Given $A \in \mathbb{R}^{n \times n}$ with spectral radius $\rho(A)<1$, is it possible for $||A||>1$ for Euclidean norm $|| \cdot ||$?
I'm working on a proof where we are given the fact that $\rho(A)<1$, and in the case of $A$ being symmetric we know that $\rho(A)=||A||$ for the Euclidean norm. I am trying to make the proof more general, for when $A$ is not symmetric. All I would need to show is that, if $\rho(A) < 1$ then $||A|| \le 1$.
My attempt here is to use the singular value decomposition and the fact that the 2-norm is invariant under orthogonal transformations. Let the SVD of $A$ be
$$U^\intercal A V = D$$
where $D$ is diagonal with singular values along the diagonal. Then
$$||U^\intercal A V|| = ||A|| = ||D||$$
Since $\rho(A)<1$ then the singular values are also less than 1, then $||Dx|| < ||x||$, hence
$$\sup_{x \ne 0}\frac{||Dx||}{||x||} < 1$$
Best Answer
Yes, it is possible. Here's a counterexample, let $$ A = \begin{pmatrix} 0 & a \\ 0 & 0 \end{pmatrix} $$ for some $a \in \mathbb{R}$. The eigenvalues of $A$ are $\{0\}$ thus $\rho(A) = 0$. However, for $x = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ we have $\|x \| = 1$ and $Ax = \begin{pmatrix} a\\0 \end{pmatrix}$. Hence $\|A\| \geq a^2$. As $a$ is arbitrary we can make the norm of $A$ arbitrarily large whilst keeping $\rho(A) =0$.