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## Zero trace implies zero product

A real positive definite matrix is a symmetric matrix whose eigenvalues are all nonnegative. Prove that if $P,Q$ are real positive semidefinite $n \times n$ matrices with $tr(PQ)=0$, then $PQ=0$.

Solution: An elegant solution to this problem can be given using the Cholesky decomposition of a symmetric positive definite real matrix. This means that for every positive semidefinite real matrix $A$ there exists a real matrix $L$ such that $A=LL^t$. We apply this for our matrices $P,Q$, and therefore $P=XX^t,\ Q=YY^t$, with $X,Y$ real matrices. Now we have $0=tr(PQ)=tr(XX^tYY^t)=tr(Y^tXX^tY)=tr((Y^tX)(Y^tX)^t)=tr(AA^t)$ where $A=Y^tX$. But we know that $tr(AA^t)=\sum_{i,j}a_{ij}^2$, and this being equal to zero is equivalent to $A=0$. Then $A^t=X^tY=0$, and finally $PQ=XX^tYY^t=0$.

We can find using the argument presented above that $tr(PQ) \geq 0$, for all positive semidefinite real matrices $P,Q$.