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If $A$ and $B$ are symmetric positive definite then $\vec{x}^TA\vec{x}>0$ and $\vec{x}^TB\vec{x}>0$ as well as $A^T=A$ and $B^T=B$. Here's where the confusion lies, when showing product of $A\cdot B$ is symmetric positive definite. I tried $(A\cdot B)^T=B^T\cdot A^T=B\cdot A=A\cdot B$, but the last equality is false, consider counterexample $A=\begin{bmatrix}1&1&1\\ 1&1&1\\ 1&1&1\end{bmatrix}$ and $B=\begin{bmatrix}1&0&1\\ 0&1&0\\ 1&0&1\end{bmatrix}$.

I also need help in showing $A\cdot B$ positive definite, does it immediately follow vacuously?

Edit: Assume both $A$ and $B$ are diagonalizable so that commutativity is allowed.

For showing product is positive definite, I'm doing this but now I'm stuck. $\vec{x}^TAB\vec{x}=\vec{x}^TA^TB\vec{x}=(A\vec{x})^T(B\vec{x})=\vec{a}^TI\vec{b}>0$ where $B\vec{x}=\vec{b}$ and $A\vec{x}=\vec{a}$ and $I_{n\times n}$ is the identity matrix and it is positive definite

user5896534
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  • What is it that you're trying to prove, exactly? If you can find symmetric matrices $A$ and $B$ such that $AB$ is not symmetric, then you won't be able to prove that $AB$ is symmetric just because $A$ and $B$ are. If you want to prove that the product $AB$ is symmetric you need more assumptions, like the matrices $A$ and $B$ being simultaneously diagonalizable (or something along those lines), otherwise the general statement will be false. – Scounged Nov 12 '21 at 23:30
  • Ok, yeah I'm supposed to show AB is symmetric positive definite if A,B are as well. Would having both properties help? If it's only the property of symmetry that I'm working with then I agree, AB will not necessarily be symmetric. – user5896534 Nov 12 '21 at 23:36
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    It isn't true that the product of two symmetric matrices is symmetric, despite both being positive definite. – hardmath Nov 12 '21 at 23:37
  • If $A$ and $B$ are simultaneously diagonalizable it's a breeze to prove your statement, but as others have pointed out, if you don't have any other assumptions than the ones you've stated in your original question then the general statement is just false, as per your counterexample. – Scounged Nov 12 '21 at 23:49
  • I'm going to assume that both matrices are diagonalizable as well, that should allow commutativity. – user5896534 Nov 12 '21 at 23:59

1 Answers1

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$A=\begin{bmatrix}1&0\\0&2\end{bmatrix},$ $B=\begin{bmatrix}2&1\\1&2\end{bmatrix}.$ $AB=\begin{bmatrix} 2 &1\\2&4 \end{bmatrix},$ $BA=(AB)^T=\begin{bmatrix}2&2\\1&4\end{bmatrix},$ so "both matrices are diagonalizable$\Rightarrow$commutativity" is false. As @Scounged point out, simultaneously diagonalizable can. Because $$AB=(PD_AP^{-1}) (PD_BP^{-1})=P D_B D_A P^{-1}=(PD_BP^{-1})(PD_AP^{-1}) =BA.$$ As diagonal matrixes are commutative.

So $AB$ may not be symmetric. Also, it may not be positive definite. Like $A=\begin{bmatrix}2&-4\\-4&10\end{bmatrix},$ $B=\begin{bmatrix}5&-4\\-4&4\end{bmatrix}.$ $AB=\begin{bmatrix} 26 &-24\\-60&56 \end{bmatrix}.$ (Just searched on Internet, many examples.)

However, if you suppose $AB=BA,$ then they can be simultaneously diagonalized, and the eigenvalue of $AB$ is just $A$'s times $B$'s, so it's positive definite. Also symmetric, of course.

DreamAR
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  • Some comments: Symmetric matrix is always diagonalized; Symmetric positive definite $\Leftrightarrow$ all eigenvalues are positive. These are basic knowledge of linear algebra, you can find them on your textbook. – DreamAR Nov 13 '21 at 05:29
  • The claim in the last paragraph is a classical exercise. Search 'simultaneously diagonalized' for help. – DreamAR Nov 13 '21 at 05:36
  • Like https://math.stackexchange.com/questions/56307/simultaneous-diagonalization-of-commuting-linear-transformations – DreamAR Nov 13 '21 at 05:56