Someone can help me with this following problem?
Let $A,B$ be symmetric matrices. If $A$ is positive definite, then $AB$ is diagonalizable.
Thanks!
P.S. The matrices are over $\mathbb{R}$
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Since $A$ is positive-definite, there exists an invertible square root of the matrix which is also symmetric. Denote this as $A^\frac{1}{2}$. Then $$A^{-\frac{1}{2}}ABA^\frac{1}{2} = A^\frac{1}{2}BA^\frac{1}{2}$$ where the latter is symmetric because $B$ and $A^\frac{1}{2}$ are both symmetric. Therefore $AB$ is similar to a symmetric matrix and hence diagonalizable.
EuYu
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2+1 for teaching a new trick, that if $X,Y$ are symmetric, so is $XYX$. – Aaron Mar 01 '13 at 00:01
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1With a similar argument, $AB$ and $BA$ are both diagonalisable under the OP's assumptions. – MathMax Jan 13 '21 at 15:18