# For real matrices, if $A$ and $B$ are both positive-definite, show that all of $AB$'s eigenvalues are positive.

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⊙)

## 1 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 $\epsilon>0$.

• @user1551 Now the matrices are not transpose congruent. From your expression it is clear that $AB$ is similar to the positive definite matrix $B^{1/2}AB^{1/2}$ via an invertible matrix. So they must have same eigenvalues. We do not need Sylvester's law of inertia here. – mrka Mar 25 '15 at 11:24
• @mrka I don't understand your comment that "now the matrices (which matrices?) are not transpose congruent". I did use matrix congruence here: $AB$ is similar to $B^{1/2}AB^{1/2}$, and $B^{1/2}AB^{1/2}$ is congruent to $A$. Sylvester's law of inertia may be an overkill, but the fact that $B^{1/2}AB^{1/2}$ is positive definite IS owing to the fact that matrix congruence preserves positive definiteness. – user1551 Mar 25 '15 at 11:31
• @Babai That's because $x^TB^{1/2}AB^{1/2}x>0$ for every nonzero $x$. Put $y=B^{1/2}x$ if you still don't see it. – user1551 Aug 8 '16 at 19:24
• @Babai That's basic linear algebra. In general, if $X,Y$ are two rectangular matrices such that the products $XY$ and $YX$ make sense, then $XY$ has the same set of nonzero eigenvalues as $YX$. In particular, if both $X$ and $Y$ are square, then $XY$ and $YX$ have identical spectra. Therefore $AB$ and $B^{1/2}AB^{1/2}$ have identical eigenvalues. These eigenvalues must be real because $B^{1/2}AB^{1/2}$ is positive semidefinite. – user1551 Aug 8 '16 at 19:46
• @Babai Surely you are correct. I didn't notice that you have changed the conditions for $A$. – user1551 Aug 8 '16 at 19:59