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If $A$, $B$ and $C$ are positive semidefinite real matrices such that $A\succeq B$, do we have: $$ AC\succeq BC\quad\text{and}\quad CA\succeq CB.\tag{i} $$ Here, $A$, $B$, and $C$ share the same dimensions so that the multiplications above make sense.

Context: originally I had to prove $\text{tr}(AC)\geq\text{tr}(BC)$, which I have by using the Cholesky decomposition $C=LL'$: $$ \text{tr}(AC)-\text{tr}(BC)=\text{tr}(ALL')-\text{tr}(BLL')=\text{tr}(L'AL)-\text{tr}(L'BL)=\text{tr}(L'(A-B)L) $$ which is nonnegative because the matrix $L'(A-B)L$ is positive semidefinite. Then, I noted that my result will also follow if either equality in (i) holds, hence this question. Thank you very much.

yurnero
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1 Answers1

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Hint: Assume that $A$ and $C$ do not commute. Is $AC$ positive semidefinite?

The following is an example taken from Is the product of symmetric positive semidefinite matrices positive definite?

$$ A=\left( \begin{array}{cc} 1 & 2\\ 2 & 5\end{array} \right)\quad \text{and}\quad C=\left( \begin{array}{rr} 1 & -1\\ -1 & 2\end{array} \right) $$ are positive semidefinite. It is easy to check that $A\geq (3-2\sqrt{2})I$. On the other hand $$ AC=\left( \begin{array}{rr} -1 & 3\\ -3 & 8\end{array} \right)$$ is not positive semidefinite as $\langle ACx,x\rangle=-1$ for $x^{T}=(1, 0)$. Thus the inequality $$ AC \geq (3-2\sqrt{2})C$$ does not have sense.

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Janko Bracic
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  • Well, one could still consider $X\succeq 0$ if $X$ had real nonnegative eigenvalues. Alternatively, since the matrices involved are symmetric, $X\succeq 0$ could as well mean that $X$ is similar to a symmetric positive semidefinite matrix. – Algebraic Pavel Jan 21 '15 at 13:40
  • @janko I read the links given above by Ofir so I understand what you are saying. However, I think the hint is too brief. Can you write a more full answer? I'll gladly accept. – yurnero Jan 21 '15 at 14:37