Beneath the valley of the noncommutative arithmetic-geometric mean inequality: conjectures, case-studies, and consequences

Introduction

Randomized sequential algorithms abound in machine learning and optimization. The most famous is the stochastic gradient method (Bottou 1998; Bertsekas 2012; Nemirovski et al. 2009; Shalev-Shwartz and Srebro 2008, see), but other popular methods include algorithms for alternating projections (Strohmer and Vershynin 2009; Leventhal and Lewis 2010, see), proximal point methods (Bertsekas 2011, see), coordinate descent (Nesterov 2010, see) and derivative free optimization (Nesterov 2011; Nemirovski and Yudin 1983, see). In all of these cases, an iterative procedure is derived where, at each iteration, an independent sample from some distribution determines the action at the next stage. This sample is selected with-replacement from a pool of possible options.

In implementations of many of these methods, however, practitioners often choose to break the independence assumption. For instance, in stochastic gradient descent, many implementations pass through each item exactly once in a random order (i.e., according to a random permutation). In randomized coordinate descent, one can cycle over the coordinates in a random order. These strategies, employing without-replacement sampling, are often easier to implement efficiently, guarantee that every item in the data set is touched at least once, and often have better empirical performance than their with-replacement counterparts (Bottou 2009; Recht and Ré 2011; Feng et al. 2012, see).

Unfortunately, the analysis of without-replacement sampling schemes are quite difficult. The independence assumption underlying with-replacement sampling provides an elegant Markovian framework for analyzing incremental algorithms. The iterates in without-replacement sampling are correlated, and studying them requires sophisticated probabilistic tools. Consequently, most of the analyses without-replacement optimization assume that the iterations are assigned deterministically. Such deterministic orders might incur exponentially worse convergence rates than randomized methods (Nedic and Bertsekas 2000), and deterministic orders still require careful estimation of accumulated errors (Luo 1991; Tseng 1998, see). The goal of this paper is to make progress towards patching the discrepancy between theory and practice of without-replacement sampling in randomized algorithms.

In particular, in many cases, we demonstrate that without-replacement sampling outperforms with-replacement sampling provided a noncommutative version of the arithmetic-geometric mean inequality holds. Namely, if $\bm{A}_{1},\ldots,\bm{A}_{n}$ are a collection of $d\times d$ positive semidefinite matrices, we define the arithmetic and (symmetrized) geometric means to be

where $S_{n}$ denotes the group of permutations. Our conjecture is that the norm of $\bm{M}_{G}$ is always less than the norm of $(\bm{M}_{A})^{n}$. Assuming this inequality, we show that without-replacement sampling leads to faster convergence for both the least mean squares and randomized Kaczmarz algorithms of Strohmer and Vershynin 2009.

Using established work in matrix analysis, we show that these noncommutative arithmetic-geometric mean inequalities hold when there are only two matrices in the pool. We also prove that the inequality is true when all of the matrices commute. We demonstrate that if we don’t symmetrize, there are deterministically ordered products of $n$ matrices whose norm exceeds $\|\bm{M}_{A}\|^{n}$ by an exponential factor. That is, symmetrization is necessary for the noncommutative arithmetic-geometric mean inequality to hold.

While we are unable to prove the noncommutative arithmetic-geometric mean inequality in full generality, we verify that it holds for many classes of random matrices. Random matrices are, in some sense, the most interesting case for machine learning applications. This is particularly evident in applications such as empirical risk minimization and online learning where the the data are conventionally assumed to be generated by some i.i.d random process. In Section 4, we show that if $\bm{A}_{1},\ldots,\bm{A}_{n}$ are generated i.i.d. from certain distributions, then the noncommutative arithmetic-geometric mean inequality holds in expectation with respect to the $\bm{A}_{i}$. Section 4.1 assumes that $\bm{A}_{i}=\bm{Z}_{i}\bm{Z}_{i}^{T}$ w.b.here $\bm{Z}_{i}$ have independent entries, identically sampled from some symmetric distribution. In Section 4.2, we analyze the random matrices that commonly arise in stochastic gradient descent and related algorithms, again proving that without-replacement sampling exhibits faster convergence than with-replacement sampling. We close with a discussion of other open conjectures that could impact machine learning theory, algorithms, and software.

Sampling in incremental gradient descent

To illustrate how with- and without-replacement sampling methods differ in randomized optimization algorithms, we focus on one core algorithm, the Incremental Gradient Method (IGM). Recall that the IGM minimizes the function

Here, $\bm{x}_{0}$ is an initial starting vector, $\gamma_{k}$ are a sequence of nonnegative step sizes, and the indices $i_{k}$ are chosen using some (possibly deterministic) sampling scheme. When $f$ is strongly convex, the IGM iteration converges to a near-optimal solution of (2.1) for any $\bm{x}_{0}$ under a variety of step-sizes protocols and sampling schemes including constant and diminishing step-sizes (Anstreicher and Wolsey 2000; Bertsekas 2012; Nemirovski et al. 2009, see). When the increments are selected uniformly at random at each iteration, IGM is equivalent to stochastic gradient descent. We use the term IGM here to emphasize that we are studying many possible orderings of the increments. In the next examples, we study the specialized case where the $f_{i}$ are quadratic and the IGM is equivalent to the least mean squares algorithm of Widrow and Hoff 1960.

First consider the following toy one-dimensional least-squares problem

where $y_{i}$ is a sequence of scalars with mean $\mu_{y}$ and variance $\sigma^{2}$. Applying (2.2) to (2.3) results in the iteration.

If we initialize the method with $x_{0}=0$ and take $n$ steps of incremental gradient with stepsize $\gamma_{k}=1/k$, we have

where ${i_{j}}$ is the index drawn at iteration $j$. If the steps are chosen using a without-replacement sampling scheme, $x_{n}=\mu_{y}$, the global minimum. On the other hand, using with-replacement sampling, we will have

Another toy example that further illustrates the discrepancy is the least-squares problem

where $\beta_{i}$ are positive weights. Here, $y$ is a scalar, and the global minimum is clearly $y$. Let’s consider the incremental gradient method with constant stepsize $\gamma_{k}=\gamma<\min\beta_{i}^{-1}$. Then after $n$ iterations we will have

If we perform without-replacement sampling, this error is given by

On the other hand, using with-replacement sampling yields

By the arithmetic-geometric mean inequality, we then have that the without-replacement sample is always closer to the optimal value in expectation. This sort of discrepancy is not simply a feature of these toy examples. We now demonstrate that similar behavior arises in multi-dimensional examples.

2 IGM in more than one dimension

We want to compare with- vs without-replacement sampling for IGD on the cost function

Suppose we walk over $k$ steps of IGD with constant stepsize $\gamma$ and we access the terms $i_{1},\ldots,i_{k}$ in that order. Then we have

Subtracting $x_{\star}$ from both sides of this equation then gives

Here, the product notation means we multiply by the matrix with smallest index first, then left multiply by the matrix with the next index and so on up to the largest index.

In this case, we need to compare the expected value of matrix products under with or without-replacement sampling schemes in order to conclude which is better. Is there a simple conjecture, analogous to the arithmetic-geometric mean inequality, that would guarantee without-replacement sampling is always better?

3 The Randomized Kaczmarz algorithm

As another high-dimensional example in the same spirit, we consider the randomized Kaczmarz algorithm of Strohmer and Vershynin 2009. The Kaczmarz algorithm is used to solve the over-determined linear system $\bm{\Phi}\bm{x}=\bm{y}$. Here $\bm{\Phi}$ is an $n\times d$ matrix with $n>d$ and we assume there exists an exact solution $\bm{x}_{\star}$ satisfying $\bm{\Phi}\bm{x}_{\star}=\bm{y}$. Kaczmarz’s method solves this system by alternating projections (Kaczmarz 1937) and was implemented in the earliest medical scanning devices (Hounsfield 1973). In computer tomography, this method is called the Algebraic Reconstruction Technique (Herman 1980; Natterer 1986) or Projection onto Convex Sets (Sezan and Stark 1987).

Kaczmarz’s algorithm consists of iterations of the form

where the rows of $\Phi$ are accessed in some deterministic order. This sequence can be interpreted as an incremental variant of Newton’s method on the least squares cost function

with step size equal to $1$ (Bertsekas 1999, see).

Establishing the convergence rate of this method proved difficult in imaging science. On the other hand, Strohmer and Vershynin 2009 proposed a randomized variant of the Kaczmarz method, choosing the next iterate with-replacement with probability proportional to the norm of $\bm{\phi}_{i}$. Strohmer and Vershynin established linear convergence rates for their iterative scheme. Expanding out (2.7) for $k$ iterations, we see that

Let us suppose that we modify Strohmer and Vershynin’s procedure to employ without-replacement sampling. After $k$ steps is the with-replacement or without-replacement model closer to the optimal solution?

Conjectures concerning the norm of geometric and arithmetic means of positive definite matrices

That is, we average the value of $f$ over all ordered tuples of elements from $(x_{1},\ldots,x_{n})$. Similarly, the with-replacement expectation is defined as

With these conventions, we can list our main conjectures as follows:

Let $\bm{A}_{1},\ldots,\bm{A}_{n}$ be a collection of positive semidefinite matrices. Then we conjecture that the following two inequalities always hold:

Assuming this conjecture holds, let us return to the analysis of the IGM (2.6). Assuming that $\bm{x}_{0}-\bm{x}_{\star}$ is an arbitrary starting vector and that (3.2) holds, we have that each term in this summation is smaller for the without-replacement sampling model than for the with-replacement sampling model. In turn, we expect the without-replacement sampling implementation will return lower risk after one pass over the data-set. Similarly, for the randomized Kaczmarz iteration (2.7), Conjecture 3.1 implies that a without-replacement sample will have lower error after $k<n$ iterations.

In the remainder of this document we provide several case studies illustrating that these noncommutative variants of the arithmetic-geometric mean inequality hold in a variety of settings, establishing along the way tools and techniques that may be useful for proving Conjecture 3.1 in full generality.

Both of the inequalities (3.1) and (3.2) are true when $n=2$. These inequalities all follow from an well-estabilished line of research in estimating the norms of products of matrices, started by the seminal work of Bhatia and Kittaneh 1990.

Proof Let $\bm{A}$ and $\bm{B}$ be positive definite matrices. Both of our arithmetic-geometric mean inequalities follow from the stronger inequality

This bound was proven by Bhatia and Kittaneh 2000. In particular, since

Here, the first inequality is the triangle inequality and the subsequent equality follows because the norm of $\bm{X}^{T}\bm{X}$ is equal to the squared norm of $\bm{X}$. The second inequality is (3.3).

in the semidefinite ordering. But this follows by observing

where $Q(p,q)=3/16$ if $p=q$ and $-1/16$ otherwise. Since $p$ and $q$ both take $4$ possible values, the matrix $Q$ is positive definite which means that $\bm{X}_{R}-\bm{X}_{L}$ can be written as a nonnegative sum of products of the form $\bm{Y}\bm{Y}^{T}$. We conclude that $\bm{X}_{R}-\bm{X}_{L}$ must be positive define and hence $\|\bm{X}_{R}\|\geq\|\bm{X}_{L}\|$, completing the proof An explicit decomposition (3.5) into Hermitian squares was initially found using the software NCSOSTools by Cafuta et al. 2011. This software finds decompositions of matrix polynomials into sums of Hermitian squares. Our argument was constructed after discovering this decomposition..

Note that this proposition actually verifies a stronger statement: for two matrices, the arithmetic-geometric mean inequality holds for deterministic orderings of two matrices. We will discuss below how symmetrization is necessary for more than two matrices. In fact, considerably stronger inequalities hold for symmetrized products of two matrices. As a striking example, the symmetrized geometric mean actually precedes the square of the arithmetic mean in the positive definite order. Let $\bm{A}$ and $\bm{B}$ be positive semidefinite. Then we have

This ordering breaks for $3$ matrices as evinced by the counterexample

The interested reader should consult Bhatia and Kittaneh 2008 for a comprehensive list of inequalities concerning pairs of positive semidefinite matrices.

Unfortunately, these techniques are specialized to the case of two matrices, and no proof currently exists for the inequalities when $n\geq 3$. There have been a varied set of attempts to extend the noncommutative arithmetic-geometric mean inequalities to more than two matrices. Much of the work in this space has focused on how to properly define the geometric mean of a collection of positive semidefinite matrices. For instance, Ando et al. 2004 demarcate a list of properties desirable by any geometric mean, with one of the properties being that the geometric mean must precede the arithmetic mean in the positive-definite ordering. Ando et al derive a geometric mean satisfying all of these properties, but the resulting mean in no way resembles the means of matrices discussed in this paper. Instead, their geometric mean is defined as a fixed point of a nonlinear map on matrix tuples. Bhatia and Holbrook 2006 and Bonnabel and Sepulchre 2009 propose geometric means based on geodesic flows on the Riemannian manifold of positive definite matrices, however these means also do not correspond to the averaged matrix products that we study in this paper.

2 When is it not necessary to symmetrize the order?

When the matrices commute, Conjecture 3.1 is a consequence of the standard arithmetic-geometric mean inequality (more precisely, a consequence of Maclaurin’s inequalities).

Let $x_{1},\ldots,x_{n}$ be positive scalars. Let

be the normalized $k$th symmetric sum. Then we have

Note that $s_{1}\geq\sqrt[n]{s_{n}}$ is the standard form of the arithmetic-geometric mean inequality. See Hardy et al. 1952 for a discussion and proof of this chain of inequalities.

To see that these inequalities immediately imply Conjecture 3.1 when the matrices $\bm{A}_{i}$ are mutually commutative, note first that when $d=1$, we have

The higher dimensional analogs follow similarly. If all of the $\bm{A}_{i}$ commute, then the matrices are mutually diagonalizable. That is, we can write $\bm{A}_{i}=\bm{U}\Lambda_{i}\bm{U}^{T}$ where $\bm{U}$ is an orthogonal matrix, and the $\Lambda_{i}=\operatorname{diag}(\lambda_{1}^{(i)},\ldots,\lambda_{d}^{(i)})$ are all diagonal matrices of the eigenvalues of $\bm{A}_{i}$ in descending order. Then we have

verifying our conjecture. In fact, in this case, any order of the matrix products will satisfy the desired arithmetic-geometric mean inequalities.

3 When is it necessary to symmetrize the order?

In contrast, symmetrizing over the order of the product is necessary for noncommutative operators. The following example, communicated to us by Aram Harrow, provides deterministic without-replacement orderings that have exponentially larger norm than the with-replacement expectation. Let $\omega_{n}=\pi/n$. For $n\geq 3$, define the collection of vectors

Note that all of the $\bm{a}_{k;n}$ have norm $1$ and, for $1\leq k<n$, $\langle\bm{a}_{k;n},\bm{a}_{k+1;n}\rangle=\cos\left(\omega_{n}\right)$. The matrices $\bm{A}_{k}:=\bm{a}_{k;n}\bm{a}_{k;n}^{T}$ are all positive semidefinite for $1\leq k\leq n$, and we have the identity

Any set of unit vectors satisfying (3.7) is called a normalized tight frame, and the vectors (3.6) form a harmonic frame due to their trigonometric origin (Hassibi et al. 2001; Goyal et al. 2001, see). The product of the $\bm{A}_{i}$ is given by

Therefore, the arithmetic mean is less than the deterministically ordered matrix product for all $n\geq 3$.

It turns out that this harmonic frame example is in some sense the worst case. The following proposition shows that the geometric mean is always within a factor of $d^{k}$ of the arithmetic mean for any ordering of the without-replacement matrix product.

Let $\bm{A}_{1},\ldots,\bm{A}_{n}$ be $d\times d$ positive semidefinite matrices. Then

Proof If we sample $j_{1},\ldots,j_{k}$ uniformly from $[n]$, then we have

Here, the first inequality follows from the triangle inequality. The second, because the operator norm is submultiplicative. The third inequality follows because the trace dominates the operator norm. The fourth inequality is Maclaurin’s. The fifth inequality follows because the trace of a $d\times d$ positive semidefinite matrix is upper bounded by $d$ times the operator norm. The final inequality is again the triangle inequality.

Note that this worst-case bound holds for deterministic orders of matrix products as well. Once we apply the submultiplicative property of the operator norm, all of the non-commutativity is washed out of the problem. Examples of deterministic matrix products saturating this upper bound can be constructed in higher dimensions using frames. If $d$ even, set

Then one can verify again using standard trigonometric identities that

and that the inner products of adjacent $\bm{f}_{i}$ are

These inner products are approximately $1-\frac{\pi^{2}(d^{2}-1)}{6n^{2}}$ for large $n$. Thus, each of these cases violate the arithmetic-geometric mean inequality for the order $(1,2,\ldots,k)$ by a factor of approximately $d^{k}$ provided $n\geq d$.

At first glance, the harmonic frames example appears to cast doubt on the validity of Conjecture 3.1. However, after symmetrizing over the symmetric group, we can show that the $d=2$ harmonic frames do obey (3.1).

Let \lambda(n)=\,_{2}F_{3}\left[\begin{array}[]{ccc}1&-n/2+1/2&-n/2\\ 1/2&-n+1\end{array};1\right]. With the $\bm{a}_{k;n}$ defined in (3.6),

This theorem additionally verifies that there is an asymptotic gap between the arithmetic and geometric means of the harmonic frames example after symmetrization. We include a full proof of this result in Appendix B. The proof treats the norm variationally using the identity that $\|\bm{X}\|_{2}$ is the maximum of $\bm{v}^{T}\bm{X}\bm{v}$ over all unit vectors $\bm{v}$. Our computation then reduces to effectively computing a Fourier transform of the function of $\bm{v}$ in an appropriately defined finite group. We show that the Fourier coefficients can be viewed as enumerating sets, and we compute them exactly using generating functions.

The combinatorial argument that we use to prove Theorem 3.5 is very specialized. To provide a broader set of examples, we now turn to show that Conjecture 3.1 does in fact hold for many classes of random matrices.

Random matrices

In this section, we show that if $\bm{A}_{1},\ldots,\bm{A}_{n}$ are generated i.i.d. from certain distributions, then Conjecture 3.1 holds in expectation with respect to the $\bm{A}_{i}$. Section 4.1 assumes that $\bm{A}_{i}=\bm{Z}_{i}\bm{Z}_{i}^{T}$ where $\bm{Z}_{i}$ have independent entries, identically sampled from some symmetric distribution. In Section 4.2, we explore when the matrices $\bm{A}_{i}$ are random rank-one perturbations of the identity as was the case in the IGM and Kaczmarz examples.

For each $i=1,\ldots,n$, suppose $\bm{A}_{i}=\bm{Z}_{i}\bm{Z}_{i}^{T}$ with $\bm{Z}_{i}$ a $d\times r$ random matrix whose entries are i.i.d. samples from some symmetric distribution. Then Conjecture 3.1 holds in expectation.

Proof Suppose the entries of each $\bm{Z}_{i}$ have finite variance $\sigma^{2}$ (the theorem would be otherwise vacuous if we assumed infinite variance). Let the $(a,b)$ entry of $\bm{Z}_{i}$ be denoted by $Z_{a,b}^{(i)}$. Also, denote by $\bm{W}$ the matrix with all of the $\bm{Z}_{i}$ stacked as columns: $\bm{W}=\sigma^{-1}[\bm{Z}_{1},\ldots,\bm{Z}_{n}]$.

Let’s first prove that (3.1) holds in expectation for these matrices. First, consider the without-replacement samples, which are considerably easy to analyze. Let $(j_{1},\ldots,j_{k})$ be a without-replacement sample from $[n]$. Then

Note that since $W_{ij}$ are iid, symmetric random variables, each term in this sum is zero if it contains an odd power of $W_{ij}$ for some $i$ and $j$. If all of the powers in a summand are even, its expected value is bounded below by $1$. A simple lower bound for this final term (4.2) thus looks only at the contribution from when all of the indices $a_{i}$ are set equal to $1$.

Here the inequality is Jensen’s. This calculation proves (3.1) for our family of random matrices. That is, we have demonstrated that the expected value of the with-replacement sample has greater norm than the expected value of the without-replacement sample.

We compute this identity in a second way that describes its combinatorics more explicitly, which we will use as to derive our lower bound.

Now consider the case that some index may be repeated (i.e., there exist $k,l$ such that $i_{j}=i_{l}$ for $j\neq l$). The key observation is the following. Let $w$ be a real-valued random variable with a finite second moment. Then,

With equality only for $p=0,1$. This is Jensen’s inequality applied to $x^{p}$ for $x\geq 0$ (since $w$ is real then $w^{2}$ is positive, and $x^{p}$ is convex on $[0,\infty)$ for $p=0,1,2,\dots$). To verify the inequality, let $n_{i}$ be the number of times index $i$ is repeated and observe

(4.5) follows from (4.3), since all terms are non-negative. (4.6) inequality is repeated application of (4.4). The final expression is precisely equal to the without-replacement average. Now, the with replacement average can be bounded as

Since each term in this last expression exceeds the without-replacement expectation, this completes the proof.

The arguments used to prove Theorem 4.1 grossly undercount the number of terms that contribute to the expectation. Bounds on the quantity (4.1) commonly arise in the theory of random matrices (see the survey by Bai 1999, for more details and an extensive list of references). Indeed, if we let $d=\delta n$ and assume that $W_{ij}$ have bounded fourth moment, we have that (4.1) tends to $(1+\sqrt{\delta})^{2k}$ almost surely a $n\rightarrow\infty$. That is, the gap between the with- and without-replacement sampling grows exponentially with $k$ in this scaling regime. Similarly, there is an asymptotic, exponential gap between the with and without-replacement expectations in (3.2). Observe that (4.4) is strict for $p\geq 2$ for $\chi$-squared random variables. Thus, for Wishart matrices, if there is even a single repeated value, i.e., $i_{j}=i_{l}$ for $j\neq l$, inequality (4.6) is strict. In Appendix A, we analyze the case where the $\bm{Z}_{i}$ are Gaussian (and hence the $\bm{A}_{i}$ are Wishart) and demonstrate that the ratio of the expectation is bounded below by $re^{\frac{1}{4k(k+1)}}\left(\frac{16k}{e^{2}r(r+d+1)}\right)^{k}$.

2 Random vectors and the incremental gradient method

We can also use a random analysis to demonstrate that for the least-squares problem (2.4), without-replacement sampling outperforms with-replacement sampling if the data is randomly generated.

because $\bm{a}_{j_{k}}$ is chosen independently from $(\bm{a}_{j_{1}},\ldots,\bm{a}_{j_{k-1}})$ On the other hand, in the with-replacement model, we have

In this case, we cannot distribute the expected value because the vector $\bm{x}-\bm{x}_{\star}$ depends on all $\bm{a}_{i}$ for $1\leq i\leq n$. To get a flavor for how these differ, consider the conditional expectation

This means that the with-replacement upper bound is worse than the without-replacement estimate with reasonably high probability on most models of $\bm{a}_{i}$. Under mild conditions on $\bm{a}_{i}$ (including Gaussianity, bounded entries, subgaussian moments, or bounded Orlicz norm), we can estimate tail bounds for the eigenvalues of $\Lambda_{n}$ and $\Delta_{n}$ (by applying the techniques of Tropp 2011, for example). These large deviation inequalities provide quantitative estimates of the gap between with- and without-replacement sampling for the least mean squares and randomized Kaczmarz algorithms. Similar, but more tedious analysis, would reveal that with-replacement sampling fares worse with diminishing step sizes as well.

Numerical Evidence

As described in the introduction, there is substantial numerical evidence that with-replacement sampling underperforms without-replacement sampling in many randomized algorithms. We invite the interested reader to consult Bottou 2009; Recht and Ré 2011; Feng et al. 2012, and many other articles in the machine learning literature to substantiate these empirical claims. However, for completeness, we provide a few examples demonstrating the gap for the examples in Section 2.

In Figure 1, we display six comparisons of with- and without-replacement sampling. In the first row, we show the discrepancy when running the randomized Kaczmarz algorithm when the rows of $\bm{\Phi}$ are the $d$-dimensional, defined by (3.8). In the second row, we plot the results for incremental gradient descent with $\bm{a}_{i}=\bm{f}_{i}$ in the same harmonic frames example. Finally, the third row plots performance when the rows of $\bm{\Phi}$ are generated i.i.d. from Haar measure on the sphere. In all three cases, without-replacement sampling converges faster than with-replacement sampling, and when $d$ and $n$ are close, the convergence rate is considerably faster.

Discussion and open problems

While i.i.d. matrices are of significant importance in machine learning, the major piece of open work is proving Conjecture 3.1 for all positive semidefinite matrix tuples or finding a counterexample for either of the assertions. As demonstrated by the harmonic frames example, symmetrized products of deterministic matrices become quickly tedious and difficult to study. Some sort of combinatorial structure might need to be exploited for a short proof to arise in general. It remains to be seen if this sort of combinatorics employed in proving Theorem 3.5 generalizes beyond this particular example, but we expect these techniques will be useful in future studies of Conjecture 3.1. In particular, it would be interesting to see if we could reduce the proof of the conjecture to verifying the conjecture on frames that arise as the orbit of the representation of some finite group. These frames have been fully classified by Hassibi et al. 2001, and would reduce Conjecture 3.1 to a finite list of cases.

The generalization of (3.3) to $n\geq 3$ asserts a stronger version of (3.1)

Certainly, (3.1) follows from (6.1) by Jensen’s inequality the triangle inequality. Moreover, using the same argument we used in proving Proposition 3.2, (3.2) also follows from (6.1). When $n\geq 3$, is it the case that (6.1) holds? It could be that for general matrices, it is easier to analyze (6.1) rather than (3.2) because the right hand side is in terms of the arithmetic mean, rather than the more complicated quadratic matrix products in (3.2).

Effect of biased orderings.

Another possible technique for solving incremental algorithms is to choose the best ordering of the increments to reach the cost function. In terms of matrices, can we find the ordering of the matrices $\bm{A}_{i}$ that achieves the minimum norm. At first glance this seems daunting. Suppose $\bm{A}_{i}=\bm{a}_{i}\bm{a}_{i}^{T}$ where the $\bm{a}_{i}$ are all unit vectors. Then for $\sigma\in S_{n}$

minimizing this expression with respect to $\sigma$ amounts to finding the minimum weight traveling salesman path in the graph with weights $\log|\langle\bm{a}_{i},\bm{a}_{j}\rangle|$. Are there simple heuristics that can get within a small constant of the optimal tour for these graphs? How do greedy heuristics fare? This sort of approach was explored with some success for the Kaczmarz method by Eldar and Needell 2011.

Nonlinear extensions

Extending even the random results in this paper to nonlinear algorithms such as the general incremental gradient descent algorithm or randomized coordinate descent would require modifying the analyses used here. However, it would be of interest to see which of the randomization tools employed in this work can be extended to the nonlinear case. For example, if we assume that the cost function (2.1) has summands which are sampled i.i.d., can we use similar tools (e.g., Jensen’s inequality, moment bounds) to show that without-replacement sampling works even in the nonlinear case?

Acknowledgements

The authors would like to thank Dimitri Bertsekas, Aram Harrow, Pablo Parrilo for many helpful conversations and suggestions. BR is generously supported by ONR award N00014-11-1-0723 and NSF award CCF-1139953. CR is generously supported by the Air Force Research Laboratory (AFRL) under prime contract no. FA8750-09-C-0181, the NSF CAREER award under IIS-1054009, ONR award N000141210041, and gifts or research awards from Google, Greenplum, Johnson Controls, Inc., LogicBlox, and Oracle. Any opinions, findings, and conclusion or recommendations expressed in this work are those of the authors and do not necessarily reflect the views of any of the above sponsors including DARPA, AFRL, or the US government.

References

Appendix A Additional calculations for random matrices

For the special case of Wishart matrices, we can show that the gap between the norm of the arithmetic and geometric means in 3.2 is quite large.

The first inequality is because all the terms are positive and we are selecting out only the self loops. The equality just groups terms. The following lower bound completes the proof.

A simple corollary is the following lower bound on the arithmetic mean

We examine the following ratio $\rho(r,k,d)$

For fixed $r,d$, $\rho$ grows exponentially with $k$.

Proof We use a very crude lower and upper bound pair that holds for all $k$ (Cormen et al. 2009, p. 55).

Appendix B Proof that harmonic frames satisfy the noncommutative arithmetic-geometric mean inequality

Let $S$ be the symmetrized geometric mean of a set of rank $1$, idempotent matrices that are parametrized by angles $\phi_{1},\dots,\phi_{n}$ (This is slightly more general than we need for our theorem above). Our goal is to compute the $2$-norm of $S$:

B.1 Cosine combinatorics

We will write this function as fourier transform (we pull out the $2^{n}$ for convenience):

To find the $c_{k}$ and $d_{k}$, we repeatedly apply the following identity:

Fix $\phi_{1},\dots,\phi_{2},\dots,\phi_{n},\dots\in[0,2\pi]$. We first consider a related form, $T_{n}$, for $n=1,2,3,\dots,$ defined by the following recurrence

We compute $T_{n}$ using the above transformation. But, first, we show the pattern by example:

In our computation above, $\psi_{1}=\phi_{v}=\psi_{n}$. And so, after writing this out, we will get two kinds of terms: even terms (corresponding to $c_{k}$) that do not depend on $\psi_{v}$ (they cancel) and odd terms that do contain $2\psi_{v}$.

With $T_{n}$ as defined above, we have for $n\geq 2$

Fix an $n$. We now count a symmetrized version of $T_{n}$ defined as follows: For $\sigma\in S_{n}$:

We now show that $S_{n}$ can be written in a form that removes the permtuation. We also assume some structure here that mimics our product above, namely that $\phi_{1}=\phi_{n}$.

Let $\phi_{1},\dots,\phi_{n}\in[0,2\pi]$ such that $\phi_{1}=\phi_{n}$. Then,

Proof To see this formula, Consider a pair of sets $X,Y\subseteq[n]$. In how many permutations $\sigma\in S_{n}$ does $(X,Y)$ contribute a term? We need to choose $|X|+|Y|$ positions for these terms to appear out of $n$ possible places in the order. Thus there are ${n\choose|X|+|Y|}$ permutations to choose the slots for $(X,Y)$. An $(|X|,|Y|)$ pair only appears in a permutation in $\sigma$ if the elements of $X$ and $Y$ can be alternated starting with $X$. This implies that $|X|=2|Y|+z_{i}$ where $z_{i}\in\{0,1\}$. Moreover, there are $|X|!|Y|!(n-(|X|+|Y||)!$ permutations that respect this structure (for any choice of $|X|+|Y|$ slots, any ordering of $X$ and $Y$ and the elements outside can occur).

Pushing the $1/n!$ factor inside completes the proof.

B.3 Counting on harmonic, finite groups

In the case we care about, the $\phi_{i}$ have more structure: the set $\{2\phi_{i}\}_{i=1}^{n}$ forms a cyclic group under addition modulo $2\pi$. Let $n$ denote the number of elements in the frame. Fix $n$. Let $\zeta$ denote a $n$th root of unity. Define a (harmonic) generating function $f$

We give a shorthand for its coefficients $q_{k,m}$ and $r_{k,m}$ as follows

$\sum_{i\in X}i-\sum_{j\in Y}j=k\mod n$ (since we inspect $\zeta^{k}$),

$|X|=|Y|=m$ (since we inpect $x^{m}y^{m}$),

For $r_{k,m}$ the only change is that $|X|=|Y|+1$ (since $x^{m+1}y^{m}$). With this notation, we can express the coefficients from Eq. B.1.

We use this representation to prove that the symmetrized geometric mean is rotationally invariant (i.e., $d_{k}=0$ for $k=0,1,\dots,n-1$). First, we show that all $d_{k}$ are equal.

Consider a frame of size $n$. For any $m$ and $k,l=0,\dots,n-1$, $\sum_{k}d_{k}\cos(\phi_{v}+2\pi k)=0$.

Proof This follow by examining the generating function above. First observe that we have congruence $f(x,x^{j}y,z)=f(x,y,z)$ for $j=0,\dots,n-1$ tells us that $[x^{j}yz^{m}]f=[yz^{m}]f$. And, the congruence that $[x^{j}yz^{m}]f=[x^{j}y^{-1}z^{m}]f$. Combining these facts, we have that $r_{k,m}=r_{l,m}$. Since this holds for all $k,l$, we can conclude that $d_{k}=d_{l}$ by summing over $m$. Finally, since $\sum_{l=0}^{n}\cos(\phi_{v}+2l\pi n^{-1})=0$ for any fixed $\phi_{v}$ we conclude the lemma.

Since the symmetrized geometric mean does not depend on $\phi_{v}$, we conclude it must be of the form $\alpha I$ for some $\alpha$. The remainder of this note is to compute that $\alpha$.

B.4 Computing the coefficients

The argument of this subsection is a generalization of that of Konvalina 1995.

Since $f(\zeta,x,y)=f(\zeta^{i},x,y)$ for any integer $n$, $F_{n}(x,y)=f(\zeta,x,y)$ is a function of $n$ alone. That is, we can write

Thus, claim boils down to $F_{n}(x,-y)=R_{n}(x,y)$.

We show that the zero sets of $F_{n}(x,-y)$ and $R_{n}$ are equal. The zero set of $F_{n}(x,-y)$ is the set of lines described by

where $\zeta$ is any $n$-th root of unity. Substituting $y$ at the root equation, we get that $xy=x\zeta+\zeta^{2}x^{2}$.

Now, we check that the following is zero:

The first equality follows from $1+4\zeta x+4\zeta^{2}x^{2}=(2x\zeta+1)^{2}$. The second is just algebra. Finaly, we use on each term that $\zeta^{n}=1$ and that $\zeta+\zeta^{2}x=-y$. This claim holds for all $\zeta$ that are roots of unity, and so the function is identically zero.

B.5 Finally, to a hypergeometric series

It is possible to get an explicit formula for $\lambda$ that is related to ${}_{3}F_{2}$. We consider the following series and show that it is hypergeometric in $k$: