The original question goes equivalently like this
For real matrices, if $A$ and $B$ are both positive-definite
Prove: all the eigenvalues of $AB$ are positive.
Facts that I know may have a role to play in the proof:
1) For real symmetrical matrices, $A$ is positive-definite $\Leftrightarrow$ all eigenvalues of $A$ are positive.
2) Any real positive-definite matrix $A$ can be diagonalized as
$$A=Q^T\Lambda Q$$
where $\Lambda=\text{diag}\{\lambda_1,\lambda_2, \cdots,\lambda_n\}$ is the diagonal matrix comprising all the eigenvalues of $A$.
3).Any real positive-definite matrix $A$ can be decomposed as
$$A=P^TP$$
where $P$ is also a real positive-definite matrix that shares the same size with $A$.
However, I failed to make any progress after trying them all. Could you please help me with the proof or drop me a hint? Best regards!
EDIT I'd be specially grateful if you could explain it in a way that a student who has only taken an elementary linear algebra can understand. 🙂 By saying "elementary", I mean I have only learnt real matrices and know very little about complex ones, to the point that I don't even really know what Hermitte is (⊙o⊙)
Best Answer
Hint. Let $B^{1/2}$ be the positive definite square root of $B$. We have $AB=B^{-1/2}(B^{1/2}AB^{1/2})B^{1/2}$. Use matrix congruence(with Sylvester's law of inertia) and matrix similarity to finish the argument.
Note that the title and the body of your question do not match. While $AB$ always possesses a full positive spectrum when $A$ and $B$ are positive definite, $AB$ is not necessarily positive definite. For a counterexample, consider $A=\pmatrix{1\\ &3}$ and $B=\pmatrix{1&1\\ 1&1+\epsilon}$ for any small $\epsilon>0$.