suppose $X=R^{n}$ and let $N \in R^{n \times n}$ be a matrix. Show that N is nonexpansive if and only if $\lambda_{max}(N^T N)\leq 1$.
For showing nonexpansiveness we must show that $(\forall x \in X) (\forall y \in X) \hspace{5mm} ||Nx-Ny|| \leq ||x-y||$. But how can I relate this definition to the largest eigenvalue.
Any help or references will be appreciated.
HINT
Assume for a second that $N$ is an orthogonal matrix, with eigenvectors $(x_n)_n$ and corresponding eigenvalues $(\lambda_n)_n$, which we can without the loss of generality assume to satisfy $$ |\lambda_1| \ge |\lambda_2| \ge \ldots \ge |\lambda_n|. $$
Therefore, for any $u \in X$, we have $$ \begin{split} \|Au\| &= \left\|A\left(\sum_{k=1}^n a_k x_k\right)\right\| \\ &= \left\|\sum_{k=1}^n a_k A x_k \right\| \\ &= \left\|\sum_{k=1}^n a_k \lambda_k x_k \right\| \\ &\le \left\|\sum_{k=1}^n a_k x_k \right\| |\lambda_1 | \\ &\le |\lambda_1| \left\|u\right\| \end{split} $$
For reference, $|\lambda_1|$ is called the operator norm of $A$.
