# Eigenvalues of product of two hermitian matrices

I know that if $A$ and $B$ are hermitian matrices, then it doesn't follow that the eigenvalues of $AB$ are real, because of the following counter-example: $A=\begin{bmatrix} 0 &1 \\ 1& 0 \end{bmatrix}$ and $B=\begin{bmatrix} 1 & 0\\ 0& -1 \end{bmatrix}$

On the other hand, I came across the following problem, which says that if $A$ is hermitian and positive definite, and $B$ is hermitian, then $AB$ has real eigenvalues. Why if we add the property "positive definite" to $A$, the eigenvalues of $AB$ become real? The proof I read in the book says: Let $\lambda$ be an eigenvalue of the hermitian matrix $AB$ with non zero eigenvector $x$. Then: $$\left \langle BABx,x \right \rangle=\left \langle ABx,Bx \right \rangle=\left \langle \lambda x,Bx \right \rangle=\lambda \left \langle x,Bx \right \rangle$$

Since $$\left \langle BABx,x \right \rangle$$ and $$\left \langle x,Bx \right \rangle,$$ then $\lambda$ is real. However, I can't see where in the proof the fact that $A$ is positive definite is used. Can anyone explain, please?

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The way I would prove the fact is that, being positive definite, we can write $A=F^*F$ for some matrix (you can take $F=A^{1/2}$ if you know functional calculus, but the point is that such an $F$ exists).
And then use the fact that the eigenvalues of a product do not change if you write the product the other way. So the eigenvalues of $AB=F^*FB$ are the same as those of $FBF^*$. This last matrix is clearly Hermitian, so it has real eigenvalues.
The eigenvalues of $F^*(FB)$ are the same as those of $(FB)F^*$; here's a proof: math.stackexchange.com/questions/124888/… – Martin Argerami Jun 6 '13 at 23:45
Generally, we write $B^{1/2}AB^{1/2}$ to explain $AB$'s eigenvalues are real. The positivity of $B$ is used in the existence of $B^{1/2}$.