Let $A$ and $B$ be real symmetric matrices with all eigenvalues strictly greater than 1. Let $\lambda$ be a real eigenvalue of matrix $AB$. Prove that $|\lambda| > 1$.
Let $a$ and $b$ be eigenvalues of $A$ and $B$ corresponding the eigenvectors $y$ and $x$, respectively.
Looking at the following dot product: $$\langle ABx,y \rangle = \langle Bx,A^Ty \rangle=\langle Bx,Ay \rangle = \langle bx,ay \rangle=ab\langle x,y \rangle=\langle abx,y \rangle$$ we get $$(AB)x=(ab)x$$ Therefore, $\lambda := ab$ is an eigenvalue of $AB$. Since $a>1$ and $b>1$, it follows that $\lambda > 1$
However, it doesn't seem ok as the problem was actually asking to prove $|\lambda|>1$. Indeed, $\lambda > 1 \implies |\lambda|>1$, but then the problem wouldn't write $|\lambda|$ in my opinion.
The given solution:
The transforms given by $A$ and $B$ strictly increase the length of every nonzero vector, this can be seen easily on a basis where the matrix is diagonal with entries greater than $1$ in the diagonal. Hence their product $AB$ also strictly increases the length of any nonzero vector, and therefore its real eigenvalues are all greater than $1$ or less than $-1$.
Any help is appreciated.