Show that $A \in \mathbb C^{n \times n}$ is diagonalizable and has real eigenvalues if and only if $A=BC$ where $B \succ 0$ and $C$ is Hermitian.

Question

Here's my work so far.

($\Leftarrow$)

Since $B \succ 0$ we know $B^{1/2} \succ 0$ (and $B^{-1/2}$ exists). Then

$A=BC=B^{1/2}(B^{1/2}CB^{1/2})B^{-1/2}$

so $BC$ is similar to $B^{1/2}CB^{1/2}$ which implies they have the same eigenvalues. Then

$(B^{1/2}CB^{1/2})^*=(B^{1/2})^*C^*(B^{1/2})^*=B^{1/2}CB^{1/2}$        (Since $B$ and $C$ are both Hermitian)

$\Rightarrow B^{1/2}CB^{1/2}$ is diagonalizable $\Rightarrow BC$ is diagonalizable $\Rightarrow A$ is diagonalizable.

The only issue I'm having is showing that A has all real eigenvalues. I'm not really sure where to start for that.

($\Rightarrow$)

I really don't know where to start with this direction. Since $A$ is diagonalizable I write $A=PDP^{-1}$ where $D$ is a diagonal matrix. My professor gave me a hint to use the fact that since $P$ is non-singular we know that $PP^* \succ 0$, but I don't really know where to take it from there.

Answer 1

1. You proved that $B^{1/2} C B^{1/2}$ is hermitian. So its eigenvalues are real. $A$ being similar to $B^{1/2} C B^{1/2}$ implies that the eigenvalues of $A$ are real.

2. $A = P D P^{-1} = P P^{*} ( P^{-*} D P^{-1} )$. So $B = P P^{*}$ and $C = P^{-*} D P^{-1}$.