If A,B are hermitian matrices and all eigenvalues of B are positive, then why all eigenvalues of AB are real?

Question

If A, and B are hermitian matrices and all eigenvalues of B are positive, then why eigenvalues of AB is real?

I can write $x^\ast Bx>0$, and $x^\ast Ax\in \mathbb{R}$, and then how about $x^\ast ABx$? I tried $x^\ast Axx^\ast Bx$ but cannot determine the $xx^\ast$ here. I also try $A=U\Lambda_A U^\ast$ and $B=V\Lambda_B U^\ast$ but still, I cannot say anything about $U^\ast V$.

Answer 1

Let $(\lambda,v)$ be an eigenpair of $AB$. Then $v^\ast BABv=v^\ast B(ABv)=v^\ast B(\lambda v)=\lambda v^\ast Bv$.

Since $B$ is positive definite, $v^\ast Bv>0$. Therefore $\lambda=\frac{1}{v^\ast Bv}v^\ast BABv$ is real.

The above argument can be extended to cover the case where $B$ is positive semidefinite. In this case, if $v^\ast Bv>0$, we proceed as before. If $v^\ast Bv=0$, then $$ 0\le(tv-Bv)^\ast B(tv-Bv)=v^\ast B^3v-2t\|Bv\|^2 $$ for every real number $t$. Hence $\|Bv\|$ must be zero, i.e., $Bv=0$. Thus $\lambda v=ABv=0$ and $\lambda$ is the real number $0$.

Answer 2

Since $B$ is positive semi-definite, $B= D^2$ for some hermitian $D$. Now, it is a fact that the eigenvalues of $A D^2$ and $DAD$ are the same. To finish, notice that $DAD$ is hermitian.