatenation of an increasing sequence and a decreasing sequence. Note that the permutation (1,2,...,n)(1,2,...,n) is in ΓBR\Gamma_{\text{BR}} since it can be viewed as the concatenation of an increasing sequence (1,2,...,n−1)(1,2,...,n-1) and a “decreasing sequence” (n)(n), and we can let i=n−1i=n-1 in the above definition to cover this case. Similarly, the permutation (n,n−1,…,1)(n,n-1,\dots,1) is also in ΓBR\Gamma_{\text{BR}} as i=1i=1 will cover this case.

The algorithm Bernoulli-randomized ADMM (BR-ADMM) is formally described below. We skip the epoch index kk since otherwise the notation would be cumbersome.

For solving linear systems of equations, the update formula is the same as (17), the update formula of RP-ADMM. The difference is that for RP-ADMM σ\sigma can be an arbitrary permuation, while for BR-ADMM there is some restriction on σ\sigma: it has to be a permuation in ΓBR\Gamma_{\text{BR}}.

Main Results

Let σi\sigma_{i} denote the permutation used in round ii of Algorithm 1, which is a uniform random variable drawn from the set of permutations Γ\Gamma. After round kk, Algorithm 1 generates a random output yk+1y^{k+1}, which depends on the observed draw of the random variable

We will show that the expected iterate (the iterate yky^{k} is defined in (16))

Assume the coefficient matrix A=[A1,…,An]A=[A_{1},\dots,A_{n}] of the constraint in (7) is a non-singular square matrix. Suppose Algorithm 1 is used to solve problem (7), then the expected output converges to the unique primal-dual optimal solution to (7), i.e.

Since the update matrix does not depend on previous iterates, we claim (and prove in Section 4.1) that Theorem 1 holds if the expected update matrix has a spectral radius less than 1, i.e. if the following Theorem 2 holds.

where the expectation is taken over the uniform random distribution over Γ\Gamma, the set of permutations of {1,2,…,n}\{1,2,\dots,n\}. Then the spectral radius of MM is smaller than 11, i.e.

For the counterexample in Chen et al. where A=[1,1,1;1,1,2;1,2,2]A=[1,1,1;1,1,2;1,2,2], it is easy to verify that ρ(Mσ)>1.02\rho(M_{\sigma})>1.02 for any permutation σ\sigma of (1,2,3)(1,2,3). Interestingly, Theorem 2 shows that even if each MσM_{\sigma} is “bad” (with spectral radius larger than 11), the average of them is always “good” (with spectral radius smaller than 11).

Theorem 2 is just a linear algebra result, and can be understood even without knowing the details of the algorithm. However, the proof of Theorem 2 is rather non-trivial. This proof will be provided in Section 4.2, and the technical results used in this proof will be proved in Section 5 and Section 6.

The convergence rate of RP-ADMM for solving linear systems of equations is closely related to the convergence rate of RP-BCD (randomly permuted BCD) for solving quadratic problems. We will discuss their relation and how our results in this paper improve our understanding for RP-BCD.

A similar convergence result holds for BR-ADMM proposed in Section 2.5, as presented below. The proof is a simple modification of the proof of Theorem 1, and can be found in Section 6.4.

Assume the coefficient matrix A=[A1,…,An]A=[A_{1},\dots,A_{n}] of the constraint in (7) is a non-singular square matrix. Suppose Algorithm 4 is used to solve problem (7), then the expected output converges to the unique primal-dual optimal solution to (7).

2 Expected Convergence Rate of RP-ADMM and RP-BCD

There is a close relation between RP-ADMM for solving linear systems and RP-CD for solving quadratic problems (see Lemma 2). Thus it is not surprising that we need to understand RP-BCD before understanding RP-ADMM. We will first present an expected convergence rate of RP-BCD (in terms of the expected iterates) for solving quadratic problems, which improves the best existing convergence rate (one type of rates, to be precise) by a factor of nn Rigorously speaking, this is not a fair comparison as the complexity of C-CD is deterministic complexity.. The result is proved via establishing a weak version of matrix AM-GM inequality. This result also establishes a large gap of O(n)O(n) between RP-BCD and C-BCD (cyclic BCD). Second, built upon the result for RP-BCD, we establish a convergence rate of RP-ADMM which is similar to RP-BCD and also nn times better than that of C-BCD.

The first result is about the expected convergence rate of RP-BCD for the case AiTAi=IA_{i}^{T}A_{i}=I. This assumption is made so that the expression is simple, and the case for general AiA_{i} is given in the next result.

(rate of RP-BCD for quadratic functions with identity diagonal blocks) Assume the coefficient matrix A=[A1,…,An]A=[A_{1},\dots,A_{n}] is a non-singular square matrix, and AiTAi=I,∀  i.A_{i}^{T}A_{i}=I,\forall\;i. Suppose RP-BCD is used to solve problem (23), where xkx^{k} denotes the variable after kk epochs (each epoch represents one cycle of updating all coordinates). Denote the unique optimal solution as x∗=A−1bx^{*}=A^{-1}b. Then

To put this convergence rate result in the context, we consider the simple case that each di=1d_{i}=1, i.e., each block consists of a single coordinate. In this case, every diagonal entry of ATAA^{T}A is 11, thus the average eigenvalue of ATAA^{T}A is 11. Throughout the paper, we consider the total computation complexity The computation complexity equals the iteration complexity times the per-iteration cost. We do not present iteration complexity since there may be confusion about whether “one iteration” means nn coordinate updates or 11 coordinate update. Presenting iteration complexity is better if one considers a general convex problem, but then one needs to discuss the per-iteration cost. We are considering quadratic problems throughout the paper, so we feel it is more clear to stick to computation complexity.; note that we assume the residual trick as described in 2.3 is always used for all methods.

Our Theorem 3 provides an expected computational complexity upper bound O(n3κCDlog⁡1ϵ)O(n^{3}\kappa_{\text{CD}}\log\frac{1}{\epsilon}) for RP-CD, since each epoch takes O(n2)O(n^{2}) time and it requires O(n2log⁡1ϵ)O(n^{2}\log\frac{1}{\epsilon}) epochs to achieve error ϵ\epsilon according to (33). It is known that the computational complexity of R-CD (randomized coordinate descent) to achieve relative accuracy ϵ\epsilon Here, the relative accuracy ϵ\epsilon means ∥E(xk)−x∗∥/∥x0−x∗∥\|E(x^{k})-x^{*}\|/\|x^{0}-x^{*}\| or ∥E(f(xk))−f∗∥/∥f(x0)−f∗∥\|E(f(x^{k}))-f^{*}\|/\|f(x^{0})-f^{*}\|. is O(n2κCDlog⁡1ϵ)O(n^{2}\kappa_{\text{CD}}\log\frac{1}{\epsilon}), where κCD=λavg(ATA)/λmin⁡(ATA)=1/λmin⁡(ATA)\kappa_{\text{CD}}=\lambda_{\text{avg}}(A^{T}A)/\lambda_{\min}(A^{T}A)=1/\lambda_{\min}(A^{T}A) is the ratio of the average eigenvalue over the minimum eigenvalue. It was recently shown that in terms of κCD\kappa_{\text{CD}} and nn only, the worst-case complexity of C-CD (cyclic CD) is O(n4κCDlog⁡1ϵ)O(n^{4}\kappa_{\text{CD}}\log\frac{1}{\epsilon}), which is n2n^{2} times worse than R-CD and nn times worse than GD. This shows a large gap between C-CD and R-CD in the worst case.

It was widely conjectured that RP-CD is at least as fast as R-CD, but this conjecture is considered to be rather difficult to prove. For a special class of matrices, recent works validated the conjecture. However, to our knowledge, even for a general quadratic function with equal diagonal entries 11, the previously best known convergence rate of RP-CD is almost the same as C-CD (see ), which can be n2n^{2} times worse than that of R-CD. Our Theorem 3 provides an expected computational complexity upper bound O(n3κCDlog⁡1ϵ)O(n^{3}\kappa_{\text{CD}}\log\frac{1}{\epsilon}) for RP-CD, which is nn times faster than C-CD and nn times slower than R-CD. This improves the best existing rate by a factor of nn Note that this “improvement” is valid when the convergence rate is characterized by only κCD\kappa_{\text{CD}} and nn. It is common to use other parameters such as the maximum eigenvalue to characterize the convergence rate (see for a detailed discussion), and our result here does not provide improvement for other kinds of convergence rate.. We summarize the comparison of the complexity for C-CD, R-CD and RP-CD in Table 1.

The following proposition generalizes Theorem 3 to the non-identity-diagonal case, i.e., AiTAiA_{i}^{T}A_{i} does not need to be an identity matrix.

(rate of RP-BCD for quadratic functions, with non-identity blocks) Assume the coefficient matrix A=[A1,…,An]A=[A_{1},\dots,A_{n}] is a non-singular square matrix. Suppose RP-BCD is used to solve problem (23). Denote D=diag(A1TA1,…,AnTAn)D=\text{diag}(A_{1}^{T}A_{1},\dots,A_{n}^{T}A_{n}) as a block-diagonal matrix, and the norm ∥z∥D=zTDz\|z\|_{D}=\sqrt{z^{T}Dz}. Then

The proof of Proposition 2 is given in Section 4.4. One can easily transform the quantity λmin⁡(D1/2ATAD−1/2)\lambda_{\min}(D^{1/2}A^{T}AD^{-1/2}) to certain quantity that only depends on the eigenvalues of AiTAiA_{i}^{T}A_{i} and ATAA^{T}A. However, as noted in , it is far from clear how tight the transformation is, thus we skip the transformation here. In fact, it is related to some open question on the so-called Jacobi-preconditioning. We refer the interested readers to for a detailed discussion of the subtle issues in the non-identity-diagonal case.

At last, we present a result on the expected convergence rate of RP-ADMM for solving linear systems, under the assumption that AiTAi=I,  ∀iA_{i}^{T}A_{i}=I,\;\forall i. Very similar to Proposition 2, we can also generalize this result to non-identity-diagonal case, i.e., AiTAi≠IA_{i}^{T}A_{i}\neq I, but to save space we skip the generalization here. The proof of Theorem 4 is given in Section 4.5.

(Expected convergence rate of RP-ADMM for linear systems) Assume the coefficient matrix A=[A1,…,An]A=[A_{1},\dots,A_{n}] of the constraint in (7) is a non-singular square matrix and AiTAi=IdiA_{i}^{T}A_{i}=I_{d_{i}}. Suppose Algorithm 1 is used to solve problem (7). Denote y∗=[A−1b0]y^{*}=\begin{bmatrix}A^{-1}b\\ 0\\ \end{bmatrix} as the unique primal-dual optimal solution to the problem (7), then

This result implies that similar to RP-CD for solving quadratic problems, the complexity of RP-ADMM in terms of the expected iterates for solving linear systems is also at most

In light of the fact that C-CD has been shown to only achieve a rate O(n4κCDlog⁡(1/ϵ))O(n^{4}\kappa_{\text{CD}}\log(1/\epsilon)) , the rate of RP-ADMM we obtain is already quite good. Nevertheless, we conjecture that this complexity upper bound can be improved to O(n2κCDlog⁡(1/ϵ))O(n^{2}\kappa_{\text{CD}}\log(1/\epsilon)), the same as the conjectured complexity for RP-CD. But an improved rate of RP-ADMM leads to an improved rate of RP-BCD (this should be clear via the comparison of (50) and (57)), thus proving this conjecture is an even more difficult problem than the long-standing open question on RP-CD.

3 Matrix AM-GM Inequality

The original version is more general: the number of matrices does not need to be the same as the dimension of the matrix. For simplicity, we just present a simpler version here.

The matrix AM-GM inequality is a generalization of the well-known AM-GM inequality: for non-negative numbers a1,…,ana_{1},\dots,a_{n}, the geometric mean (a1a2…an)1/n(a_{1}a_{2}\dots a_{n})^{1/n} is no more than the algebraic mean 1n∑i=1nai\frac{1}{n}\sum_{i=1}^{n}a_{i}. When extending this inequality to matrix domain, the non-commutative nature of matrix multiplication makes the problem rather difficult to prove.

We observe that we only need to prove a matrix AM-GM inequality for projection matrices. We conjecture that the following matrix AM-GM inequality holds.

Compared with (36), our conjecture makes a stronger claim on the relation, but it only applies to projection matrices. We have found examples to show that (37) does not hold for general positive semi-definite matrices, but it holds for projection matrices in all of our experiments.

We are not able to prove the new conjecture – that would solve the open question of the best convergence rate of RP-CD for quadratic problem. Nevertheless, inspired by the new conjecture, we prove a weaker version (see Lemma 3), which can lead to an improved convergence rate estimate for RP-CD.

Proof of Main Results

Denote σk\sigma_{k} as the permutation used in round kk, and define ξk\xi_{k} as in (28). Rewrite the update equation (17) below (replacing σ\sigma by σk\sigma_{k}):

We first prove (30) for the case b=0b=0. By (18) we have bˉ=0\bar{b}=0, then (38) is simplified to yk+1=Lˉσk−1Rˉσkyky^{k+1}=\bar{L}_{\sigma_{k}}^{-1}\bar{R}_{\sigma_{k}}y^{k}. Taking the expectation of both sides of this equation in ξk\xi_{k} (see its definition in (28)), and note that yky^{k} is independent of σk\sigma_{k}, we get

Since the spectral radius of MM is less than 1 by Theorem 2, we have that {ϕk}→0\{\phi^{k}\}\rightarrow 0, i.e. (30).

We then prove (30) for general bb. Let y∗=[A−1b;0]y^{*}=[A^{-1}b;0] denote the optimal solution. Then it is easy to verify that

for all σk∈Γ\sigma_{k}\in\Gamma (i.e. the optimal solution is the fixed point of the update equation for any order). Compute the difference between this equation and (38) and letting y^k=yk−y∗\hat{y}^{k}=y^{k}-y^{*} , we get y^k+1=Lˉσk−1Rˉσky^k\hat{y}^{k+1}=\bar{L}_{\sigma_{k}}^{-1}\bar{R}_{\sigma_{k}}\hat{y}^{k}. According to the proof for the case b=0b=0, we have E(y^k)⟶0E(\hat{y}^{k})\longrightarrow 0, which implies E(yk)⟶y∗E(y^{k})\longrightarrow y^{*}.

2 Proof of Theorem 2

The difficulty of proving Theorem 2 (bounding the spectral radius of MM defined in (31)) is two-fold. First, MM is a non-symmetric matrix, and there are very few tools to bound the spectral radius of a non-symmetric matrix. In fact, spectral radius is neither subadditive nor submultiplicative (see, e.g. Kittaneh ). Note that the spectral norm of MM can be much larger than 11 (there are examples that ∥M∥>2\|M\|>2), thus we cannot bound the spectral radius simply by the spectral norm. Second, although it is possible to explicitly write each entry of MM as a function of the entries of ATAA^{T}A, these functions are very complicated (nn-th order polynomials) and it is not clear how to utilize this explicit expression.

The proof outline of Theorem 2 and the main techniques are described below. In Step 0, we provide an expression of the expected update matrix MM. In Step 1, we establish the relationship between the eigenvalues of MM and the eigenvalues of a simple symmetric matrix AQATAQA^{T}, where QQ is defined in (39). As a consequence, the spectral radius of MM is smaller than one iff the eigenvalues of AQATAQA^{T} lie in the region (0,4/3)(0,4/3). This step partially resolves the first difficulty, i.e. how to deal with the spectral radius of a non-symmetric matrix. In Step 2, we show that the eigenvalues of AQATAQA^{T} do lie in (0,4/3)(0,4/3) using mathematical induction. The induction analysis circumvents the second difficulty, i.e. how to utilize the relation between MM and AA.

Step 0: compute the expression of the expected update matrix MM. Define

It is easy to prove that QQ defined by (39) is symmetric. In fact, note that LσT=Lσˉ,∀σ∈ΓL_{\sigma}^{T}=L_{\bar{\sigma}},\forall\sigma\in\Gamma, where σˉ\bar{\sigma} is a reverse permutation of σ\sigma satisfying σˉ(i)=σ(n+1−i),∀ i\bar{\sigma}(i)=\sigma(n+1-i),\forall\ i, thus Q=1n!∑σQσ=(1n!∑σQσˉ)T=QT,Q=\frac{1}{n!}\sum_{\sigma}Q_{\sigma}=(\frac{1}{n!}\sum_{\sigma}Q_{\bar{\sigma}})^{T}=Q^{T}, where the last step is because the sum of all QσˉQ_{\bar{\sigma}} is the same as the sum of all QσQ_{\sigma}.

Substituting the expression of Lˉσ−1\bar{L}_{\sigma}^{-1} into the above relation, and replacing RσR_{\sigma} by Lσ−ATAL_{\sigma}-A^{T}A, we obtain

Since MσM_{\sigma} is linear in Lσ−1L_{\sigma}^{-1}, we have

Step 1: relate MM to a simple symmetric matrix. The main result of Step 1 is given below, and the proof of this result is relegated to Section 5.

Furthermore, when QQ is symmetric, we have

Remark: For our problem, the matrix QQ as defined by (39) is symmetric (see the argument after equation (39)), thus the relation (45) indeed holds according to Lemma 1. For a general non-symmetric QQ, (45) does not need to hold, but the first conclusion (44) still holds.

Step 2: Bound the eigenvalues of QATAQA^{T}A. The main result of Step 2 is summarized in the following Lemma 2. The proof of Lemma 2 is given in Section 6.

in which LσL_{\sigma} is defined by (21) and Γ\Gamma is defined by (4). Then all eigenvalues of QATAQA^{T}A lie in (0,4/3)(0,4/3), i.e.

Remark: The upper bound 43\frac{4}{3} in (47) is probably tight, since we have found numerical examples with eig(QATA)>1.3333\text{eig}(QA^{T}A)>1.3333. Now the expected convergence of RP-ADMM seems to be a pleasant coincidence: Lemma 1 shows that to prove the expected convergence we need to prove sup⁡Aeig(QATA)\sup_{A}\text{eig}(QA^{T}A), a quantity that can be defined without knowing ADMM, is bounded by 4/34/3; Lemma 2 and numerical experiments show that this quantity happens to be exactly 4/34/3 so that RP-ADMM can converge (in expectation).

Theorem 2 follows immediately from Lemma 1 and Lemma 2.

3 Proof of Theorem 3

We first describe the outline of the proof. The expected update matrix of RP-BCD is I−QATAI-QA^{T}A, and the eigenvalues of this matrix lie in (−1,1)(-1,1). The expected convergence speed of RP-BCD depends on the distance between the eigenvalues and the two extremes −1-1 and 11. Lemma 2 shows that the distance to −1-1 is at least 1/31/3, which is a constant. We will show that the distance to 11 is at least λmin⁡(ATA)/n\lambda_{\min}(A^{T}A)/n, by proving a weaker version of matrix AM-GM inequality. Combining the two results, we obtain the expected convergence speed of RP-BCD.

According to (24), we have xk+1−x∗=(I−Lσ−1ATA)(xk−x∗)x^{k+1}-x^{*}=(I-L_{\sigma}^{-1}A^{T}A)(x^{k}-x^{*}), where σ\sigma is the randomly picked permutation at the kk-th epoch. Therefore, the expected update formula of RP-BCD for solving the least squares problem is

Suppose the eigenvalues of QATAQA^{T}A are η1≥η2≥⋯≥ηn\eta_{1}\geq\eta_{2}\geq\dots\geq\eta_{n}, then according to Lemma 2,

thus the spectral radius of I−QATAI-QA^{T}A is

An interesting phenomenon occurs here. The spectral radius is either 1−ηn1-\eta_{n} or ∣1−η1∣|1-\eta_{1}|. In the latter case, ρ(I−QATA)=∣1−η1∣≤1/3\rho(I-QA^{T}A)=|1-\eta_{1}|\leq 1/3, implying that ∥E(xk)−x∗∥≤13k∥E(x0)−x∗∥\|E(x^{k})-x^{*}\|\leq\frac{1}{3^{k}}\|E(x^{0})-x^{*}\|, or equivalently, the relative error ∣E(xk)−x∗∥/∣E(x0)−x∗∥|E(x^{k})-x^{*}\|/|E(x^{0})-x^{*}\| achieves ϵ\epsilon in log⁡3log⁡(1/ϵ)\log 3\log(1/\epsilon) epochs. We do not even need to compute η1\eta_{1} since it will only affect the convergence speed when the speed is already very fast. From a theoretical perspective, the improvement from log⁡3\log 3 to log⁡(1/(1−∣1−η1∣))\log(1/(1-|1-\eta_{1}|)) is just an improvment in the constant. Therefore, it is reasonable to ignore η1\eta_{1} and focus on the estimate of 1−ηn1-\eta_{n}.

To estimate the maximum eigenvalue of I−QATAI-QA^{T}A (or equivalently, that of I−AQATI-AQA^{T}), we first provide a useful identity that connects I−AQATI-AQA^{T} and projection matrices Pi=I−AiAiTP_{i}=I-A_{i}A_{i}^{T}.

Suppose A=[A1,…,An]A=[A_{1},\dots,A_{n}] is a non-singular square matrix, and AiTAi=I,∀  i.A_{i}^{T}A_{i}=I,\forall\;i. For a permutation σ=(σ1,…,σn)∈Γ\sigma=(\sigma_{1},\dots,\sigma_{n})\in\Gamma, LσL_{\sigma} is defined as in (21), and Qσ=Lσ−1Q_{\sigma}=L_{\sigma}^{-1}. Denote Pi=I−AiAiTP_{i}=I-A_{i}A_{i}^{T}, i=1,…,ni=1,\dots,n. Then we have

The proof of Claim 4.1 is given at the end of this subsection. Claim 4.1 states that I−AQATI-AQA^{T} is exactly equal to 1n!∑σ=(σ1,…,σn)∈ΓPσnPσn−1…Pσ1\frac{1}{n!}\sum_{\sigma=(\sigma_{1},\dots,\sigma_{n})\in\Gamma}P_{\sigma_{n}}P_{\sigma_{n-1}}\dots P_{\sigma_{1}}, thus we only need to estimate the maximal eigenvalue of the latter expression. This is achieved by the following lemma (the proof is given in Section 7.3).

The above Lemma 3 and Claim 4.1 immediately lead to the following corollary.

Suppose A=[A1,…,An]A=[A_{1},\dots,A_{n}] is a non-singular square matrix, and AiTAi=I,∀  i.A_{i}^{T}A_{i}=I,\forall\;i. Suppose Pi=I−AiAiT,  ∀  i.P_{i}=I-A_{i}A_{i}^{T},\;\forall\;i. LσL_{\sigma} is defined as in (21), and Q=Eσ(Lσ−1)Q=E_{\sigma}(L_{\sigma}^{-1}). Then

Note that 1n∑iPi=1n(nI−∑iAiAiT)=I−1nAAT\frac{1}{n}\sum_{i}P_{i}=\frac{1}{n}(nI-\sum_{i}A_{i}A_{i}^{T})=I-\frac{1}{n}AA^{T}, thus (53) implies

Substituting this relation into (49), we obatain

Remark: There is a coefficient 1/n1/n in front of λmin⁡(AAT)\lambda_{\min}(AA^{T}) in (54), and this is why the complexity of RP-CD we establish is nn times worse than the conjectured one in Table 1. If Conjecture 3.1 holds, then this factor of 1/n1/n would be removed and the conjectured (expected) complexity of RP-CD in Table 1 would hold.

We prove (51a) by induction on nn. Without loss of generality, we can assume σ=(1,2,…,n)\sigma=(1,2,\dots,n), then Lσ=[A1TA10…0A2TA1A2TA2…0⋮⋮⋱⋮AnTA1AnTA2…AnTAn].L_{\sigma}=\begin{bmatrix}A_{1}^{T}A_{1}&0&\dots&0\\ A_{2}^{T}A_{1}&A_{2}^{T}A_{2}&\dots&0\\ \vdots&\vdots&\ddots&\vdots\\ A_{n}^{T}A_{1}&A_{n}^{T}A_{2}&\dots&A_{n}^{T}A_{n}\\ \end{bmatrix}. In this case, (51a) becomes

The expression obviously holds for n=1n=1. Suppose the expression holds for n−1n-1, i.e., for A^=[A1,…,An−1]\hat{A}=[A_{1},\dots,A_{n-1}], we have

where σ^=(1,2,…,n−1)\hat{\sigma}=(1,2,\dots,n-1) is a permutation of n−1n-1 elements and L^σ^\hat{L}_{\hat{\sigma}} is the counterpart of LσL_{\sigma} for n−1n-1 blocks defined as

The two matrices LσL_{\sigma} and L^σ′\hat{L}_{\sigma^{\prime}} are related by

where in the last step we use the induction hypothesis (55). Thus we have proved (51a). Summing up (51a) for all possible permutations σ\sigma and divide by n!n!, we obtain (51b). □\Box

4 Proof of Proposition 2

According to (48), the (expected) update equation of RP-BCD is given by E(xk+1)−x∗=(I−QATA)(E(xk)−x∗)=Z(E(xk)−x∗)E(x^{k+1})-x^{*}=(I-QA^{T}A)(E(x^{k})-x^{*})=Z(E(x^{k})-x^{*}), where Z=I−QATA=I−E(Lσ−1ATA)Z=I-QA^{T}A=I-E(L_{\sigma}^{-1}A^{T}A).

5 Proof of Theorem 4

Now we consider the expected convergence rate of RP-ADMM. The difference with the analysis for RP-BCD is that here we need to consider the distance between the eigenvalues of I−AQATI-AQA^{T} with −1/3-1/3 while for RP-BCD what matters is the distance between the eigenvalues of I−AQATI-AQA^{T} and −1-1 which is at least 2/32/3 and thus can be ignored.

Suppose the minimum and maximum eigenvalues of QATAQA^{T}A are 0<τmin⁡≤τmax⁡<4/30<\tau_{\min}\leq\tau_{\max}<4/3. Then

where z+=max{z,0}z_{+}=max\{z,0\}. Furthermore, we have

The proof of Claim 4.2 is given in Section 7.1. The next lemma provides a universal estimate of the maximum eigenvalules of QATAQA^{T}A.

The maximum eigenvalues of QATAQA^{T}A is at most 43−491n+1\frac{4}{3}-\frac{4}{9}\frac{1}{n+1}, i.e.,

The proof of Lemma 4 is given in Section 7.2

According to (54), which is established in the proof of the expected convergence rate of RP-BCD, we have

Substituting the bounds (58) and (59) into (57), we obtain

Since λmin⁡(ATA)≤1\lambda_{\min}(A^{T}A)\leq 1, 12n≤1n+1\frac{1}{2n}\leq\frac{1}{n+1}, this bound can be simplified to

Remark: The eigenvalues of QATAQA^{T}A lie in the region (0,4/3)(0,4/3), which guarantees the expected convergence of RP-ADMM. To obtain the expected convergence rate, we need to know the distance of the spectrum to the two extremes and 4/34/3. We conjecture that the bound can be improved to ρ(M)≤1−12λmin⁡(ATA)\rho(M)\leq 1-\frac{1}{2}\lambda_{\min}(A^{T}A). This requires more effort than the conjecture of RP-CD: besides showing τmin⁡≥O(λmin⁡(ATA)),\tau_{\min}\geq O(\lambda_{\min}(A^{T}A)), we also need to show τmax⁡≤43−O(λmin⁡(ATA))\tau_{\max}\leq\frac{4}{3}-O(\lambda_{\min}(A^{T}A)). This is left as future work.

Proof of Lemma 1

The proof of Lemma 1 relies on two simple techniques. The first technique, as elaborated in the Step 1 below, is to factorize MM and rearrange the factors. The second technique, as elaborated in the Step 2 below, is to reduce the dimension by eliminating a variable from the eigenvalue equation.

Step 1: Factorizing MM and rearranging the order of multiplication. The following observation is crucial: the matrix MM defined by (43) can be factorized as

Switching the order of the products by moving the first component to the last, we get a new matrix

Note that eig(XY)=eig(YX)\text{eig}(XY)=\text{eig}(YX) for any two square matrices, thus

Step 2: Relate the eigenvalues of M′M^{\prime} to the eigenvalues of QATAQA^{T}A, i.e. prove (62). This step is simple as we only use the definition of eigenvalues. However, note that, without Step 1, just applying the definition of eigenvalues of the original matrix MM may not lead to a simple relationship as (62).

We claim that (63) holds when v1=0v_{1}=0. In fact, in this case we must have v0≠0v_{0}\neq 0 (otherwise v=0v=0 cannot be an eigenvector). By (64b) we have λv0=v0\lambda v_{0}=v_{0}, thus λ=1\lambda=1. By (64a) we have 0=QATv0=QATA(A−1v0)0=QA^{T}v_{0}=QA^{T}A(A^{-1}v_{0}), which implies (1−λ)21−2λ=0∈eig(QATA)\frac{(1-\lambda)^{2}}{1-2\lambda}=0\in\text{eig}(QA^{T}A), therefore (63) holds in this case.

The equation (64b) implies (1−λ)v0=Av1(1-\lambda)v_{0}=Av_{1}. Multiplying both sides of (64a) by (1−λ)(1-\lambda) and invoking this equation, we get

We must have λ≠12\lambda\neq\frac{1}{2}; otherwise, the above relation implies v1=0v_{1}=0, which contradicts (65). Then (66) becomes

Therefore, (1−λ)21−2λ\frac{(1-\lambda)^{2}}{1-2\lambda} is an eigenvalue of QATAQA^{T}A, with the corresponding eigenvector v1≠0v_{1}\neq 0, which finishes the proof of (63).

The other direction For the purpose of proving Theorem 2, we do not need to prove this direction. Here we present the proof since it is quite straightforward and makes the result more comprehensive.

is easy to prove. Suppose (1−λ)21−2λ∈eig(QATA)\frac{(1-\lambda)^{2}}{1-2\lambda}\in\text{eig}(QA^{T}A). We consider two cases.

Case 2: (1−λ)21−2λ≠0\frac{(1-\lambda)^{2}}{1-2\lambda}\neq 0, then λ≠1\lambda\neq 1. Let v1v_{1} be the eigenvector corresponding to (1−λ)21−2λ\frac{(1-\lambda)^{2}}{1-2\lambda} (i.e. pick v1v_{1} that satisfies (67)), and define v0=v1/(1−λ)v_{0}=v_{1}/(1-\lambda). It is easy to verify that v=[v1v0]v=\begin{bmatrix}v_{1}\\ v_{0}\end{bmatrix} satisfies Mv=λvMv=\lambda v, which implies λ∈eig(M)\lambda\in\text{eig}(M).

Step 3: When QQ is symmetric, prove (45) by simple algebraic computation.

Note that when τ(τ−1)<0\tau(\tau-1)<0, the expression τ(τ−1)\sqrt{\tau(\tau-1)} denotes a complex number iτ(1−τ)i\sqrt{\tau(1-\tau)}, where ii is the imaginary unit. To prove (45), we only need to prove

Case 1: τ<0\tau<0. Then τ(τ−1)=∣τ∣(∣τ∣+1)>0\tau(\tau-1)=|\tau|(|\tau|+1)>0. In this case, λ1=1+∣τ∣+∣τ∣(∣τ∣+1)>1.\lambda_{1}=1+|\tau|+\sqrt{|\tau|(|\tau|+1)}>1.

Case 2: 0<τ<10<\tau<1. Then τ(τ−1)<0\tau(\tau-1)<0, and (69) can be rewritten as

which implies ∣λ1∣=∣λ2∣=(1−τ)2+τ(1−τ)=1−τ<1|\lambda_{1}|=|\lambda_{2}|=\sqrt{(1-\tau)^{2}+\tau(1-\tau)}=\sqrt{1-\tau}<1.

Case 3: τ>1\tau>1. Then τ(τ−1)>0\tau(\tau-1)>0. According to (69), it is easy to verify λ1>0>λ2\lambda_{1}>0>\lambda_{2} and

Combining the conclusions of the three cases immediately leads to (70).

Proof of Lemma 2

This section is devoted to the proof of Lemma 2. We first give a proof overview in Section 6.1. The formal proof of Lemma 2 is given in Section 6.2. The proofs of the technical results involved in the proof are given in the subsequent subsections.

Without loss of generality, we can assume

In the proof overview, we discuss a few issues one may encounter when proving the result, and how we resolve these issues.

The simulations show that ∥QATA∥<43≪∥Q∥∥ATA∥\|QA^{T}A\|<\frac{4}{3}\ll\|Q\|\|A^{T}A\|, thus we cannot relax ∥QATA∥\|QA^{T}A\| to the product of ∥Q∥\|Q\| and ∥ATA∥\|A^{T}A\|, and have to treat QATAQA^{T}A as a single subject. However, each entry of QATAQA^{T}A is a complicated function (in fact, a high order polynomial) of the entries of ATAA^{T}A. In other words, QQ is like a black box. To open the “black box”, we use a simple expression of Z=I−AQATZ=I-AQA^{T} proved in Claim 4.1, i.e., Z=Eσ(Pσ1…,Pσn),Z=E_{\sigma}(P_{\sigma_{1}}\dots,P_{\sigma_{n}}), where Pi=I−AiAiTP_{i}=I-A_{i}A_{i}^{T} is directly related to AiA_{i}. The problem becomes how to connect the eigenvalues of Eσ(Pσ1…,Pσn)E_{\sigma}(P_{\sigma_{1}}\dots,P_{\sigma_{n}}) with those of AAT=∑iAiAiT=n−∑iPiAA^{T}=\sum_{i}A_{i}A_{i}^{T}=n-\sum_{i}P_{i}.

Although this is a clear linear algebra problem, it is not easy to obtain a lower bound of Eσ(Pσ1…,Pσn)E_{\sigma}(P_{\sigma_{1}}\dots,P_{\sigma_{n}}). In fact, even though we know the eigenvalues of Z=Eσ(Pσ1…,Pσn)Z=E_{\sigma}(P_{\sigma_{1}}\dots,P_{\sigma_{n}}) are lower bounded by −1-1 because RP-CD converges, it is not clear how to prove this lower bound directly from a linear algebra perspective.

In our solution, we apply two tricks. The first trick is to view Eσ(Pσ1…,Pσn)E_{\sigma}(P_{\sigma_{1}}\dots,P_{\sigma_{n}}) as an induction formula that connects it and its lower dimensional analogs. This is based on a simple observation that any permutation (σ1σ2…σn)(\sigma_{1}\sigma_{2}\dots\sigma_{n}) can be written as the concatenation of (σ1σ2…σn−1)(\sigma_{1}\sigma_{2}\dots\sigma_{n-1}) and σn\sigma_{n}, thus the expression of Z=Eσ(Pσ1…,Pσn)Z=E_{\sigma}(P_{\sigma_{1}}\dots,P_{\sigma_{n}}) can be decomposed accordingly. We then reduce the problem to bounding the eigenvalues of a Jordan product PnZ^+Z^PnP_{n}\hat{Z}+\hat{Z}P_{n}, where PnP_{n} is a projection matrix and Z^\hat{Z} is the lower dimensional analog of ZZ. The second trick is to apply a formula on the eigenvalues of Jordan product developed by Strang in 1962 . Somewhat surprisingly, his formula exactly leads to the desired lower bound of −1/3-1/3.

2 Proof of Lemma 2

The proof can be divided into three steps: first provide an alternative expression of AQATAQA^{T}, then prove an induction formula, and finally apply Strang’s formula to perform mathematical induction. This subsection contains the major part of the proof, and the intermediate technical results will be proved in later subsections.

Step 0: Expression of I−AQATI-AQA^{T}. As proved in Claim 4.1, we have a simple expression of the update matrix I−AQATI-AQA^{T}

Define WkW_{k} as the kk-th block-column of ATAA^{T}A excluding the block AkTAkA_{k}^{T}A_{k}, i.e.

Based on the expression of I−AQATI-AQA^{T} presented before, we build a connection between the update matrix I−AQATI-AQA^{T} and its lower dimensional analogs. The proof of Proposition 3 is given in Section 6.3.

where QQ is defined as in (39), A^k=[A1,…,Ak−1,Ak+1,…,An]\hat{A}_{k}=[A_{1},\dots,A_{k-1},A_{k+1},\dots,A_{n}], and Q^k\hat{Q}_{k} is defined in (73), and Pk=I−AkAkTP_{k}=I-A_{k}A_{k}^{T}. Then we have

Step 2: Applying Strang’s result on Jordan product to perform mathematical induction.

It is obvious that the product of two symmetric matrices is not necessarily symmetric, so it is common to encounter the symmetrized product XY+YXXY+YX, which is called Jordan product of two matrices XX and YY. Our induction formula basically states that ZZ is the average of the Jordan product of the lower dimensional analog and PkP_{k}.

The eigenvalues of the Jordan product of two matrices have been studied before. The following result is proved in Strang .

([44, Theorem 1]; eigenvalues of Jordan product) Suppose two symmetric positive-semidefinite matrices XX and YY satisfy

then the maximal (resp. minimal) eigenvalue of the Jordan product XY+YXXY+YX are the largest (resp. smallest) of the set

Let us come back to the proof of Lemma 2. We use mathematical induction to prove Lemma 2. For the basis of the induction (n=1n=1), Lemma 2 holds since QATA=Id1×d1QA^{T}A=I_{d_{1}\times d_{1}}. Assume Lemma 2 holds for n−1n-1, we will prove Lemma 2 for nn.

Consider one term of (75) PkZ^k+Z^kPkP_{k}\hat{Z}_{k}+\hat{Z}_{k}P_{k}. Note that Pk=I−AkAkTP_{k}=I-A_{k}A_{k}^{T} is a projection matrix, since we have assumed AkTAk=IA_{k}^{T}A_{k}=I. Combining with the induction hypothesis, we have

Let α1=0,αn=1,β1=−1/3,βn=1\alpha_{1}=0,\alpha_{n}=1,\beta_{1}=-1/3,\beta_{n}=1, then the set (76) becomes (keep the repeated values)

Note that since by the induction hypothesis the eigenvalues of Z^k\hat{Z}_{k} cannot achieve the extreme values of region (−1/3,1)(-1/3,1), the eigenvalues of 12(PkZ^k+Z^kPk)\frac{1}{2}(P_{k}\hat{Z}_{k}+\hat{Z}_{k}P_{k}) also cannot A more detailed argument is as follows. Since −I/3⪯Z^k-I/3\preceq\hat{Z}_{k}, we can let β1=−1/3+ϵ\beta_{1}=-1/3+\epsilon for a sufficiently small positive number ϵ\epsilon, while keeping α1=0,αn=1,βn=1\alpha_{1}=0,\alpha_{n}=1,\beta_{n}=1. The set (76) now becomes {0,0,−2/3+2ϵ,2,−(4/3−ϵ)24(2/3+ϵ)}.\{0,0,-2/3+2\epsilon,2,-\frac{(4/3-\epsilon)^{2}}{4(2/3+\epsilon)}\}. Both −2/3+2ϵ-2/3+2\epsilon and −(4/3−ϵ)24(2/3+ϵ)-\frac{(4/3-\epsilon)^{2}}{4(2/3+\epsilon)} are strictly larger than 2/32/3, thus the extreme value −2/3-2/3 cannot be achieved. By a similar argument the other extreme value 22 also cannot be achieved. . So we have

Remark: Where does the magical number −1/3-1/3 come from? It is actually the strange and complicated term 16α1αnβ1βn−(β1−βn)2(α1−αn2)4(α1+αn)(β1+βn)\frac{16\alpha_{1}\alpha_{n}\beta_{1}\beta_{n}-(\beta_{1}-\beta_{n})^{2}(\alpha_{1}-\alpha_{n}^{2})}{4(\alpha_{1}+\alpha_{n})(\beta_{1}+\beta_{n})} in Strang’s result (76), which occurs due to the special structure of the Jordan product.

3 Proof of Proposition 3 (the induction formula)

It is easy to build an induction formula from the expression (51b). For example, when n=3n=3, the matrix ∑σPσ1Pσ2Pσ3\sum_{\sigma}P_{\sigma_{1}}P_{\sigma_{2}}P_{\sigma_{3}} can be decomposed as the sum of P1(P2P3+P3P2)+(P2P3+P3P2)P1P_{1}(P_{2}P_{3}+P_{3}P_{2})+(P_{2}P_{3}+P_{3}P_{2})P_{1} and two other similar terms (changing the outside part P1P_{1} to P2,P3P_{2},P_{3} and the inside part P2P3+P3P2P_{2}P_{3}+P_{3}P_{2} correspondingly). The inside part P2P3+P3P2P_{2}P_{3}+P_{3}P_{2} only involves two matrices, thus is a lower-dimensional analog. To make this even easier to see, denote X=P1,Y=P2,Z=P3,X=P_{1},Y=P_{2},Z=P_{3}, then

A rigorous argument based on the above intuition is given as follows. Applying the formula (51b) to the matrix P1,…,Pk−1,Pk+1,…,PnP_{1},\dots,P_{k-1},P_{k+1},\dots,P_{n}, and by the definition A^k=[A1,…,Ak−1,Ak+1,…,An]\hat{A}_{k}=[A_{1},\dots,A_{k-1},A_{k+1},\dots,A_{n}] and the definition of Q^k\hat{Q}_{k} in (73), we have

4 Proof of Proposition 1

We provide the proof of the expected convergence of BR-ADMM here, as this proof is a slightly smaller subset of the proof of Theorem 1. We will just describe the necessary modifications.

We only need to prove a similar version of Theorem 2, i.e., the spectral radius of the expected update matrix of BR-ADMM is less than 1. Throughout the proof, we need to change the matrix Q=1∣Γ∣∑σ∈ΓQσQ=\frac{1}{|\Gamma|}\sum_{\sigma\in\Gamma}Q_{\sigma} to another one defined as

where ΓBR\Gamma^{\text{BR}} denotes the set of all possible permutations according to the Bernoulli randomization rule. It is easy to see that ∣ΓBR∣=2n|\Gamma^{\text{BR}}|=2^{n}. Other matrices such as MM should be changed accordingly.

The proof of Theorem 2 mainly consists of Lemma 1 and Lemma 2. Since Lemma 1 has nothing to do with the specific expression of QQ, so we only need to prove Lemma 2 for BR-ADMM, i.e., the matrix AQBRATAQ^{\text{BR}}A^{T} has all eigenvalues in the region (0,4/3)(0,4/3). Following the proof of Lemma 2, we divide the proof into three steps.

Step 0: Expression of ZBR≜I−AQBRATZ^{\text{BR}}\triangleq I-AQ^{\text{BR}}A^{T}. In Claim 4.1, we have prove the expression (51a) that I−AQσAT=PσnPσn−1…Pσ1I-AQ_{\sigma}A^{T}=P_{\sigma_{n}}P_{\sigma_{n-1}}\dots P_{\sigma_{1}} for any permutation σ\sigma, which implies

Step 1: Induction formula. Notice that a characteristic of the Bernoulli randomization rule is: the first block is either updated first or updated last. For instance, when n=4n=4, (1,3,4,2)(1,3,4,2) is a feasible permutation in ΓBR\Gamma^{\text{BR}} and (3,4,2,1)(3,4,2,1) is also a feasible permutation, but (3,1,4,2)(3,1,4,2) is not feasible. After removing the first block, the rest n−1n-1 blocks form a permutation in Γ^BR,\hat{\Gamma}_{\text{BR}}, where Γ^BR\hat{\Gamma}_{\text{BR}} is the set of all permutation of 2,3,…,n2,3,\dots,n according to the Bernoulli randomization rule. In other words, we have ΓBR={(1,σ^),(σ^,1), where σ^∈Γ^BR}\Gamma^{\text{BR}}=\{(1,\hat{\sigma}),(\hat{\sigma},1),\text{ where }\hat{\sigma}\in\hat{\Gamma}^{\text{BR}}\}. Thus we have an induction formula

where Z^BR\hat{Z}^{\text{BR}} is the lower dimensional analog of ZBRZ^{\text{BR}} for the rest n−1n-1 blocks (after removing the first block).

Step 2: Applying mathematical induction. This step is almost the same as Step 2 of the proof of Lemma 2. More specifically, combining the induction hypothesis that eig(Z^BR)∈(−1/3,1),\text{eig}(\hat{Z}^{\text{BR}})\in(-1/3,1), Strang’s result Lemma 5 and (78), we obtain the desired result eig(ZBR)∈(−1/3,1).\text{eig}(Z^{\text{BR}})\in(-1/3,1). This finishes the proof.

Proof of Technical Results for Expected Convergence Rates

Suppose all the distinct eigenvalues of I−QATAI-QA^{T}A are 0<τN′<⋯<τ1<4/30<\tau_{N^{\prime}}<\dots<\tau_{1}<4/3, where 1≤N′≤N1\leq N^{\prime}\leq N. Denote τmin⁡=τN′,τmax⁡=τ1.\tau_{\min}=\tau_{N^{\prime}},\tau_{\max}=\tau_{1}. According to Lemma 1, the expected update matrix of RP-ADMM MM has 2N′2N^{\prime} distinct eigenvalues λk,1,λk,2\lambda_{k,1},\lambda_{k,2} given by

Suppose the integer m∈[1,N′+1]m\in[1,N^{\prime}+1] satisfies τm≤1<τm−1\tau_{m}\leq 1<\tau_{m-1}. When m=1m=1, every τk≤1\tau_{k}\leq 1; when m=N′+1m=N^{\prime}+1, every τk>1\tau_{k}>1.

For N′≥k≥mN^{\prime}\geq k\geq m, i.e., τk≤1\tau_{k}\leq 1, we have τk(τk−1)≤0\tau_{k}(\tau_{k}-1)\leq 0, thus the two corresponding eigenvalues of MM are

which implies ∣λk,1∣=∣λk,2∣=(1−τk)2+τk(1−τk)=1−τk|\lambda_{k,1}|=|\lambda_{k,2}|=\sqrt{(1-\tau_{k})^{2}+\tau_{k}(1-\tau_{k})}=\sqrt{1-\tau_{k}}. Thus ρ1=max⁡N′≥k≥m{∣λk,1∣,∣λk,2∣}=1−τN′=1−τmin⁡\rho_{1}=\max_{N^{\prime}\geq k\geq m}\{|\lambda_{k,1}|,|\lambda_{k,2}|\}=\sqrt{1-\tau_{N^{\prime}}}=\sqrt{1-\tau_{\min}} if such kk exists; when such kk does not exist, i.e., τk>1  ∀  k\tau_{k}>1\;\forall\;k we denote ρ1=0\rho_{1}=0 which equals (1−τmin⁡)+\sqrt{(1-\tau_{\min})_{+}}. In summary, we have ρ1=(1−τmin⁡)+\rho_{1}=\sqrt{(1-\tau_{\min})_{+}}.

For m−1≥k≥1m-1\geq k\geq 1, i.e., τk>1\tau_{k}>1, we have τk(τk−1)>0\tau_{k}(\tau_{k}-1)>0. It is easy to verify λk,1>0>λk,2\lambda_{k,1}>0>\lambda_{k,2} and

Denote ρ2=max⁡m−1≥k≥1{∣λk,1∣,∣λk,2∣}\rho_{2}=\max_{m-1\geq k\geq 1}\{|\lambda_{k,1}|,|\lambda_{k,2}|\}, then ρ2=max⁡m−1≥k≥1{∣λk,2∣}=max⁡m−1≥k≥1{τk−1+τk(τk−1)}=τmax⁡−1+τmax⁡(τmax⁡−1)\rho_{2}=\max_{m-1\geq k\geq 1}\{|\lambda_{k,2}|\}=\max_{m-1\geq k\geq 1}\{\tau_{k}-1+\sqrt{\tau_{k}(\tau_{k}-1)}\}=\tau_{\max}-1+\sqrt{\tau_{\max}(\tau_{\max}-1)} if such kk exists; when such kk does not exist, i.e., τk≤1  ∀  k\tau_{k}\leq 1\;\forall\;k, we denote ρ2=0\rho_{2}=0 which equals (τmax⁡−1)++τmax⁡((τmax⁡−1)+)(\tau_{\max}-1)_{+}+\sqrt{\tau_{\max}((\tau_{\max}-1)_{+})}.

Combining the two scenarios, we have ρ(M)=max⁡N′≥k≥1{∣λk,1∣,∣λk,2∣}=max⁡{ρ1,ρ2}=max⁡{(1−τmin⁡)+,  (τmax⁡−1)++τmax⁡((τmax⁡−1)+)}.\rho(M)=\max_{N^{\prime}\geq k\geq 1}\{|\lambda_{k,1}|,|\lambda_{k,2}|\}=\max\{\rho_{1},\rho_{2}\}=\max\{\sqrt{(1-\tau_{\min})_{+}},\;(\tau_{\max}-1)_{+}+\sqrt{\tau_{\max}((\tau_{\max}-1)_{+})}\}.

In fact, when 4/3≥τ≥14/3\geq\tau\geq 1, we have 1−(τ−1+τ(τ−1))=2−τ−τ(τ−1)=(2−τ)2−τ(τ−1)2−τ+τ(τ−1)=3−4τ2−τ+τ(τ−1)≥34(3−4τ)1-(\tau-1+\sqrt{\tau(\tau-1)})=2-\tau-\sqrt{\tau(\tau-1)}=\frac{(2-\tau)^{2}-\tau(\tau-1)}{2-\tau+\sqrt{\tau(\tau-1)}}=\frac{3-4\tau}{2-\tau+\sqrt{\tau(\tau-1)}}\geq\frac{3}{4}(3-4\tau), thus τ−1+τ(τ−1)≤1−34(3−4τ).\tau-1+\sqrt{\tau(\tau-1)}\leq 1-\frac{3}{4}(3-4\tau). When τ<1\tau<1, clearly τ−1+τ(τ−1)=0\tau-1+\sqrt{\tau(\tau-1)}=0. Thus (τ−1)++τ(τ−1)+≤max⁡{0,1−34(4−3τ)}.(\tau-1)_{+}+\sqrt{\tau(\tau-1)_{+}}\leq\max\{0,1-\frac{3}{4}(4-3\tau)\}. For the second relation, if 0≤τ<10\leq\tau<1 then 1−τ=1−τ1+1−τ≤1−τ2\sqrt{1-\tau}=1-\frac{\tau}{1+\sqrt{1-\tau}}\leq 1-\frac{\tau}{2}; if 1≤τ≤4/31\leq\tau\leq 4/3 then (1−τ)+=0<1−12τ.\sqrt{(1-\tau)_{+}}=0<1-\frac{1}{2}\tau. Thus (1−τ)+≤1−12τ\sqrt{(1-\tau)_{+}}\leq 1-\frac{1}{2}\tau holds for any τ∈[0,4/3]\tau\in[0,4/3].

Substituting (79) into the expression of ρ(M)\rho(M), we obtain the desired inequality

2 Proof of Lemma 4

This is one of the two main lemmas of proving the expected convergence rate of RP-ADMM (the other is the expected convergence rate of RP-CD).

The proof outline of Lemma 4 and the main techniques are described below. The previous proof for the expected convergence of RP-ADMM in Section 6 is not strong enough to prove a convergence rate. We have to obtain a more refined estimate of the spectral radius of AQATAQA^{T}. To do so, we transform the induction formula in Proposition 3 to a “dual” form: instead of AQATAQA^{T}, we consider a similar matrix QATAQA^{T}A. We then apply the two simple techniques used in the proof of Lemma 1: factorize and rearrange, and reduce the dimension by eliminating a variable from the eigenvalue equation. We obtain a somewhat complicated inequality relating λmax⁡(QATA)\lambda_{\max}(QA^{T}A) and its lower-dimensional analog λmax⁡(Q^A^TA^)\lambda_{\max}(\hat{Q}\hat{A}^{T}\hat{A}). Finally, we perform a detailed analysis of the inequality to prove the desired bound.

Define a sequence {αk}k=1∞\{\alpha_{k}\}_{k=1}^{\infty} such that

It is easy to verify that 0<αk+1<αk≤1/30<\alpha_{k+1}<\alpha_{k}\leq 1/3 for all kk. The following claim provides a bound of αk\alpha_{k} (the proof will be given in Section 7.2.4).

Suppose the sequence {αk}k=1∞\{\alpha_{k}\}_{k=1}^{\infty} satisfies (80), then αk≥49(k+1),∀  k≥1.\alpha_{k}\geq\frac{4}{9(k+1)},\forall\;k\geq 1.

According to this claim, to prove the desired result λmax⁡(AQAT)≤43−49(k+1)\lambda_{\max}(AQA^{T})\leq\frac{4}{3}-\frac{4}{9(k+1)}, we only need to prove the following result:

We prove this result by mathematical induction. When n=1n=1, since ATA=A1TA1=IA^{T}A=A_{1}^{T}A_{1}=I, we have λmin⁡(AQAT)=λmax⁡(AQAT)=1=43−α1\lambda_{\min}(AQA^{T})=\lambda_{\max}(AQA^{T})=1=\frac{4}{3}-\alpha_{1}.

Suppose the result holds for n−1n-1, i.e., for a problem with n−1n-1 blocks, the eigenvalues of the corresponding matrix A^Q^A^T\hat{A}\hat{Q}\hat{A}^{T} lie in the region (0,43−αn−1)(0,\frac{4}{3}-\alpha_{n-1}).

Next, we build the induction formula, which is the dual form of the one we derived before. According to (75), we have

where in the last step we use the definitions

Sum up (85) for k=1,…,nk=1,\dots,n and applying (82), we have

To prove eig(AQAT)⊆(0,43−αn]\text{eig}(AQA^{T})\subseteq(0,\frac{4}{3}-\alpha_{n}], we only need to prove for any k=1,…,nk=1,\dots,n,

where {αk}\{\alpha_{k}\} is defined in (80). Define

Then eig(AQnAT)⊆(0,43−αn]\text{eig}(AQ_{n}A^{T})\subseteq(0,\frac{4}{3}-\alpha_{n}].

The proof of Proposition 4 will be divided into two parts, and given in Section 7.2.2 and Section 7.2.3.

We claim that (89) follows from the induction hypothesis (90) and the expressions of Aˉk\bar{A}_{k} and QkQ_{k} in (86). In fact, the above proposition directly proves (89) for k=nk=n. If we replace A,A^n,An,Q^n,QnA,\hat{A}_{n},A_{n},\hat{Q}_{n},Q_{n} by Aˉk,Ak^,Ak,Q^k,Qk\bar{A}_{k},\hat{A_{k}},A_{k},\hat{Q}_{k},Q_{k} respectively in the following proposition, we will obtain (89) for any kk. Finally, as mentioned earlier, the desired result eig(AQAT)⊆(0,34−αn]\text{eig}(AQA^{T})\subseteq(0,\frac{3}{4}-\alpha_{n}] in Lemma 2 follows immediately from (89) and (88).

In this subsection, we provide a proof of a weaker result eig(AQnAT)⊆(0,43)\text{eig}(AQ_{n}A^{T})\subseteq(0,\frac{4}{3}) under the conditions of Prop. 4; the proof of the desired result eig(AQnAT)⊆(0,43−αn]\text{eig}(AQ_{n}A^{T})\subseteq(0,\frac{4}{3}-\alpha_{n}] will be provided in the next subsection.

For simplicity, throughout this proof, we denote

According to the assumption of Prop. 4, we have

Since eig(Q^A^TA^)⊆(0,∞)\text{eig}(\hat{Q}\hat{A}^{T}\hat{A})\subseteq(0,\infty) and A^\hat{A} is non-singular, thus Q^≻0\hat{Q}\succ 0. Then we have Θ=WTQ^W⪰0\Theta=W^{T}\hat{Q}W\succeq 0, which proves the first relation of (94). By the definition W=A^TAnW=\hat{A}^{T}A_{n} we have

where the last equality is due to the assumption AnTAn=IA_{n}^{T}A_{n}=I, and the last inequality is due to the assumption (91). By (95) we have Θ≺43I\Theta\prec\frac{4}{3}I, thus (94) is proved.

We apply a trick that we have previously used: factorize QnQ_{n} and change the order of multiplication. To be specific, QnQ_{n} defined in (92) can be factorized as

where J≜[I0−12WTI]J\triangleq\begin{bmatrix}I&0\\ -\frac{1}{2}W^{T}&I\\ \end{bmatrix}, II in the upper left block denotes the (N−dn)(N-d_{n})-dimensional identity matrix, II in the lower right block denotes the dnd_{n}-dim identity matrix, and

In fact, we only need to prove Qn≻0Q_{n}\succ 0. According to (96), we only need to prove [Q^00C]≻0.\begin{bmatrix}\hat{Q}&0\\ 0&C\\ \end{bmatrix}\succ 0. This follows from Q^≻0\hat{Q}\succ 0 and the fact C=I−14WTQ^W≻\eqrefthetadef,n−blockI−13I≻0.C=I-\frac{1}{4}W^{T}\hat{Q}W\overset{\eqref{theta def, n-block}}{\succ}I-\frac{1}{3}I\succ 0. Thus (98) is proved.

We simplify the expression of ρ(AQnAT)\rho(AQ_{n}A^{T}) as follows:

Suppose λ>0\lambda>0 is the maximal eigenvalue of YY. According to (101) that ρ(AQnAT)=ρ(Y)\rho(AQ_{n}A^{T})=\rho(Y), we also have λ=λmax⁡(AQnAT)\lambda=\lambda_{\max}(AQ_{n}A^{T}). To prove (99), we only need to prove

If λI−C\lambda I-C is singular, i.e. λ\lambda is an eigenvalue of CC, then by (94) we have 23I≺C=1−14Θ⪯I\frac{2}{3}I\prec C=1-\frac{1}{4}\Theta\preceq I, which implies λ≤1\lambda\leq 1, thus (104) holds. In the following, we assume

since otherwise (105b) implies Cv0=λv0Cv_{0}=\lambda v_{0}, which combined with (106) leads to v0=0v_{0}=0 and thus v=0v=0, a contradiction.

Here we have used the definition C=I−14WTQ^W=I−14ΘC=I-\frac{1}{4}W^{T}\hat{Q}W=I-\frac{1}{4}\Theta. Since Θ\Theta is a symmetric matrix, Φ\Phi is also a symmetric matrix.

It is well-known that if αI+Θ\alpha I+\Theta is invertible, then Θ\Theta has an eigenvalue θ\theta iff (αI+Θ)−1(\alpha I+\Theta)^{-1} has an eigevalue (α+θ)−1(\alpha+\theta)^{-1}, and the corresponding eigen-vectors are the same. Similarly, since we already assumed (4λ−4)I+Θ(4\lambda-4)I+\Theta is invertible, θ\theta is an eigenvalue of Θ\Theta iff H=−Θ+λI−λ(4λ−4)[(4λ−4)I+Θ]−1H=-\Theta+\lambda I-\lambda(4\lambda-4)[(4\lambda-4)I+\Theta]^{-1} has an eigenvalue −θ+λ−λ(4λ−4)[(4λ−4)+θ]−1-\theta+\lambda-\lambda(4\lambda-4)[(4\lambda-4)+\theta]^{-1}. Recall that Θ=WTQ^W\Theta=W^{T}\hat{Q}W satisfies 0⪯Θ⪯λ^I0\preceq\Theta\preceq\hat{\lambda}I, thus any eigenvalue θ\theta satisfies 0≤θ≤λ^0\leq\theta\leq\hat{\lambda}. Therefore

Since v1≠0v_{1}\neq 0, without loss of generality, we can assume ∥v1∥=1\|v_{1}\|=1. We have

Case 1: max⁡θ∈[0,λ^]g(θ)≤0.\max_{\theta\in[0,\hat{\lambda}]}g(\theta)\leq 0. In this case, λ≤λ^<4/3\lambda\leq\hat{\lambda}<4/3, where the first inequality is due to (111), and the second inequality is due to the induction hypothesis. Thus in Case 1 (104) holds.

Case 2: max⁡θ∈[0,λ^]g(θ)>0.\max_{\theta\in[0,\hat{\lambda}]}g(\theta)>0. Then there exists some θ≥0\theta\geq 0 such that g(θ)>0g(\theta)>0. Note that g(θ)g(\theta) can also be expressed as g(θ)=θ(−1+λ(4λ−4)+θ)g(\theta)=\theta(-1+\frac{\lambda}{(4\lambda-4)+\theta}), thus

If λ<1\lambda<1, then (104) already holds; so we can assume λ>1\lambda>1. Thus (112) implies 1<λ(4λ−4)+θ≤λ4λ−41<\frac{\lambda}{(4\lambda-4)+\theta}\leq\frac{\lambda}{4\lambda-4}, which leads to λ<43\lambda<\frac{4}{3}. Thus in Case 2 (104) also holds. This finishes the proof of (104).

Remark: The proof of this subsection can lead to an alternative proof of Lemma 2. In particular, the induction step (Step 2) of Section 6.2 can be replaced by the proof here. The proof presented here is more complicated and less intuitive than the one in Section 6.2 (which is just a straightforward application of Strang’s result Lemma 5, but the benefit is that it can help establish a stronger bound of λ\lambda, as done in the next subsection.

2.3 Step 3: More Precise Bound of λ𝜆\lambda

We will continue the proof in Section 7.2.2, to further prove

If λ<1\lambda<1, then we are done since 1≤4/3−αn1\leq 4/3-\alpha_{n}. Assume 1≤λ<4/31\leq\lambda<4/3 from now on.

We first analyze the function g(θ)g(\theta). Taking the derivative of gg, we get

Since λ>1\lambda>1 and θ≥0\theta\geq 0, the term in the first bracket in the numerator is positive. Define

where the inequality holds due to λ<4/3.\lambda<4/3. Then we have

Therefore, g(θ)g(\theta) is increasing in [0,θ∗][0,\theta^{*}] and decreasing in [θ∗,∞)[\theta^{*},\infty). This implies

According to 0<λ<4/30<\lambda<4/3, we have λ>λ(4λ−4)=4λ−4+θ∗⇒−1+λ4λ−4+θ∗>0⇒g(θ∗)>0.\lambda>\sqrt{\lambda(4\lambda-4)}=4\lambda-4+\theta^{*}\Rightarrow-1+\frac{\lambda}{4\lambda-4+\theta^{*}}>0\Rightarrow g(\theta^{*})>0. Together with (115) we obtain max⁡{0,max⁡θ∈[0,λ^]g(θ)}≤g(θ∗)\max\{0,\max_{\theta\in[0,\hat{\lambda}]}g(\theta)\}\leq g(\theta^{*}). Substituting into (114), we obtain

We will derive an inequality on λ\lambda and λ^\hat{\lambda} from the above relation as below. Substituting the expression of g(⋅)g(\cdot) into the relation, we obtain

It is easy to verify that h(t)h(t) is increasing in t∈[0,4/3]t\in[0,4/3]; in fact, h′(t)=36(2+3t)2−1=(8+3t)(4−3t)(2+3t)2≥0h^{\prime}(t)=\frac{36}{(2+3t)^{2}}-1=\frac{(8+3t)(4-3t)}{(2+3t)^{2}}\geq 0 for t∈[0,4/3]t\in[0,4/3]. According to (93), we have δ^=4/3−λ^≥αn−1\hat{\delta}=4/3-\hat{\lambda}\geq\alpha_{n-1}. Applying the monotonicity of hh, we have

which combined with (117) leads to (113). This finishes the proof of Proposition 4.

2.4 Proof of Claim 7.1

Define another sequence as ωk=163αk−9k\omega_{k}=\frac{16}{3\alpha_{k}}-9k. Then αk=16319k+ωk\alpha_{k}=\frac{16}{3}\frac{1}{9k+\omega_{k}} and ω1=7\omega_{1}=7, ω2=38/5.\omega_{2}=38/5. We then derive the recurrence equation of ωk\omega_{k}. According to (80), we have

It is easy to see that ωk>0⇒ωk+1>ωk>0\omega_{k}>0\Rightarrow\omega_{k+1}>\omega_{k}>0, thus

Furthermore, ωk+1=ωk+99k+ωk−1≤ωk+1k,\omega_{k+1}=\omega_{k}+\frac{9}{9k+\omega_{k}-1}\leq\omega_{k}+\frac{1}{k}, thus

The lower bound and upper bound on ωk\omega_{k} imply upper and lower bounds on αk\alpha_{k}:

As a side comment, this implies that lim⁡k→∞αk=1627k≈0.59k.\lim_{k\rightarrow\infty}\alpha_{k}=\frac{16}{27k}\approx\frac{0.59}{k}. For our purpose, we need a universal lower bound on αk\alpha_{k}. When k≥3k\geq 3, we have 3k≥8+log⁡(k−1)3k\geq 8+\log(k-1), thus 12k≥9k+8+log⁡(k−1)12k\geq 9k+8+\log(k-1), which further implies

Combining with the bound (118), we obtain

Notice that α1=13>49⋅12,\alpha_{1}=\frac{1}{3}>\frac{4}{9}\cdot\frac{1}{2}, and α2=524>49⋅13\alpha_{2}=\frac{5}{24}>\frac{4}{9}\cdot\frac{1}{3}, we have αk>49(k+1)\alpha_{k}>\frac{4}{9(k+1)} for any k≥1k\geq 1. This finishes the proof of the claim.

3 Proof of Lemma 3

We first prove the case n=2n=2, n=3n=3 and n=4n=4, then prove the general case n=2kn=2k and n=2k+1n=2k+1 separately.

When n=2n=2, (119) reduces to P1P2+P2P1⪯P1+P2P_{1}P_{2}+P_{2}P_{1}\preceq P_{1}+P_{2}. Notice that Pi=Pi2P_{i}=P_{i}^{2} since PiP_{i} is a projection matrix, we have P1+P2−P1P2+P2P1=P12+P22−P1P2+P2P1=(P1−P2)2=(P1−P2)T(P1−P2)⪰0P_{1}+P_{2}-P_{1}P_{2}+P_{2}P_{1}=P_{1}^{2}+P_{2}^{2}-P_{1}P_{2}+P_{2}P_{1}=(P_{1}-P_{2})^{2}=(P_{1}-P_{2})^{T}(P_{1}-P_{2})\succeq 0.

When n=3n=3, (119) reduces to 16∑i,j,k are distinctPiPjPk⪯13(P1+P2+P3)\frac{1}{6}\sum_{i,j,k\text{ are distinct}}P_{i}P_{j}P_{k}\preceq\frac{1}{3}(P_{1}+P_{2}+P_{3}). Note that (Pi−Pk)Pj(Pi−Pk)⪰0(P_{i}-P_{k})P_{j}(P_{i}-P_{k})\succeq 0, thus

Summing up the above inequality for all possible triples (i,j,k)(i,j,k), we get

We then need to bound the left-hand-side of the above inequality. Since I−Pj⪰0I-P_{j}\succeq 0, we have Pi(I−Pj)Pi⪰0P_{i}(I-P_{j})P_{i}\succeq 0, which implies Pi⪰PiPjPi.P_{i}\succeq P_{i}P_{j}P_{i}. Summing up this inequality for all pairs i≠ji\neq j, we obtain 16∑i≠jPiPjPi⪯13(P1+P2+P3)\frac{1}{6}\sum_{i\neq j}P_{i}P_{j}P_{i}\preceq\frac{1}{3}(P_{1}+P_{2}+P_{3}). Combining with (120), we obtain the desired inequality 16∑i,j,k are distinctPiPjPk⪯13(P1+P2+P3)\frac{1}{6}\sum_{i,j,k\text{ are distinct}}P_{i}P_{j}P_{k}\preceq\frac{1}{3}(P_{1}+P_{2}+P_{3}).

The proof for n=4n=4 illustrates partially the gist of a general proof, so we present this proof. When n=4n=4, (119) reduces to 124∑i,j,k,l are distinctPiPjPkPl⪯14(P1+P2+P3+P4)\frac{1}{24}\sum_{i,j,k,l\text{ are distinct}}P_{i}P_{j}P_{k}P_{l}\preceq\frac{1}{4}(P_{1}+P_{2}+P_{3}+P_{4}). Similar to (120) in the n=3n=3 case, we first prove

To prove this inequality, we need the following two basic inequalities:

Summing up this inequality for all possible (i,j,k,l)(i,j,k,l) that are distinct, we obtain (121). Similar to the proof of n=3n=3 case, we have 112∑i≠jPiPjPi≤14(P1+P2+P3+P4)\frac{1}{12}\sum_{i\neq j}P_{i}P_{j}P_{i}\leq\frac{1}{4}(P_{1}+P_{2}+P_{3}+P_{4}), thus combining with (121) we obtain the desired result.

We next prove the case n=2kn=2k, where k≥2k\geq 2 is a positive integer. We will prove that

where Γk\Gamma_{k} is the set of kk-permutations of 1,2,…,n1,2,\dots,n (here, a kk-permutation is a permutation of kk distinct numbers chosen from 1,2,…,n1,2,\dots,n), and Eσ∈ΓE_{\sigma\in\Gamma} and Eπ∈ΓkE_{\pi\in\Gamma_{k}} denote the expectation over a uniform distribution on Γ\Gamma and Γk\Gamma_{k} respectively.

To prove (122), we need the following fact: for any ϵ=(ϵ1,…,ϵk)∈{1,−1}k\epsilon=(\epsilon_{1},\dots,\epsilon_{k})\in\{1,-1\}^{k}, we have

This relation holds because for any positive-semidefinite matrix XX and any symmetric matrix YY, we have YXY=YTXY⪰0YXY=Y^{T}XY\succeq 0. Applying this fact kk times leads to (123).

The expression of Gσ,ϵG_{\sigma,\epsilon} in (123) involves 2k2^{k} terms in the form of Pi1Pi2…PinP_{i_{1}}P_{i_{2}}\dots P_{i_{n}}. To prove (122), only two terms are of interest to us. The strategy is to pick ϵi\epsilon_{i}’s properly so that summing up a bunch of relations of the form (123) will eliminate all but the two desired terms. We elaborate this strategy below.

For example, when k=3k=3, Λ3={(−1,1,1),(1,−1,1),(1,1,−1),(−1,−1,−1)}\Lambda_{3}=\{(-1,1,1),(1,-1,1),(1,1,-1),(-1,-1,-1)\}, and the complement Λ3c={(1,1,1),(−1,−1,1),(−1,1,−1),(1,−1,−1)}\Lambda_{3}^{c}=\{(1,1,1),(-1,-1,1),(-1,1,-1),(1,-1,-1)\}. As a well-known fact,

This matrix Gσ,ϵG_{\sigma,\epsilon} can be expressed as the sum of 2k2^{k} terms, and each term is of the form ±Pπ1…Pπn\pm P_{\pi_{1}}\dots P_{\pi_{n}}, where πi∈{σi,σn+1−i}\pi_{i}\in\{\sigma_{i},\sigma_{n+1-i}\}. For the fixed permutation σ\sigma, define a set

For most of the proof, we will use the abbreviation Ωt=Ωt(σ),t=0,1,2.\Omega_{t}=\Omega_{t}(\sigma),t=0,1,2. For any π=(π1,…,πn)∈Ω\pi=(\pi_{1},\dots,\pi_{n})\in\Omega, define an indicator vector of π\pi as δ(π)=(δ1,…,δk),\delta(\pi)=(\delta_{1},\dots,\delta_{k}), where each δi\delta_{i} is determined by

In the expression of Gσ,ϵG_{\sigma,\epsilon}, half of the terms have coefficient 11 and the other half have coefficient −1-1. To understand which terms have coefficient 11 and which have coefficient −1-1, consider a special ϵ=(−1,1,…,1)\epsilon=(-1,1,\dots,1), i.e., ϵ1=−1\epsilon_{1}=-1 and all other ϵi=1\epsilon_{i}=1. A term with coefficient −1-1 has the form Pσ1Pπ2…Pπn−1PσnP_{\sigma_{1}}P_{\pi_{2}}\dots P_{\pi_{n-1}}P_{\sigma_{n}} or PσnPπ2…Pπn−1Pσ1P_{\sigma_{n}}P_{\pi_{2}}\dots P_{\pi_{n-1}}P_{\sigma_{1}}, i.e., with an indicator vector whose first element δ1=1\delta_{1}=1, and a term with coefficient 11 has the form Pσ1Pπn−1…Pπ2Pσ1P_{\sigma_{1}}P_{\pi_{n-1}}\dots P_{\pi_{2}}P_{\sigma_{1}} or PσnPπn−1…Pπ2PσnP_{\sigma_{n}}P_{\pi_{n-1}}\dots P_{\pi_{2}}P_{\sigma_{n}}, i.e., with an indicator vector whose first element δ1=0\delta_{1}=0. We can see that the coefficient is in fact ϵ1δ1\epsilon_{1}^{\delta_{1}}. For general ϵ∈Λ\epsilon\in\Lambda and π∈Ω\pi\in\Omega, the coefficient of PπnPπn−1…Pπ2Pπ1P_{\pi_{n}}P_{\pi_{n-1}}\dots P_{\pi_{2}}P_{\pi_{1}} in Gσ,ϵG_{\sigma,\epsilon} is (ϵ1)δ1…(ϵk)δk(\epsilon_{1})^{\delta_{1}}\dots(\epsilon_{k})^{\delta_{k}}, where δ=δ(π)\delta=\delta(\pi) is defined as in (125). We can then write the expression of Gσ,ϵG_{\sigma,\epsilon} as

Summing up this relation for all ϵ\epsilon in Λk\Lambda_{k}, we have

Note that in this expression, δ1,…,δk\delta_{1},\dots,\delta_{k} depend on π\pi.

For any δ≠0k\delta\neq 0_{k}, we have gk(δ)+hk(δ)=∑ϵ∈{1,−1}kϵ1δ1…ϵkδk=(1δ1+(−1)δ1)…(1δk+(−1)δk)=0,g_{k}(\delta)+h_{k}(\delta)=\sum_{\epsilon\in\{1,-1\}^{k}}\epsilon_{1}^{\delta_{1}}\dots\epsilon_{k}^{\delta_{k}}=(1^{\delta_{1}}+(-1)^{\delta_{1}})\dots(1^{\delta_{k}}+(-1)^{\delta_{k}})=0, thus

We will prove: for any δ∉{0k,1k},\delta\notin\{0_{k},1_{k}\},

We prove (130) by induction on kk. When k=2k=2, Λ2={(−1,1),(1,−1)}\Lambda_{2}=\{(-1,1),(1,-1)\}, we have:

Now consider kk. Since δ≠0k\delta\neq 0_{k}, there must exist some jj such that δj=1\delta_{j}=1; without loss of generality, we assume

If ϵ\epsilon contains an odd number of −1-1 and the last element ϵk=1\epsilon_{k}=1 (or ϵk=−1\epsilon_{k}=-1), then the first k−1k-1 elements contain an odd (or even) number of −1-1. Thus

Split gk(δ)g_{k}(\delta) into two parts gk(δ)=gk,1(δ)+gk,2(δ),g_{k}(\delta)=g_{k,1}(\delta)+g_{k,2}(\delta), where

Denote δ^=(δ1,…,δk−1)\hat{\delta}=(\delta_{1},\dots,\delta_{k-1}). We already assume δ≠1k\delta\neq 1_{k} and δk=1\delta_{k}=1, so we know

But it is possible that δ^=0k−1\hat{\delta}=0_{k-1}. Consider two cases.

Case 1: δ^=0k−1\hat{\delta}=0_{k-1}, i.e., δ=(0k−1,1)\delta=(0_{k-1},1).

Case 2: δ^≠0k−1\hat{\delta}\neq 0_{k-1}. Together with (135), we have

which enables us to apply the induction hypothesis (131) and its corollary (132). In fact,

Thus gk(δ)=gk,1(δ)+gk,2(δ)=0g_{k}(\delta)=g_{k,1}(\delta)+g_{k,2}(\delta)=0.

In both cases, we have proved gk(δ)=0g_{k}(\delta)=0, which finishes the induction step. Therefore (130) holds for any kk.

Next, we analyze the sum ∑ϵ∈ΛkGσ,ϵ.\sum_{\epsilon\in\Lambda_{k}}G_{\sigma,\epsilon}. According to (127), we have

where (i) is due to (126) and (ii) is due to (129), (130). According to (123), any Gσ,ϵ⪰0G_{\sigma,\epsilon}\succeq 0, thus the above relation implies the following important relation

Note that this relation holds for a fixed permutation σ\sigma and the corresponding set Ω0=Ω(σ)\Omega_{0}=\Omega(\sigma) and Ω1(σ)\Omega_{1}(\sigma). Each π∈Ω0\pi\in\Omega_{0} corresponds to a kk-permutation χ\chi of (12…n)(12\dots n) determined by π=(χ1…χk−1χkχkχk−1…χ1)\pi=(\chi_{1}\dots\chi_{k-1}\chi_{k}\chi_{k}\chi_{k-1}\dots\chi_{1}) and each π∈Ω1\pi\in\Omega_{1} corresponds to a permutation of (12…n)(12\dots n). We rewrite (136) as

and summing up this relation for all possible permutations σ∈Γ\sigma\in\Gamma leads to

In fact, for any positive-semidefinite matrix XX and any symmetric matrix YY, we have YXY=YTXY⪰0YXY=Y^{T}XY\succeq 0. Applying this fact k−1k-1 times leads to (137).

Combining (122) and (137), we immediately obtain the desired result (119) for the case n=2kn=2k.

The case that n=2k−1n=2k-1 is an odd number is almost the same, except that the key quantity Gσ,ϵG_{\sigma,\epsilon} is now defined as

In words, we pair PσiP_{\sigma_{i}} with Pσn+1−iP_{\sigma_{n+1-i}} for i=1,…,k−1i=1,\dots,k-1 and leave PσkP_{\sigma_{k}} alone (following the same rule it would have been paired with itself). The rest of the proof is almost the same as the even case, so we skip it. Q.E.D.

Numerical Experiments

In the numerical experiments, we set b=0b=0, thus the unique optimal solution is x∗=0x^{*}=0. The coefficient matrix AA will be generated according to one of the random distributions below:

Gauss: independent Gaussian entries Ai,j∼N(0,1)A_{i,j}\sim\mathcal{N}(0,1).

Log-normal: independent log-normal entries Ai,j∼exp(N(0,1))A_{i,j}\sim\text{exp}(\mathcal{N}(0,1)).

Uniform: each entry is drawn independently from a uniform distribution on $$.

Circulant Hankel: circulant Hankel matrix with independent standard Gaussian entries. More specifically, generate δ1,δ2,…,δN∼N(0,1)\delta_{1},\delta_{2},\dots,\delta_{N}\sim\mathcal{N}(0,1) and let Ai,j=δi+j−1A_{i,j}=\delta_{i+j-1} (define δk=δk−N\delta_{k}=\delta_{k-N} if k>Nk>N). Note that the entries of the circulant Hankel matrix are not independent since one δi\delta_{i} can appear in multiple positions.

For the two ADMM algorithms, we only consider the nn-coordinate versions, i.e. each block consists of only one coordinate. We let the three tested algorithms start from the same random initial point y0=[x0;λ0]y^{0}=[x^{0};\lambda^{0}] (GD will start from x0x^{0}). To measure the performance, we define the epoch complexity kk to be the minimum kk so that the relative error

where ϵ\epsilon is a desired accuracy (we consider 10−210^{-2} and 10−310^{-3}For high accuracy such as ϵ=10−6\epsilon=10^{-6}, it takes too many epochs for the algorithms to converge when n=100n=100 as most matrices we generated are highly ill-conditioned, so we do not report the results. Based on the limited experiments for high accuracy, similar gaps between RP-ADMM and GD are observed. ). For the two ADMM algorithms, one epoch refers to one round of primal and dual steps; for GD, one epoch refers to one gradient step. The total computation time should be proportional to the epoch complexity since GD and the two ADMM variants have similar per-epoch costIn matlab simulations each epoch of GD takes much less time than a round of ADMM because matlab implements matrix operations much faster than a “for” loop. For a more fair CPU time comparison, one should use other programming languages such as C. : a gradient descent step xk+1=xk−αAT(Ax−b)x^{k+1}=x^{k}-\alpha A^{T}(Ax-b) contains two matrix-vector multiplications and thus takes time 2N2+O(N)2N^{2}+O(N), and an ADMM round also takes time 2N2+O(N)2N^{2}+O(N) (the primal update step of ADMM takes time 2N2+O(N)2N^{2}+O(N) and the dual update step of ADMM takes time O(N)O(N)). We test 1000 random instances for N∈{3,10}N\in\{3,10\} and 300300 random instances for N=100N=100, and record the geometric mean of the number of epochs. In the table, “Diverg. Ratio” represents the percentage of tested instances for which cyclic ADMM diverges and “CycADMM” represents “cyclic ADMM” (note that RP-ADMM converges in all instances we tested, so its divergence ratio is 0). Note that for cyclic ADMM we only report the epoch complexity when it converges, while for RPADMM and GD we report the epoch complexity in all tested instances. If restricting to the successful instances of cyclic ADMM, we find that the epoch complexity of RPADMM does not change too much, while the epoch complexity of GD will be reduced (significantly in some settings).

The simulation results are summarized in Table 2. The main observations from the simulation are:

For all random distributions of AA we tested, cyclic ADMM does not always converge even when NN is fixed to be 33. For N=100N=100 and many random distributions, cyclic ADMM diverges with probability 11. This means that the divergence of cyclic ADMM is not merely a “worst-case” phenomenon, but actually quite common. When the dimension increases, the divergence ratio will increase.

For standard Gaussian entries, cyclic ADMM converges with high probability. When cyclic ADMM converges, it converges faster than RP-ADMM and sometimes much faster.

RPADMM typically converges faster than the basic gradient descent method and sometimes more than 1010 times faster.

We have also tested BR-ADMM for solving the same problems, though the simulation results are not listed in the above table. As expected, BR-ADMM also always converges for solving these linear systems. The convergence speed is usually slower than RP-ADMM. Nevertheless, BR-ADMM can save some sampling time compared to RP-ADMM, and may be more favorable if random permutation is not available due to system architecture constraint. The detailed comparison of BR-ADMM and RP-ADMM, and the design of other randomized schemes or even deterministic schemes that outperform RP schemes are left as future work.

Concluding Remarks

In this paper, we prove the expected convergence of randomly permuted ADMM (RP-ADMM) for solving a non-singular square system of equations (extension to non-square systems is straightforward). We also prove a bound on the expected convergence rate of RP-ADMM for solving linear systems and the expected convergence rate of RP-BCD for solving quadratic problems. The motivation is to resolve the divergence issue of cyclic multi-block ADMM. Our result shows that RP-ADMM may serve as a simple remedy, and we expect RP-ADMM to be one of the important solvers in large-scale optimization. One interesting finding along the path is that the update matrix of RP-BCD has spectrum lying in (−1/3,1)(-1/3,1) instead of the commonly seen (−1,1)(-1,1).

Randomly permutation is widely known to be empirically better than independently randomized versions, but little was known about its theoretical properties in general. Note that most existing analyses of BCD (e.g. ) are applicable to both the cyclic update rule and the random permutation update rule. However, in light of a recent study which established an up to O(n2)O(n^{2}) gap between cyclic CD and R-CD , it is unlikely that RP-CD will have the same rate as cyclic CD. Our result in this paper established, for the first time, an O(n)O(n) gap between RP-CD and cyclic-CD for general quadratic problems, making some progress towards the conjecture that RP-CD is faster than R-CD.

We emphasize that the convergence speed analysis of large-scale optimization has mostly been limited to independently randomized update order in the past decade. Going beyond independent randomized order is an important topic for enlarging the scope of large-scale optimization. Not only the analysis of random permutation is quite challenging, even the analysis of the most classical cyclic order is highly nontrivial . There are quite a few open questions regarding the convergence rate of non-independent-randomized order. Regarding the random permutation order, a very interesting open question is the worst-case convergence rate of RP-BCD for quadratic problems. Due to the close relation with matrix AM-GM inequality, this problem seems to be a quite fundamental problem. Moving to ADMM, the similar questions about the convergence rate of various variants of ADMM, including RP-ADMM and BR-ADMM, are also open.

Acknowledgment

We thank an anonymous reviewer for many helpful comments on the manuscript, which enabled us to improve the presentation of the paper.

References