Operator norms of three basic matrices

Idea. In each case we find the largest value of $|Ax|$ on the unit circle, by an upper bound plus an example attaining it.

Case 1: $N=\begin{pmatrix}0&1\\0&0\end{pmatrix}$. For $x=(x_1,x_2)$, $Nx=(x_2,0)$, so $|Nx|=|x_2|\le|x|$. Equality holds for $x=e_2$, where $Ne_2=e_1$. Hence $\|N\|=1$. (Note $Ne_1=0$: the matrix kills a direction but still has norm $1$, and $\rho(N)=0<\|N\|$.)

Case 2: $D=\operatorname{diag}(1,-3)$. $|Dx|^2=x_1^2+9x_2^2\le9(x_1^2+x_2^2)=9|x|^2$, with equality at $x=e_2$. So $\|D\|=3$, the largest absolute value of a diagonal entry.

Case 3: rotation. $|R_\theta x|=|x|$ for all $x$ since a rotation preserves lengths (indeed $R_\theta^TR_\theta=I$, so $|R_\theta x|^2=x^TR_\theta^TR_\theta x=|x|^2$). So $\|R_\theta\|=1$.

Remark. Case 1 shows that the operator norm can be positive for a nilpotent matrix, so $\|A\|$ is not the spectral radius. This phenomenon is the reason for the Gelfand-type bound $\rho(A)\le\|A^k\|^{1/k}$ in a later problem. In Case 3 the Frobenius norm is $\sqrt2$, so $\|R_\theta\|_F=\sqrt2\|R_\theta\|$: the extremal case of $\|A\|_F\le\sqrt n\|A\|$.