5
$\begingroup$

I want to maximize $\text{Tr}(XY)$ over $X$ for fixed $Y$, where $X$ and $Y$ are both hermitian (but doesn't necessarily positive) operators, and $X$ is constrained by its p-norm bounded by $1$, i.e. $|X|_p \leq 1$. I want to prove that optimized operators, the one when the norm constraint is $p=1$ and the other one when $p=2$, are related to each other.

Update: My question turns out to be very misleading, so please ignore the q! Apologize for the inconvenience.

Improve this question
$\endgroup$
4
  • $\begingroup$ Isn't $\vert X_{opt,2}\vert_2 = 1$ without the factor of $\sqrt{d}$? That is, if $X_{opt,p}$ is a rank-1 projection onto $Y$'s maximum eigenvalue (I assume finite-dimensional), then $\vert X_{opt,p}\vert_p=1$ for all $p$, right? And it will saturate the bound you found, which seems to apply for all $p$ as well. $\endgroup$
    – Sam Jaques
    Feb 24, 2022 at 15:21
  • $\begingroup$ @SamJaques $|X_{opt,2}|_2$ is 1. Since $max \text{Tr}(XY) = |X|_1 |Y|_\infty \leq \sqrt{d} |X|_2 |Y|_\infty$ (now, just for general $X$, and using $|X|_2 \leq \sqrt{d}|X|_1$ inequaility), if we optimize with $p=2$-norm constraint, we end up getting $\sqrt{d} |X_{opt,2}|_2 |Y|_\infty = \sqrt{d} |Y|_\infty$. Does this make senes? I could be entirely wrong though! that's why I want discussions in the forum $\endgroup$
    – Jon Megan
    Feb 24, 2022 at 17:26
  • $\begingroup$ @SamJaques Now I think about it again, and I think $X_{opt,p}$ should not be always rank-1 projection onto $Y$'s maximum eigenvalue. Rather, $X_{opt,p}$ can be a sum of projections to + eigenspace of $Y$ and - eigenspace of $Y$ with -1 factor, then divide by 2, i.e. $X_{opt,p} = \frac{|Y+><Y+| - |Y-><Y-|}{2}$ where $|Y+>, |Y->$ are eigenspace for largest and smallest eigenvalues respectively. $\endgroup$
    – Jon Megan
    Feb 24, 2022 at 17:45
  • $\begingroup$ When you say $\vert A\vert_p$, do you mean $\max_{x:\Vert x\vert_p=1}\{\Vert Ax\Vert_p\}$ or do you mean $\left(\text{tr}(A^p)\right)^{1/p}$? For the second (the Schatten $p$-norm) there is a general Holder inequality (mathoverflow.net/questions/158881/…) and I think this could be used directly in your case (take $X=Y/\vert Y\vert_2$). For the first I'm not sure. But if you're thinking of the first, $\vert Y\vert_\infty$ isn't the same as it's maximum eigenvalue anyway, is it? $\endgroup$
    – Sam Jaques
    Feb 25, 2022 at 10:24

2 Answers 2

Reset to default
4
$\begingroup$

If I understand your question correctly, you're trying to prove something that is false. Consider the operator $$ Y = \begin{pmatrix} 2 & 0 & 0\\ 0 & -1 & 0\\ 0 & 0 & -1 \end{pmatrix}. $$ In the case $p = 2$, the optimizer is $X_{\mathrm{opt},2} = Y/\sqrt{6}$, but for $p = 1$ we have $$ X_{\mathrm{opt},1} = \begin{pmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}. $$ These optimizers are not related in the way you have expressed.

Let me point out a mistake in the second paragraph of the question, in case it is causing confusion: the maximum of $\operatorname{Tr}(XY)$ over all $X$ with $\|X\|_2 \leq 1$ is not equal to $\|X\|_1 \|Y\|_{\infty}$ in general. It is equal to $\|Y\|_2$, and the optimizer is $X = Y/\|Y\|_2$. You can prove this using Cauchy-Schwarz (for the Hilbert-Schmidt inner product).

Improve this answer
$\endgroup$
1
$\begingroup$

They are related when $Y$ is a rank-2 matrix with eigenvalues. (say $\pm \lambda$)

Also, you need to prove the equality condition for the Holder's inequality in each case. You are using the equality condition without proving it.

Suppose $Y$ is a rank-2 matrix. Then equality case for Holder's inequality $Tr(XY) \leq |X|_1|Y|_\infty$ holds when $|Y|$ has an eigenvalue of $|Y|_\infty$, which is easy to show. So $max Tr(XY) = |Y|_\infty =\lambda$, and you're right that $|X|_{opt,1}$ is one of the suggested solutions.

The equality case for Holder's inequality $Tr(XY) \leq |X|_2|Y|_2 = \sqrt{2}\lambda|X|_2$ holds when $|X|^2 = a|Y|^2$ for some scaler $a>0$. When you think about maximizing trace, a natural choice would be aligning $X$ to $Y$, so that $X = a' Y$ for some scalar $a'$. This covers the equality condition trivially. Therefore, since $max Tr(XY) = |X_{opt,2}|_2|Y|_2 = \sqrt{2}\lambda $, $X_{opt,2}$ is just $\sqrt{2}$ multiplied by one of the $|X|_{opt,1}$ solutions suggested earlier. So they are related.

But in general case of $Y$ being a random Hermitian operator, this is not true at all.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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