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

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 $\|AB\|\le \|A \|\, \|B\|$. 
