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Some one can help me with this problem?

I have two real positive-semidefinite matrices $P$ and $Z$, $P \succeq 0$, $Z \succeq 0$, and they are both symmetric ($P^T = P$ and $Z^T = Z$). Also trace$ (P\cdot Z) = 0$. Does that mean $P \cdot Z = 0$ ? Assume they are both $n \times n$ square matrices. Thanks!

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Since $P$ and $Z$ are semidefinite, $P=S^TS$ and $Z=YY^T$ for some $S$ and $Y$. Then $$ \mathrm{tr}(PZ)=\mathrm{tr}(S^TSYY^T)=\mathrm{tr}(Y^TS^TSY)=\mathrm{tr}[(SY)^T(SY)]=\|SY\|_F^2. $$ Hence $SY=0$ and $PZ=S^T(SY)Y^T=S^T(0)Y^T=0$.

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  • $\begingroup$ Thanks! That solved all the problem! $\endgroup$ – Pew Apr 10 '15 at 2:00
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    $\begingroup$ Does $\operatorname{tr}(C^T C) = || C ||_F^2$ because if $c_i$ denotes the $i$th column of $C$, the $i$th diagonal entry of $C^TC = \langle c_i , c_i \rangle = \sum_{k=1}^n c_{ki}^2$, so $\operatorname{tr}(C^TC ) = \sum_{i=1}^n \sum_{k=1}^n c_{ki}^2 = \text{sum of squares of all entries of $C$ }=||C||_F^2$? $\endgroup$ – Chill2Macht Sep 15 '18 at 3:46

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