Show that $ ‖ A^2 ‖_2 = ‖ A ‖^2_2$ for a symmetric matrix $A$

Question

I want to show that $ ‖ A^2 ‖_2 = ‖ A ‖^2_2$ for a symmetric matrix $A$. So far, I got $$‖ A^2 ‖_2 = \underset{x\neq0}{\max} \frac{‖ A^2x ‖}{‖ x ‖} = \underset{x\neq0, ‖ x ‖ = 1}{\max} ‖ A^2x ‖$$ $$ ‖ A ‖^2_2 = (\underset{x\neq0}{\max} \frac{‖ Ax ‖}{‖ x ‖})^2 = \underset{x\neq0}{\max} \frac{‖ Ax ‖^2}{‖ x ‖^2} = \underset{x\neq0, ‖ x ‖ = 1}{\max} ‖Ax‖^2.$$ How should I proceed?

Answer 1

Note that $\| A^2 \| \leq \| A \| \| A \| = \| A \|^2$ is clear by the definition of the operator norm. The other direction now follows from

$$\| Ax\|^2 = \langle Ax, Ax \rangle = \langle A^2 x, x \rangle \leq \|A^2x\| \|x\| \leq \|A^2 \| \|x \|^2.$$

Answer 2

You should use the hypothesis.

Hint: $\|Ax\|^2 =x^TA^2x\le \|A^2x\|\le\|A^2\|$ if $\|x\|\le 1$,
and for any two matrices $A,B$ we have $\|AB\|\le \|A \|\, \|B\|$.

Answer 3

The $2$-norm of a matrix is given by $$ \|A\|_2 = \sqrt{\lambda_{\max}(A^*A)} = \sigma(A), $$ where $\sigma(A)$ denotes the spectral radius of $A$, $\lambda_{\max}$ denotes the largest eigenvalue of $A$, and $A^*$ denotes the Hermitian adjoint (i.e. conjugate transpose) of $A$. Also note that if $\lambda$ is an eigenvalue of $A$, then $\lambda^2$ is an eigenvalue of $A^2$ (since $Av=\lambda v$ implies $A^2v = A(Av) = A(\lambda v) = \lambda (Av) = \lambda^2 v$). So if $A$ is Hermitian, i.e. $a_{ij} = \overline{a_{ji}}$ for all $i,j$, then $A^*A = AA^* = A^2$ and hence $$ \|A\|_2^2 = \lambda_{\max}(A^2) = \sigma(A^2) = \|A^2\|_2. $$