# Is function $X \mapsto \mbox{Tr}\left(X X^T X X^T\right)$ convex?

1. Is $$\mbox{Tr}\left(X X^T X X^T\right)$$ a convex function of arbitrary real matrix $$X$$?

2. More generally, is $$\mbox{Tr}\left(\left(X X^{\dagger}\right)^m\right)$$ a convex function of arbitrary complex matrix $$X$$ for any integer $$m \ge 1$$?

Any advice or suggestions would be greatly appreciated.

The proof hint:

Let us apply SVD to a matrix $$X$$: $$X$$ = $$U D V^{\dagger}$$. Every matrix has SVD with non-negative singular values on the main diagonal of $$D$$. Next:

$$\left(X X^{\dagger}\right)^m$$ = $$U D V^{\dagger} V D U^{\dagger} U D V^{\dagger} V D U^{\dagger} \ldots U D V^{\dagger} V D U^{\dagger}$$ = $$U D^{2 m} U^{\dagger}$$.

Note, all unitary matrices are cancelled in between, because $$U^{\dagger}U = V^{\dagger}V = I$$.

$$\mbox{Tr} \left(\left(X X^{\dagger}\right)^m\right)$$ = $$\mbox{Tr} \left(U D^{2 m} U^{\dagger}\right)$$ = $$\mbox{Tr} \left(D^{2 m} U^{\dagger} U\right)$$ = $$\mbox{Tr} \left(D^{2 m}\right)$$ = $$\sum_i \sigma_i^{2 m}$$, where we used the cyclic property of trace operation, and $$\{\sigma_i\}$$ are the singular values of matrix $$X$$.

Let $$\{x_i\}$$, $$\{y_i\}$$ and $$\{z_i\}$$ be the singular values of arbitrary matrices $$X$$, $$Y$$ and their convex combination $$Z$$ = $$\alpha X + (1 - \alpha) Y$$ respectively. It was shown below by @PSL that for the Frobenius norm ($$m = 1$$) the following holds true:

$$\alpha \sum_i x_i^2 + (1 - \alpha) \sum_i y_i^2 \ge \sum_i z_i^2$$.

Considering that the function $$\phi: x \rightarrow x^m, x \in R^+$$ is convex, would it be possible to show that the case $$m > 1$$ is also satisfied:

$$\alpha \sum_i x_i^{2 m} + (1 - \alpha) \sum_i y_i^{2 m} \ge \sum_i z_i^{2 m}$$ ?

Update: by numerical simulation I found that $$\sum_i x_i^{2} \ge \sum_i z_i^{2}$$ does not necessarily entails $$\sum_i x_i^{4} \ge \sum_i z_i^{4}$$ on roughly 9% of random configurations. Seems like this line of thoughts does not work. However, the extended brute force simulation still succeeds for $$m$$ 2 to 5.

Brute force approach to answer the questions. Here I literally check convexity on random matrices. The Python code speaks for itself:

import numpy as np

tol = 10.0 * np.finfo(float).eps
count_ok, count_fail = int(0), int(0)

for m in range(2, 5 + 1):
print("m:", m)
for dim in range(2, 10 + 1):
print(f"matrix size: {dim}x{dim}")
for test in range(100000):
X = 2 * np.random.rand(dim, dim) - 1
Y = 2 * np.random.rand(dim, dim) - 1
XXt = X @ X.T
YYt = Y @ Y.T
for t in np.linspace(0.01, 0.99, 20):
Z = X * t + Y * (1 - t)
ZZt = Z @ Z.T
ok = (np.trace(np.linalg.matrix_power(ZZt, m)) <=
np.trace(np.linalg.matrix_power(XXt, m)) * t +
np.trace(np.linalg.matrix_power(YYt, m)) * (1 - t) + tol)
if ok:
count_ok += 1
else:
count_fail += 1

print(f"succeeded: {count_ok} times")
print(f"failed: {count_fail} times")
print("")
.......
succeeded: 72,000,000 times
failed: 0 times

• What did you try to solve the problem? Nov 14, 2021 at 21:33
• Sorry, I did not get the question. Nov 14, 2021 at 21:40
• The question of @Arctic Char is crystal clear : Have you made some previous Web searching, have you made some computational attempts, for example computations entrywise in dimension 2 or 3 ? etc. Nov 14, 2021 at 22:02
• What I tried beforehand was quite naive. For now, I realised which direction to move on, thanks to @Bananach. Nov 14, 2021 at 22:35
• Brute force simulation with random matrices of size 2x2 to 10x10 favours the "yes" answer to the first question (1.800.000 trials have been made in total). Nov 14, 2021 at 23:57

Your problem reduces to showing that $$(XX^T)^m$$ is convex in the matrix sense. Recall that a matrix valued function $$f$$ is convex, if $$\alpha f(X) + (1-\alpha) f(Y) \succeq f(\alpha X + (1-\alpha)Y)$$.

This is because $$A \succeq B$$ implies $$\textrm{Tr}(A) \geq \textrm{Tr}(B)$$.

For $$m = 1$$ this is not too difficult:

Claim 1. $$AA^T\succeq 0$$.

Proof. $$v^TAA^Tv = \|Av\|^2_2 \geq 0$$ for every $$v$$.

Claim 2. $$f(X) = XX^T$$ is convex in the matrix sense.

Proof. Compute $$\alpha f(X) + (1-\alpha) f(Y) - f(\alpha X + (1-\alpha)Y) = \alpha(1-\alpha)XX^T + \alpha(1-\alpha)YY^T - \alpha(1-\alpha)(XY^T + YX^T).$$ We want to show that the above is $$\succeq 0$$, which happens if and only if $$XX^T + YY^T - (XY^T + YX^T) \succeq 0$$, as $$\alpha \in (0,1)$$. But this is just $$(X - Y)(X - Y)^T$$, which is positive semidefinite by Claim 1. Hence, $$\alpha f(X) + (1-\alpha) f(Y) \succeq f(\alpha X + (1-\alpha)Y)$$.

•  Didactic proof ! See here for other approaches. Nov 15, 2021 at 8:43

By https://en.m.wikipedia.org/wiki/Trace_inequality the function $$A\mapsto \text{tr} f(A)$$ is convex if $$f\colon \mathbb{R}\to\mathbb{R}$$ is.

• But here $f$ isn't a function $\mathbb R \to \mathbb R$... Nov 14, 2021 at 22:04
• @JeanMarie ah, if $X$ were symmetric it would be though, by spectral calculus. Not sure what's salvageable for asymmetric matrices Nov 14, 2021 at 22:14
• @Bananach, thank you for suggestion. This is a really good starting point for me. Nov 14, 2021 at 22:36

Yes, $$X\mapsto\operatorname{tr}((XX^\ast)^m)$$ is convex for any integer $$m\ge1$$.

The function is a composition of the inner function $$h:M_n(\mathbb C)\ni X\mapsto A=XX^\ast\in\mathbb S_+$$ (where $$\mathbb S_+$$ denotes the set of all positive semidefinite matrices) and the outer function $$g:\mathbb S_+\ni A\mapsto\operatorname{tr}(A^m)\in\mathbb R$$. The matrix-valued inner function $$h$$ is convex because $$\theta h(X)+(1-\theta)h(Y)-h\left(\theta X+(1-\theta)Y\right) =\theta(1-\theta)(X-Y)(X-Y)^\ast$$ is positive semidefinite. The outer function $$g$$ is convex because it is in the form of $$g(A)=\operatorname{tr}(f(A))$$ (see footnote below), where $$f(x)=x^m$$ is a continuous convex function on the positive reals. Clearly $$g(A)=\operatorname{tr}(A^m)$$ is also (weakly) increasing on $$\mathbb S_+$$ (although $$A\mapsto A^m$$ is not increasing on $$\mathbb S_+$$ in general). Therefore $$g\circ h$$ is a composition of a convex inner function and a convex increasing outer function. Hence it is convex.

Footnote.

As pointed out by the Wikipedia article linked by Bananach's answer here, a proof of the result that $$f$$ is continuous and convex implies $$A\mapsto\operatorname{tr} f(A)$$ is convex can be found, for instance, in the 2009 paper Trace inequalities and quantum entropy: an introductory course by Eric Carlen. Although the domain and codomain of the $$f$$ in this paper are different from ours (reals vs positive reals and Hermitian vs PSD), the same proof applies.

• Thank you very much @user1551. Very concise and clear explanation. I tried a similar way but was unable to show that $g(A)$ is increasing. Thanks for the reference. For anyone who is interested why $g(A)$ should be increasing see, for example, link this answer, where inequalities should be replaced by matrix definiteness relations. Nov 18, 2021 at 22:41