Eigenvalues Bounded by Singular Values Proof

Question

I'm trying to prove the inequality for a real square matrix $M$

$\sigma_\min(M)\leq|\lambda_i|\leq\sigma_\max(M)$ for all $i$

We have that:

$\lambda_\max(M)=\sup_{x\neq0}\frac{x^TMx}{||x||_2^2}$

$\lambda_\min(M)=\inf_{x\neq0}\frac{x^TMx}{||x||_2^2}$

and

$\|M\|_2=\sqrt{\lambda_{\max}(M^{T}M)}=\sigma_{\max}(M)$

though I'm trying to more thoroughly prove the above statement but even then I'm having trouble putting the pieces together.

I am also able to use the results I've proved previously of

$\sigma_\max(M)||x||_2\geq||Mx||_2\geq\sigma_\min(M)||x||_2$

I think I've just hit a mental block because no matter how I put this together it doesn't seem sufficient enough of a proof. Thanks in advance.

Answer 1

I am a bit confused by your confusion. $\sigma_\max (M)$ is the maximal norm of $M x,$ for a unit vector $x.$ While $\lambda_\max$ is the maximal value of the inner product of a unit vector $x$ with $M x$. It is clear that the latter cannot be greater than the former.