Given commuting, positive definite A and B, why are the eigenvalues of AB strictly positive? I am asked to prove that if A and B are positive definite, and AB = BA, then AB is positive definite.
I would first need to prove that AB has positive eigenvalues.
From there, the rest of the proof is easy:
(AB)* = B* A* = BA = AB. Therefor AB is Hermitian. Therefor, having only positive eigenvalues, it is positive definite.
I have taken a look at this proof, but I don't know what congruent matrices are.
I was wondering if there is another proof, or, alternatively, if anyone can explain the above proof in greater detail.
I am new to Stack Exchange, apologies for not using appropriate formatting.
I am also new to Linear Algebra so go easy on me:

*

*I am somewhat familiar with the square root of matrices.

*I understand eigenvalues and eigenvectors basic theory.Namely, I understand that
if two matrices are similar, they have the same eigenvalues.

*I have just been introduced to positive definite matrices, but understand the very basic theory.

Thanks,
Miguel
 A: I have since found a pretty cool proof (not made by me).
$Proof$ that if $A$ and $B$ are positive definite, $AB$ has only positive eigenvalues:
($A^*$ denotes the conjugate transpose of $A$. According to this definition, a positive definite matrix is necessarily hermitian. $ \langle ·,· \rangle $ denotes the standard inner product, dot product.)

Because $A$ is positive definite, there exists invertible $\sqrt A$ such that $A = \sqrt A\sqrt A$.
Because $B$ is positive definite, there exists invertible $\sqrt B$ such that $B = \sqrt B\sqrt B$.
Note that $(\sqrt A)^* = \sqrt A.$ (The same is true for $B$)

Let $\lambda$ be an eigenvalue of $AB$, and $v$ a corresponding eigenvector.
Let $u = \sqrt A^{-1}v$. (Then, $v = \sqrt A u$)
$ABv = \lambda v \iff \sqrt A \sqrt A B \sqrt A u = \lambda \sqrt A u \iff \sqrt A^{-1} \sqrt A \sqrt A B \sqrt A u = \sqrt A^{-1} \sqrt A \lambda u \iff  \sqrt A B \sqrt A u = \lambda u$
Therefor, $AB$ and $\sqrt A B \sqrt A$ have the same eigenvalues.

I'll now show that all eigenvalues of $\sqrt A B \sqrt A$ are positive. Let $\lambda$ be an eigenvalue of $\sqrt A B \sqrt A$, and $u$ a corresponding eigenvector.
$\lambda \langle u,u\rangle = \langle\lambda u,u\rangle = \langle\sqrt A B \sqrt A u,u\rangle = \langle (\sqrt A \sqrt B)(\sqrt B \sqrt A) u, u\rangle = \langle (\sqrt B \sqrt A) u, (\sqrt A \sqrt B)^* u\rangle = \langle \sqrt B \sqrt A u, \sqrt B^* \sqrt A^* u\rangle = \langle \sqrt B \sqrt A u, \sqrt B \sqrt A u\rangle > 0$
, by the positive-definite property of the inner prooduct.
(Note that $\sqrt B \sqrt A u \neq  0$, as that would mean $\sqrt B \sqrt A$ is not invertible.)
Therefor, $\lambda \langle u,u\rangle > 0 \Rightarrow \lambda > 0$, which concludes this proof.
