Two norms of the identity matrix

Idea. The two norms measure different things: the Frobenius norm adds up all the entries, whereas the operator norm only measures the largest stretching factor.

Step 1: Frobenius. The identity has $n$ entries equal to $1$ (the diagonal) and all the others $0$. Therefore
$$\|I_n\|_F^2=\sum_{i,j}\delta_{ij}^2=n,\qquad \|I_n\|_F=\sqrt n.$$

Step 2: operator norm. For every unit vector $x$ we have $|I_nx|=|x|=1$, so the supremum is $\|I_n\|=1$.

Remark. This shows that the two norms are really different when $n\ge2$: $\|I_n\|_F/\|I_n\|=\sqrt n$. This ratio is the largest possible one (see the problem on the best equivalence constants): the identity, and more generally every orthogonal matrix, is the extremal case where the Frobenius norm is $\sqrt n$ times the operator norm. A common mistake is to say "$\|I\|=1$ for every norm"; this is true for operator norms but false for the Frobenius norm.