A doubt on compression of norm of a vector upon action by a matrix

Question

Check whether the following statement is true.

Suppose that $A \in M_n(\mathbb{R}),$ whose all eigenvalues have absolute values less than $1.$ Then for any $v\in \mathbb{R}^n,\lVert Av \rVert \leq \lVert v \rVert .$

The answer for this is given to be false.

I think that it is true.

My Attempt:

We can write $\lVert Av \rVert \leq \lVert A \rVert\lVert v \rVert \leq \rho(A)\lVert v \rVert < \lVert v \rVert,$ where $\rho(A)$ is the spectral radius of A, which is given to be less than $1.$ From this we can conclude that $\lVert Av \rVert \leq \lVert v \rVert,$ for all $v \in \mathbb{R}^n.$
I have used the fact that $\rho(A)=\lVert A \rVert,$ which is true for normal matrices. Is my logic wrong because I used a result which is not true for all matrices?

Any ideas would be of great help. Thanks.

Answer 1

Consider $A=\pmatrix{\frac23&\frac23\\0&\frac23}$. It has eigenvalues less than $1$, yet applying it to $\pmatrix{0\\1}$ increases the norm.

Answer 2

Yes, the counterexamples exist in non-normal matrices.

For example, let $A=\begin{pmatrix} 1/2 & 1 \\ 0 & 1/2 \end{pmatrix}$ and $v=\begin{pmatrix} 0 \\ 1 \end{pmatrix}$, then $Av=\begin{pmatrix} 1 \\ 1/2 \end{pmatrix}$. The eigenvalues of $A$ are two $1/2$'s, but $\lVert Av \rVert = \sqrt{5}/2 > 1 = \lVert v \rVert$.

In general, the spectral radius of a matrix $A$ is the largest absolute value of its eigenvalue. So for a non-zero nilpotent matrix, it has spectral radius $0$ but non-zero norm (whatever norm you choose) due to the definition of norm.