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I have a diagonal matrix D with all positive entries on diagonal and a quadratic matrix $A=B^{T}B$, is the product of the two matrices $C=DA$ also positive definite? or does the product has other properties? Thanks!

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  • $\begingroup$ why don't you try some very simple ($2\times 2$) examples for $D$ when $A=\begin{bmatrix} 1 & 1\\ 1 & 1 \end{bmatrix}$ (i.e. the all ones matrix) $\endgroup$ Feb 14, 2021 at 18:45
  • $\begingroup$ $A$ considered here is positive definite, is the result also positive definite and I'm wondering is there any simple proof for that? Thanks! $\endgroup$ Feb 14, 2021 at 20:25

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The product doesn't have to be positive definite. For a simple counterexample take any positive semidefinite $A$ (any positive semidefinite $A$ can be written as $B^T B$). Then take $D$ to be the identity matrix and $$ C = DA = IA = A $$ is not positive definite.

Edit The product need not be positive semidefinite either. For the product to be positive semidefinite one would require $DA$ to be a symmetric matrix. However, this is the case iff $A$ and $D$ commute, i.e. $[A,D]=0$. Which will not be the case for most choices of $A$ and $D$. You will however still have that the eigenvalues of $DA$ are all non-negative.

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  • $\begingroup$ Thanks for the answer. How about the conclusion that the product is positive semi-definite? $\endgroup$ Feb 22, 2021 at 7:42
  • $\begingroup$ @ChasonYoung No, the product need not be PSD either. I've updated the answer to add some details. $\endgroup$
    – Rammus
    Feb 22, 2021 at 16:05
  • $\begingroup$ Thanks so much for the updated detailed answer. Yes, I can draw that the eigenvalues of the product are all non-negative. Then, I want to ask for this case can I say $u^{T} DA u \geq 0$? Or if I give the additional condition that $B$ is of full rank, can I say that the resultant product is positive definite? $\endgroup$ Feb 24, 2021 at 8:34
  • $\begingroup$ No for the product to be PD you also need it to be symmetric so the same argument holds. $\endgroup$
    – Rammus
    Feb 24, 2021 at 10:08
  • $\begingroup$ How about the matrix form is $$BAB^{T}$$ where A is the diagonal matrix with all positive entries? Thanks! $\endgroup$ May 2, 2021 at 6:05

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