The following statement is true or false.
Suppose $A$ is an $n \times n$-real matrix, all whose eigenvalues have absolute value less than 1. Then, for any $v \in \mathbb{R}^n$, $||Av||\leq||v||$.
It is an entrance test question and the answer is given false. I am failing to understand why. Let $\lambda$ be an eigen value of $A$. Then it has a corresponding eigen vector $v$ and subsequently $Av = \lambda v$. Therefore, $||A v || \leq ||\lambda v||$. We can also conclude $||Av|| \leq |\lambda| ~||v||$. Is the answer is given false because of the fact that $|\lambda| < 1$ instead of $|\lambda| \leq 1$?
your reasoning would be sound, if all vectors were Eigenvectors. But consider for example $$A=\begin{pmatrix}0 & 2 \\ 0 & 0\end{pmatrix}.$$ We have $A e_2 = 2 e_1$, hence $\vert \vert A v \vert \vert > \vert \vert v \vert \vert$. Also, all Eigenvalues of $A$ are at most $1$ (it has only one).