# trace norm of AGB, where G is Gaussian random matrix

Suppose that $A$, $B$ are deterministic $n\times n$ matrices and $G$ a Gaussian matrix of i.i.d. entries $N(0,1)$.

I'd like to establish an upper bound of the trace norm of $AGB$ as $$\mathbb{E}\|AGB\|_\ast \leq \max\{\|A\|_\ast \|B\|_F, \|A\|_F\|B\|_\ast \},$$ where $\|\cdot\|_\ast$ denotes the trace norm and $\|\cdot\|_F$ the Frobenius norm. I can show the inequality with an additional $\log n$ factor on the right-hand side but I do not think the $\log n$ factor is necessary...

(First question: is it true to replace $\max$ with $\min$?)

[Update] As answered by Mikael de la Salle below, it is possible to replace $\max$ with $\min$ for $p\in [1,2]$, to obtain that $$\mathbb{E}\|AGB\|_p \leq \min\{\|A\|_p \|B\|_F, \|A\|_F \|B\|_p\}.$$

For $p>2$, it is necessary to have $\max$. Take $A$ to be identity matrix and $B = e_1e_1^T$ (zero matrix except the top-left entry being $1$), then $\|AGB\|_p \sim \sqrt{n}$ while $\|A\|_p \|B\|_F = n^{1/p} < \sqrt{n}$.

Second question: Does the following hold for $p>2$? $$\mathbb{E}\|AGB\|_p \leq C\max\{\|A\|_p \|B\|_F, \|A\|_F \|B\|_p\}.$$ for some constant $C$ (which can depend on $p$)? I could get the inequality with an extra $\log n$ factor on the right-hand side, but I am not sure if it is necessary. For $p=\infty$, it is too easy to get $\sqrt{n}\|A\|_{op}\|B\|_{op}$, which is bigger than the bound I want.

Third question: How about the lower bound on $\mathbb{E}\|AGB\|_p$? Do we have $$\mathbb{E}\|AGB\|_{p} \geq c\max\{\|A\|_p \|B\|_F, \|A\|_F \|B\|_p\}$$ for some constant $c$?

I can show that $$\mathbb{E}\|AGB\|_{op} \geq \max\{\|A\|_{op} \|B\|_F, \|A\|_F \|B\|_{op}\}$$ So I think the lower bound inequality holds for $p\geq 2$. For $p = 1$, if we take $A$ to be identity matrix and $B=e_1e_1^T$ then $\mathbb{E}\|AGB\|_{p}\approx \sqrt{n}$, so the inequality cannot hold with $\max$. So the question is, will it hold $$\mathbb{E}\|AGB\|_{p} \geq c\min\{\|A\|_p \|B\|_F, \|A\|_F \|B\|_p\}$$ for $p\in [1,2)$? Again I can show it with an extra $1/\log n$ factor on the right but I am not sure if it is necessary.

[Edit: Now I answer all questions.]

The answer to the first question is yes, the answer to the second question is no, and the answer to the third question is if and only if $p \geq 2$ (only a guess in the case $p<2$).

• First question

The inequality $$\mathbb E \|A G B\|_* \leq \min( \|A\|_* \|B\|_F,\|A\|_F \|B\|_*)$$ always holds. By symmetry, it is enough to prove the inequality $\mathbb E \|A G B\|_* \leq \|A\|_* \|B\|_F$. And by writing $A$ as a sum of rank one operators ("the unit ball of the trace class is the convex hull of the norm $1$ rank $1$ matrices"), we can assume that $A$ has rank one.

In that case, $A G B$ has rank one for every $G$. Using (1) that for a rank $1$ matrix, the trace norm and Frobenius norm coincide, and (2) that the $L^1$ norm is less than the $L^2$ norm, we get $$\mathbb E \|A G B\|_* = \mathbb E \|A G B\|_F \leq (\mathbb E \|A G B\|_F^{2})^{\frac 1 2} = \|A\|_F \|B\|_F=\|A\|_* \|B\|_F.$$ The second equality is a straighforward computation, at least when $A$ and $B$ are diagonal.

• Second question, and third question when $p \geq 2$.

It follows from the non-commutative Khintchine inequalities of Françoise Lust-Picard that, for every $2 \leq p < \infty$, $$\mathbb E \|A G B\|_p \simeq \max( \|A\|_p \|B\|_2,\|A\|_2 \|B\|_p)\ if\ p \geq 2$$ (up to constants depending on $p$, whose growth rate is known as $p \to \infty$).

Indeed, the non-commutative Khintchine inequalities states that if $(C_k)$ is a family of matrices and $g_k$ are iid $N(0,1)$ random variables, $$\mathbb E \| \sum_k g_k C_k\|_p \simeq \max ( \|(\sum C_k C_k^*)^{\frac 1 2}\|_p, \|(\sum C_k^* C_k)^{\frac 1 2}\|_p).$$ (the inequality is sometimes stated in terms of Bernoulli random variables instead of gaussians, and with $L^p$-norm, of $L^2$ norm instead of $L^1$-norm on the left-hand side, but all versions are equivalent by Kahane's inequalities and standard probabilistic arguments). Applying this to $C_{i,j}$, the product of the $i$-th column of $A$ and and $j$-th row of $B$, one gets the answer because $\sum_{i,j} C_{i,j} C_{i,j}^* = \|B\|_2^2 A A^*$ and $\sum_{i,j} C_{i,j}^* C_{i,j}=\|A\|_2^2 B^*B$.

• Third question if $p \leq 2$.

Here also, the non-commutative Khintchine inequalities (due to Lust-Picard and Pisier when $p=1$) allow, in principle, to give an explicit equivalent of $\mathbb E\|A G B\|_p$. Indeed, when $p \leq 2$, the inequality reads $$\mathbb E \| \sum_k g_k C_k\|_p \simeq \inf \|(\sum D_k D_k^*)^{\frac 1 2}\|_p+ \|(\sum E_k^* E_k)^{\frac 1 2}\|_p),$$ (up to universal constants) where the infimum is over all families $(D_k,E_k)$ such that $C_k=D_k+E_k$ for all $k$.

This gives a very involved proof of the inequality of the first question. I believe that it can be deduced from this inequality that the inequality $$\mathbb E \|A G B\|_p \geq \min( \|A\|_p \|B\|_2,\|A\|_2 \|B\|_p)$$ does not hold if $p<2$, but this deserves to be checked.

• Is there anything similar to "the unit ball of the trace class is the convex hull of the norm 1 rank 1 matrices" for Schatten-p class? That is, if I look at the Schatten p-norm of $AGB$ instead of the trace norm (which is Schatten 1-norm)? – user58955 Jan 14 '17 at 4:08
• No, there is no useful description of the extreme points of the unit ball for the Schatten classes if $p \neq 1,\infty$. However, you can use interpolation to get $(\mathbb E \|A G B\|_p^2)^{\frac 1 2} \leq \min(\|A\|_p \|B\|_2, \|A\|_2 \|B\|_p)$ for every $1\leq p\leq 2$ (the case $p=2$ being an equality, as shown above). – Mikael de la Salle Jan 14 '17 at 6:18
• Ah, non-commutative Khintchine -- I didn't think of that -- thanks! – user58955 Jan 16 '17 at 8:54
• For non-commutative Khintchine, it's just the lower bound for which you need $p\geq 2$, isn't it? The upper bound holds for all $p\geq 1$. (The classical Khintchine does hold for $p<2$ though). – user58955 Jan 16 '17 at 11:31
• Non-commutative Khintchine inequalities hold for every $0<p<\infty$, it is just that the formula are different for $p>2$ and $p<2$. But you are right, the inequality $\mathbb E \| \sum_k g_k C_k\|_p \lesssim \max ( \|(\sum C_k C_k^*)^{\frac 1 2}\|_p, \|(\sum C_k^* C_k)^{\frac 1 2}\|_p)$ is true for all $p$ (and trivial if $1 \leq p<2$). – Mikael de la Salle Jan 16 '17 at 12:01