Show that all eigenvalues of the matrix $AB$ are real

Question

Suppose that the matrices $A, B \in \Bbb R^{n \times n}$ are symmetric and that $A$ is also non-negative definite. Show that all eigenvalues of the matrix $AB$ are real.

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I haven't find this question on StackMath. Suppose that $A$ and $B$ are two $n \times n$ real symmetric matrices and $A$ is also non-negative definite. Let $\lambda$ be an eigenvalue of $AB$, and let $v$ be a corresponding eigenvector. Then $ABv = \lambda v$, so $v$ is also an eigenvector of $A$ with eigenvalue $\lambda/B$. Since $A$ is non-negative definite, all of its eigenvalues are non-negative.

Now how to approach? Please help.

Answer 1

Hints:

• If the use of matrix square root is allowed, consider the relationship between $AB$ and $A^{1/2}BA^{1/2}$.
• Without using matrix square root: let $\lambda$ be any eigenvalue of $AB$ over $\mathbb C$ and let $x$ be a corresponding eigenvector. Then $\langle Bx,ABx\rangle=\lambda\langle Bx,x\rangle$. The result follows immediately if $\langle Bx,x\rangle\ne0$. If $\langle Bx,x\rangle=0$ instead, show that $ABx=0$.