# Spectral Norm Proof

I don't really understand the question below:

A matrix can have small eigenvalues but large spectral norm (i.e., largest singular value). Find such a matrix A of proper dimension so that all of its eigenvalues λ satisfy |λ| < 0.5, but its spectral norm satisfies ||A|| ≥ K for some K > 0 that can be made arbitrarily large.

I feel like I need to start with the definition of the spectral norm: $$||A||=\sqrt{\lambda_{\max}(A^TA)}$$

Not really sure if that's the right place to start or where to go from there. Any and all help is appreciated!

Thanks.

Start with $A=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ and find $\|A\| = 1$. Then $\| K A \| = |K|$. All eigenvalues are zero.
• $||KA|| =|K|$ but how does that prove that $||A||\geq K$? Thank you for your help – TryingToLearnMath Mar 22 '16 at 17:48
• Pick $K$. Then choose the matrix $KA$, which has eigenvalues that satisfy $|\lambda| < {1 \over 2}$ and $\|KA\| = K$. If you would rather, pick $K$ then define $A=\begin{bmatrix} 0 & K \\ 0 & 0 \end{bmatrix}$. Same effect. – copper.hat Mar 22 '16 at 18:00