9
$\begingroup$

Let $A_i$ with $i=1,\dots,N$ and $p$ be real $M\times M$ matrices. Further, let $p$ be positive definite, i.e., $p\succ 0$, with $Tr(p)=1$. Let $0< a_i<1$ and $\sum_{i=1}^N a_i = 1$.

Claim: $Tr( A_1 p^{a_1} A_2 p^{a_2} \dots A_N p^{a_N} ) \leq \|A_1\| \|A_2\|\dots \|A_N\|$

Here, \|X\| denotes the operator norm of $X$ (= largest singular value).

Can you prove this claim (at least for symmetric matrices A_i)? It is trivial for N=1: $Tr(A p)=\sum_i A_{i,i} p_i\leq \|A\|\sum_i p_i = \|A\|$, where I used the representation of $A$ in the eigenbasis of $p$.

The claim even seems to hold if one uses N different matrices $p_i\succ 0$, $Tr(p_i)=1$ on the left-hand side of the claim.

thanks a lot, Tom

$\endgroup$
10
$\begingroup$

The claim is true. More generally the Hölder inequality holds for the Schatten $p$-norms. The statement is here without proof in wikipedia, and by induction it implies that for matrices $X_1,\dots, X_K$ and $p_1,\dots,p_K \in [1,\infty]$ with $\sum_i 1/p_i=1$, $$ |Tr(X_1\dots X_K)| \leq \prod \|X_i\|_{p_i}.$$ In your setting, take $K=2N$, $X_{2k} = p^{a_k}$, $X_{2k-1} = A_k$, $p_{2k} = 1/a_k$ and $p_{2k-1} = \infty$.

I guess that all this is clearly explained (and with no mistake?) in Barry Simon's book Trace ideals and their applications.

$\endgroup$
7
$\begingroup$

I include some information about Hölder's inequality just for completion of details for Mikael's nice answer.

The Schatten-$p$ norm of a matrix $X$ is defined as $$\|X\|_p := \Bigl(\sum\nolimits_i \sigma_i^p(X)\Bigr)^{1/p},$$ where $\sigma_i(X)$ denotes the $i$-th singular value of $X$.

From this definition, we see that $\|X\|_p = \|\sigma(X)\|_p$, where $\sigma(X)$ is the vector of singular values.

From von Neumann's trace inequality we know that $$|\mbox{trace}(XY)| \le \langle \sigma(X), \sigma(Y)\rangle,$$

while from the Hölder inequality for vectors, we have $$\langle \sigma(X), \sigma(Y)\rangle \le \|\sigma(X)\|_p\|\sigma(Y)\|_q,$$ where $1/p + 1/q = 1$.

On combining the above two, we immediately get alleged Schatten-p norm Hölder inequality.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.