# On a trace condition for positive definite $2\times 2$ block matrices

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.

• By $X>0$, I guess you mean positive definite, not positive entries. – T. Amdeberhan Sep 18 '16 at 15:50
• Yes, I just edited the question in order to specify it. – Ludwig Sep 18 '16 at 15:55

• But, the desired inequality is supposed to be sharp $<$ and not $\leq$. Am I right? – T. Amdeberhan Sep 18 '16 at 16:46
• 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. – Suvrit Sep 18 '16 at 17:22
• 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. – Ruy Sep 18 '16 at 18:01