I was asked the following question:
Let $A,B \in \mbox{Mat}_n(\mathbb R)$ be symmetric positive definite matrices. Show that $AB$ does not have negative eigenvalues.
My answer
I'm getting something very weird. According to my answer, $AB$ has to be positive definite. which is weird because we don't know that $AB$ is symmetric.
$B$ is symmetric so there is a basis $v_1,...,v_n$ of eigenvectors such that $Bv_i=\lambda_i$ and $\lambda_i>0$ for all $i$, and $<v_i,v_j>=0$ when $i\neq j$
let's look at $<ABx,x>$ where $x$ is some vector in $\mathbb R^n$
$<ABx,x>=<AB\sum_{i=1}^n \alpha_i v_i,\sum_{i=1}^n \alpha_i v_i> = <AB\alpha_1 v_1,\alpha_1 v_1>+<AB\alpha_2 v_2,\alpha_2 v_2>+...+<AB\alpha_n v_n,\alpha_n v_n> = \alpha_1 ^2 \lambda_1<Av_1,v_1>+\alpha_2 ^2\lambda_2<Av_2,v_2>+...+\alpha_n ^2 \lambda_n<Av_n,v_n>$
obviously for all $i$ $\alpha_i ^2$ is positive.
Because $B$ is positive definite, for all $i : \lambda_i >0$.
Because $A$ is positive definite, for all $i: <Av_i,v_i>$ is positive.
so overall, for all $i: \alpha_i ^2 \lambda_i <Av_i,v_i>$ is positive.
So we got that $<ABx,x>$ is larger than zero for any vector $x$, and so I infer that $AB$ is positive definite. And so it doesn't have negative eigenvalues.
But how can we say that it's positive definite if it's not symmetric? I did something wrong I'm sure.
