# 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?

• With "Hermitian", do you perhaps mean real and symmetric? It seems like this uses $B^T=B$ everywhere. – punctured dusk Feb 14 '17 at 23:10

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.

Let $A$ be positive definite and $B$ be Hermitian. If $x$ is an eigenvector of $AB$ with eigenvalue $\lambda$, then $$\langle ABx,Bx \rangle = \lambda \langle x,Bx \rangle$$ But if we denote $v = Bx$ we can see that since $A$ is positive definite, either $v=0$ and $ABx = Av = 0$ meaning $\lambda = 0$, or $$\langle ABx,Bx \rangle = \langle Av,v \rangle > 0 \text{ and real}$$ This tells us two things: $$\lambda \neq 0 \text{ , } \langle x,Bx \rangle \neq 0$$ and if $\langle x,Bx \rangle$ is real then $\lambda$ is real. But since B is Hermitian, $\langle x,Bx \rangle$ is real! This can be seen by decomposing x into the eigenbasis $\{v_n\}$ of B: $$x= \sum_n{c_nv_n}$$ $$\langle x,Bx \rangle = \sum_n{|c_n|^2\lambda_nv_n}$$ which is real since each component is real. Thus every eigenvalue $\lambda$ of $AB$ is real.

• +1. A minor issue: $\lambda$ and $\langle x,Bx\rangle$ can be zero (because $Bx$ may be zero), but surely, a zero eigenvalue is real too. – user1551 Feb 18 '17 at 21:53
• Ah yes of course – The Yomster Feb 19 '17 at 23:34

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 $B^{1/2}$.