If $|a_{ij}| \leq L$ for all $i,j$, then $\| A \| \leq L\,$?

Question

This is my attempt: Let $A \in M(n, \mathbb R)$, $v \in \mathbb R^n$. Note that $\| A \| = \sup\limits_{v, \|v\| \leq 1} \|Av\|$.
I may assume WLOG that $\|v\| = 1$.
Take any $v \in \mathbb R^n$. Then, $Av = y$, where $y_i = \sum\limits_{j=1}^n a_{ij}v_j$. Then, \begin{align*} \|y\|^2 &= \Big(\sum\limits_{j=1}^n a_{1j}v_j\Big)^2 + \cdots + \Big(\sum\limits_{j=1}^n a_{nj}v_j\Big)^2 \\ &\leq\Big(\sum_{j=1}^n a_{1j}^2\Big) \cdot \sum_{j=1}^n v_j^2 + \cdots + \Big(\sum_{j=1}^n a_{nj}^2\Big) \cdot \sum_{j=1}^n v_j^2 \quad \text{(by Cauchy-Schwarz)}\\ &\leq\sum_{j=1}^n a_{1j}^2 + \cdots + \sum_{j=1}^n a_{nj}^2 \\ &= nL^2 + \cdots +nL^2 \\ &= (nL)^2 \end{align*} Hence, $\|A\| \leq nL$. However, my lecture notes state that $\|A\| \leq L$, and I am wondering what should be done to achieve that inequality.

Answer 1

Your question is answered with a No.

Please calculate the norm of $\begin{pmatrix}1 &1\\0&1\end{pmatrix}$. Or consider its action on the vector $\begin{pmatrix}1 \\1\end{pmatrix}$.

A symmetric or self-adjoint example would be the orthogonal projection $\begin{pmatrix}0.5 &0.5\\ 0.5 & 0.5\end{pmatrix}$ which turns into $\begin{pmatrix}1 & 0\\0 & 0\end{pmatrix}$ upon diagonalisation.