I have question about optimize problem need to proof: suppose that $M$ is a symmetric matrix, then the following are equivalent:
(1) $||M^{-1}|| < \frac{1}{h}$, where $h > 0$
(2) $||Mv|| \geq h||v||$, where $h>0$, $v \in \mathbb{R}^n$
proof from (1) to (2) and then (2) to (1)
Thank you very much
$M$ is symmetric so it is diagonalizable by an orthonormal eigenbasis. $M=BDB^\top$.
Both (1) and (2) are equivalent to (3) "the smallest absolute eigenvalue of $M$ is $\ge h$."
(3) $\iff$ (1): Note that (1) can be restated as "the largest absolute eigenvalue of $M^{-1}$ is $\le 1/h$. Since $M=BD^{-1}B^\top$, the equivalence follows.
(3) $\implies$ (2): $\|Mv\|^2 = v^\top B D^2 B^\top v \ge v^\top B (h^2 I) B^\top v = h^2 \|B^\top v\|^2 = h^2 \|v\|^2$.
(2) $\implies$ (3): if $v$ is a $\lambda$-eigenvector of $M$, then $h\|v\| \le \|Mv\| = |\lambda| v $ so $|\lambda| \ge h$.