How to show det $A$ is positive from the following question Let $A$, $B$ are two  real symmetric matrix of order $n$ and all eigenvalues of $A$, $B$ are strictly greater than 1. If $\lambda$ is some real eigenvalue of $AB$ then show that $\vert\lambda\vert>1$.
I have tried using the diagonalisation concept but not able to proceed further.
 A: Let $\{e_1, \ldots, e_n\}$ be an orthonormal basis such that $Ae_i = \lambda_i e_i$.
We have
$$\|Ax\|^2 = \left\|\sum_{i=0}^n \langle x, e_i\rangle Ae_i\right\| ^2 = \left\|\sum_{i=0}^n \lambda_i \langle x, e_i\rangle e_i\right\| ^2= \sum_{i=0}^n \lambda_i^2 |\langle x, e_i\rangle|^2 > \sum_{i=0}^n |\langle x, e_i\rangle|^2 = \|x\|^2$$
so $\|Ax\| > \|x\|$ for all $x \ne 0$. Similarly we get $\|Bx\| > \|x\|, \forall x \ne 0$.
If $ABx = \lambda x$ for some $x \ne 0$ then 
$$|\lambda|\|x\| = \|ABx\| > \|Bx\| > \|x\|$$
so $|\lambda| > 1$.
A: Hint - any real symmetric matrix is diagonalizable. If you diagonalize them, it's especially easy to calculate their determinant. If all eigenvalues are real and strictly greater than 1, what does this tell you about the determinant? Remember that the determinant of the product is the product of the determinants.
A: We have that for all $x$ such that $\|x\|=1$
$$x^TAx> 1\quad x^TBx> 1$$
let consider any $x=\sum a_i y_i$ with $y_i$ a basis of orthonormal eigenvectors of $B$ hence $\|x\|^2=\sum a_i^2=1$ and
$$x^TABx=\sum \lambda_ia_i^2y_i^TAy_i>\sum \lambda_ia_i^2>\sum a_i^2=1$$
therefore $|\lambda_{AB}|>1$.
