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In the paper On a product of positive semidefinite matrices, page 6, for the proof of (1) - (2): if A and B are PSD, how to draw the conclusion that AB is PSD? Can someone explain in detail?

Lynn
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For a matrix $X$, let $\sigma (X)$ is the set all eigenvalues of $X$.

First note that, $AB=A^{1/2}(A^{1/2}B)$. Thus $\sigma (AB)=\sigma ((A^{1/2}B)A^{1/2})$. (Since for two matrices X, Y, we have $\sigma(XY)=\sigma(YX)$.)

Since $A^{1/2}BA^{1/2}$ is PSD, it follows that eigenvalues of $AB$ are non-negative.

Finally, since $AB$ is normal, by Spectral theorem $AB=UDU^*$, for some unitary matrix $U$ and diagonal matrix $D$. Since $D$ has non-negative diagonal entries, we conclude that $AB$ is PSD.

Black-horse
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  • Thank you but I'm still confused about some conclusions relevant to PSD. How do you know that $A^{1/2}BA^{1/2}$ and $UDU^* $ are PSD? – Lynn Oct 23 '18 at 06:24
  • If $X$ is PSD and $Y$ is any matrix, then $YXY^$ is always PSD. This follows from the equality $YXY^=(YX^{1/2})(YX^{1/2})^*$. Let me know if you still have some confusion. Thanks.. – Black-horse Oct 23 '18 at 06:34
  • So for $A^{1/2}BA^{1/2}$, did you use $A^=A$? But I think A is only PSD, why does it imply that $A^=A$? – Lynn Oct 23 '18 at 06:45
  • Can you tell me what is your definition of a PSD matrix? – Black-horse Oct 23 '18 at 06:48
  • For any vector $x$, $x^TAx>=0$, then A is PSD. – Lynn Oct 23 '18 at 06:55
  • If the matrix is over the filed of the complex numbers (which is the case in the paper you have referred), then $x^TAx\geq 0$ for all $x$ implies $A=A^*$. See https://math.stackexchange.com/questions/433006/proof-complex-positive-definite-self-adjoint – Black-horse Oct 23 '18 at 09:00
  • For a PSD matrix $A$, $A = A^*$ means that $A = (\bar{A})^T$. If A only contains real numbers, does that mean $A = A^T$? But a PSD matrix A doesn't have to be symmetric. That's what I'm confused about. – Lynn Oct 23 '18 at 23:32