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?
Best Answer
I cannot follow the argument from your book, at least the way it's written here.
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.
Edit: I noticed that I never explained what happens with the reasoning in the question. What happens, if one actually use the $A,B$ given in the question with $\lambda=i$, is that $$ \langle BABx,x\rangle=\langle ABx,Bx\rangle=\lambda\,\langle x,Bx\rangle=0. $$ So one cannot conclude that $\lambda$ is real.