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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.

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    $\begingroup$ Knowing the eigenvalues have a lower bound like that is not sufficient for what you want, because $x^T M x$ is unrelated to the eigenvalues of $M$ for $M$ not symmetric. In particular, can't the left side be negative already for $n=2$? $\endgroup$
    – Will Sawin
    Commented Jan 21, 2020 at 23:26
  • $\begingroup$ Oh good point. I'm not sure. In the $n=2$ case the eigenvalues are real and positive and so I thought that implied $x^T T_1 T_2 x > 0$, but I guess that may not be true. $\endgroup$
    – Mido
    Commented Jan 21, 2020 at 23:50
  • $\begingroup$ Yes indeed, perhaps you can find a vector $x$ making the left-hand side negative. $\endgroup$
    – Mido
    Commented Jan 21, 2020 at 23:59

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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.

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