# Bounding $\Vert M^{-1} \Vert$ and eigenvalues of $M$.

We work in $\mathbb{R}^n$. Let $M$ be an $n\times n$ matrix with $$x^TMx \geq k\Vert x\Vert^2$$ for all $x \in \mathbb{R}^n$, where $k>0$.

I want to show that $\Vert M^{-1} \Vert \leq \frac{1}{k}$ and that the real parts of the (possibly complex) eigenvalues of $M$ are at least $k$.

Attempt so far: It is easy to see that $M$ is invertible since otherwise there would be an $x\neq 0$ with $Mx=0$ and hence $x^TMx = 0$ which would violate the condition above.

Furthermore if $\lambda$ is an eigenvalue, then $|\lambda| \geq k$ by similar reasoning. I don't see how to deal with the complex eigenvalues at all though, since this bound doesn't ignore the imaginary part.

I know that $\Vert M^{-1} \Vert = \sqrt{\rho((M^{-1})^T(M^{-1})}$ is the square root of the norm of the maximum eigenvalue of $(M^{-1})^TM^{-1}$. But since $M$ isn't necessarily symmetric I don't know how much help that will be.

Any help would be greatly appreciated.

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I suppose that $\alpha$ and $k$ are meant to be the same here. – Geoff Robinson Mar 3 '12 at 16:34
Yes, thank you. I changed it. – nullUser Mar 3 '12 at 16:41
(Oops. Wrong comment removed. The software won't let med delete it.) – Harald Hanche-Olsen Mar 3 '12 at 16:59

We have writing the hypothesis for $M^{-1}x$ that $x^TM^{-1}x\geq k\lVert M^{-1}x\rVert^2$ so $$k\lVert M^{-1}x\rVert^2\leq x^TM^{-1}x=\langle x,M^{—1}x\rangle\leq \lVert x\rVert \cdot\lVert M^{—1}x\rVert$$ so for $x\neq 0$ $\lVert M^{—1}x\rVert\leq \frac{\lVert x\rVert}k$ which gives the result.
I was so close to this in my scratchwork! What can I do about the complex eigenvalues? Say an eigenvalue is $c+di$, the real part of $x^*Mx$ is $c\Vert x\Vert^2$, can I somehow extract a real vector $v$ for which $v^TMv = c\Vert v \Vert^2$? – nullUser Mar 3 '12 at 17:41