Let $T_1, \ldots, T_n$ by real symmetric positive definite matrices, with eigenvalues bounded below by $\mu > 0$.
Can I say
$$
\frac{x^T T_1 T_2 \ldots T_n x}{x^T x} \geq \mu^n
$$
If these matrices commute the result is straightforward, but I'm interested in the case where these matrices don't necessarily commute.
Edit: Not sure that you can say this for the $n=2$ case either.
Best Answer
It is true that the product $M=T_1T_2$ of two positive definite symmetric matrices has real and positive eigenvalues. And conversely, every matrix $M$ with real positive eigenvalues can be factored $M=T_1T_2$ as above. But $x^TMx$ does not need to be positive. Here is an example: $$M=\begin{pmatrix} 3 & a \\ -a & -1 \end{pmatrix}, \qquad \sqrt3<a<2.$$ The eigenvalues, roots of $X^2-2X-3+a^2$, are real and positive, while $x^TMx=3x_1^2-x_2^2$ is indefinite.