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I have the following (real) matrix which I need to be positive semi-definite,

$P = \begin{bmatrix} P_1 & -\frac{1}{2}(P_1+P_2)\\-\frac{1}{2}(P_1+P_2) & P_2\end{bmatrix} \succeq 0$.

I think this is only the case when, $P_1 = P_2 \succeq 0$, but I couldn't find a way to prove this. I was therefore wondering if this is even the case and if so how to prove this.

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Indeed your claim is correct.

The characteristic polynomial of the matrix is:

$$\lambda^2-(P_1+P_2)\lambda-\frac{1}{4}(P_1-P_2)^2,$$ so that the eigenvalues are: $$ \lambda_\pm=\frac{P_1+P_2\pm\sqrt{2(P_1^2+P_2^2)}}{2}. $$ To ensure that both eigenvalues are non-negative it should hold: $$ P_1+P_2\ge0;\quad (P_1+P_2)^2\ge 2(P_1^2+P_2^2)\Rightarrow 0\ge (P_1-P_2)^2. $$ The last inequality can however hold only if $P_1=P_2$.

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  • $\begingroup$ I forgot to mention that P1 and P2 are symmetric matrices of an arbitrary size, so not necessarily scalar. Will it still hold in that case? $\endgroup$ – Patbott Apr 13 '18 at 8:35
  • $\begingroup$ @Patbott 0f course the proof is valid only for numbers, not for matrices. You should reformulate the question. Probably the best way is to write a new one. $\endgroup$ – user Apr 13 '18 at 8:59

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