Let $lambda$ be a real eigenvalue of matrix $AB$. Prove that $|lambda| 1$.

Question

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 $|lambda| > 1$.

My solution:

Let $a$ and $b$ be eigenvalues of $A$ and $B$ corresponding the eigenvectors $y$ and $x$, respectively.
Looking at the following dot product:
$$langle ABx,y rangle = langle Bx,A^Ty rangle=langle Bx,Ay rangle = langle bx,ay rangle=ablangle x,y rangle=langle abx,y rangle$$
we get $$(AB)x=(ab)x$$ Therefore, $lambda := ab$ is an eigenvalue of $AB$. Since $a>1$ and $b>1$, it follows that $lambda > 1$

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.

user1551 · Answer

$AB$ is similar to $A^{1/2}BA^{1/2}$. Hence its eigenvalues are positive.
Full solution:
Since $A$ and $B$ are positive definite, $AB$ is similar to the positive definite matrix $A^{1/2}BA^{1/2}$. Hence $AB$ has a positive spectrum.
Furthermore, since $A$ and $B$ are unitarily diagonalisable and their eigenvalues are greater than $1$, we have $|Ax|_2,|Bx|_2>|x|_2$ for all nonzero vector $x$. It follows that $|ABx|_2=|A(Bx)|_2>|Bx|_2>|x|_2$ for all nonzero $x$. As $AB$ has a positive spectrum, the eigenvalues of $AB$ must be positive numbers greater than $1$.
Alternative solution (that uses the induced $2$-norm for matrices). Since $A,Bsucc I$, we have $0prec A^{-1},B^{-1}prec I$ and $|A^{-1/2}B^{-1}A^{-1/2}|_2le|A^{-1/2}|_2^2|B^{-1}|_2=|A^{-1}|_2|B^{-1}|_2<1$. Hence $0prec A^{-1/2}B^{-1}A^{-1/2}prec I$ and $A^{1/2}BA^{1/2}succ I$. Since $A^{1/2}BA^{1/2}$ is similar to $AB$, all eigenvalues of $AB$ are positive numbers greater than $1$.

Let $lambda$ be a real eigenvalue of matrix $AB$. Prove that $|lambda| > 1$.

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