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Consider the following block matrix $$ X = \begin{bmatrix} A & C \\ C^\top & B\end{bmatrix}, $$ where $A\in\mathbb{R}^{n\times n}$, $B\in\mathbb{R}^{m\times m}$, and $C\in\mathbb{R}^{n\times m}$.

To Prove (or disprove): If $X$ is positive definite, i.e. $X>0$, then the following trace inequality holds $$ \left[\mathrm{tr}(CC^\top)\right]^2< \mathrm{tr}(A^2)\mathrm{tr}(B^2). $$

Some comments. Based on Theorem 2.3 of Horn and Mathias. "Cauchy-Schwarz inequalities associated with positive semidefinite matrices." Linear Algebra and Its Applications 142 (1990): 63-82, I think it is possible to prove the previous fact if we replace strict inequalities with non-strict ones.

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    $\begingroup$ By $X>0$, I guess you mean positive definite, not positive entries. $\endgroup$
    – T. Amdeberhan
    Commented Sep 18, 2016 at 15:50
  • $\begingroup$ Yes, I just edited the question in order to specify it. $\endgroup$
    – Ludwig
    Commented Sep 18, 2016 at 15:55

1 Answer 1

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For any unitarily invariant norm it can be shown that \begin{equation*} \|X\| = \left\Vert \begin{bmatrix} A & C\\ C^* & B \end{bmatrix} \right\Vert \le \|A\| + \|B\|. \end{equation*} Thus, using the squared Frobenius norm on both sides and cancelling, we obtain \begin{equation*} \|C\|_F^2 \le \|A\|_F\|B\|_F, \end{equation*} as desired.

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    $\begingroup$ But, the desired inequality is supposed to be sharp $<$ and not $\leq$. Am I right? $\endgroup$
    – T. Amdeberhan
    Commented Sep 18, 2016 at 16:46
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    $\begingroup$ You are right, the OP asks for a strict inequality; I believe that it should be possible to make it strict without much trouble. $\endgroup$
    – Suvrit
    Commented Sep 18, 2016 at 17:08
  • $\begingroup$ Yes, indeed, I'm wondering whether this still holds true with strict inequality. $\endgroup$
    – Ludwig
    Commented Sep 18, 2016 at 17:18
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    $\begingroup$ Alternatively, we know that $X > 0$ iff there exists a contractive matrix $K$ (i.e., $\|K\| < 1$) such that $C=A^{1/2}KB^{1/2}$. Thus, we see that \begin{equation*} \|C\|_F^2 = \mathrm{tr}(KAKB) < \text{rhs}, \end{equation*} because $K$ is a strict contraction. $\endgroup$
    – Suvrit
    Commented Sep 18, 2016 at 17:22
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    $\begingroup$ To get the strict inequality it is enough to apply the above solution to $X-\epsilon I$, where $\epsilon$ is small enough so as to make $X-\epsilon I$ positive. $\endgroup$
    – Ruy
    Commented Sep 18, 2016 at 18:01

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