Operator norm of a symmetric matrix

Idea. A symmetric matrix is a diagonal matrix in a well-chosen orthonormal basis, and orthonormal changes of basis do not change the operator norm.

Step 1: reduction. By the spectral theorem, $A=Q\Lambda Q^T$ with $Q\in O(n)$ and $\Lambda=\operatorname{diag}(\lambda_1,\dots,\lambda_n)$. By the previous problem, $\|A\|=\|Q\Lambda Q^T\|=\|\Lambda\|$.

Step 2: diagonal matrices. For $x\in\mathbb{R}^n$,
$$|\Lambda x|^2=\sum_i\lambda_i^2x_i^2\le\max_i\lambda_i^2\sum_ix_i^2=\max_i\lambda_i^2\,|x|^2,$$
with equality for $x=e_{i_0}$, where $i_0$ attains the max. So $\|\Lambda\|=\max_i|\lambda_i|$.

Conclusion. $\|A\|=\max_i|\lambda_i|=\rho(A)$.

Remark. This is false for non-symmetric matrices: $\begin{pmatrix}0&1\\0&0\end{pmatrix}$ has $\rho=0$ but $\|\cdot\|=1$. The equality $\|A\|=\rho(A)$ holds more generally for *normal* matrices (those with $AA^T=A^TA$). Note also that $\max|\lambda_i|$ may be attained at a *negative* eigenvalue, e.g. $\|\operatorname{diag}(1,-3)\|=3$.