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I read the following statement:

$\text{tr} [ A(B+C)] \leq \left\Vert B+C \right\Vert$. Since $B+C \geq 0$, one can always find $A$ s.t. the inequality is saturated.

Here, $A$ is a hermitian PSD matrix with $\text{tr}(A) = 1$, whereas $B,C \geq 0$ and also hermitian.

Regarding that, I have the following questions:

  1. Is it obvious what operator norm they are using?
  2. Why is $\text{tr} [ A(B+C)] \leq \left\Vert B+C \right\Vert$ true? Could I show that by saying $\text{tr} [ A(B+C)] \leq \left\Vert A \right\Vert \left\Vert B+C \right\Vert$, and as $\text{tr}(A) = 1 \Longrightarrow \left\Vert A \right\Vert \geq 1$, then the statement is true?
  3. Why $B+C \geq 0$ must be true so that we can find $A$ that saturates the inequality?
  4. How can I find such $A$ that saturates the bound? Is it related to the eigenvectors of $(B+C)$?

P.s.: I only know introductory linear algebra, so I'm sorry if these questions are obvious.

cab20
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    It seems from the context that the authors are referring to the induced $2$-norm. With that, it seems that the inequality holds as a consequence of this general statement. – Ben Grossmann Feb 02 '21 at 19:55
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    It also seems from the context that PSD matrices are, in this context, symmetric (or perhaps Hermitian) by definition. With that: if $\max_{|x| = 1}|(B + C)x|_2$ is attained with $x = v$, then taking $A = vv^T$ saturates the inequality. – Ben Grossmann Feb 02 '21 at 19:57
  • The first comment solves the first question. For the second, you're correct, the matrices should be Hermitian. I've edited this into the question. For the answer on the saturation, is $v$ somehow related to any property of $B+C$, perhaps being an eigenvector of it? Thank you both. – cab20 Feb 02 '21 at 20:11
  • Yes, $v$ is the eigenvector of $B+C$ associated with the maximal eigenvalue – Ben Grossmann Feb 02 '21 at 20:28

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