Someone can help me with this following problem?

Let $A,B$ be symmetric matrices. If $A$ is positive definite, then $AB$ is diagonalizable.


P.S. The matrices are over $\mathbb{R}$


1 Answer 1


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.

  • 1
    $\begingroup$ +1 for teaching a new trick, that if $X,Y$ are symmetric, so is $XYX$. $\endgroup$
    – Aaron
    Mar 1, 2013 at 0:01
  • $\begingroup$ With a similar argument, $AB$ and $BA$ are both diagonalisable under the OP's assumptions. $\endgroup$
    – MathMax
    Jan 13, 2021 at 15:18

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