14
$\begingroup$

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 $\mathbb{R}$

$\endgroup$
0

1 Answer 1

Reset to default
29
$\begingroup$

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.

$\endgroup$
2
  • 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

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .