# Prove/disprove $||A^2|| \leq ||A||^2$ for a nxn matrix

The question is: Prove/disprove $||A^2|| \leq ||A||^2$ where A is some nxn matrix.

I've played around with a while few matrices and I'm pretty sure that this is correct but I can't quite figure out how to prove it. My first guess is just define some arbitrary matrix A, and show that for any vector $x$ of size 1, $||A^2 x|| \leq ||Ax||^2$ however this method seems rather involved.

Any tips would be great! :)

Edit: $||A||= \max ||Ax||$, where $|x| = 1$

Edit: Wow this was rather fast. Thanks everyone.

• Knowing about eigenvalues maybe provides a shortcut. Have you done eigenvalues yet? – mathreadler Jan 21 '16 at 1:25

Recall that by the definition of $\|A\|$ we have $$\|Ay\| \leq \|A\|\|y\|.$$
For any $x \in \mathbb{R}^n$, using this inequality twice (on $y = Ax$ and on $y = x$), we have $$\|A^2x\| = \|A(Ax)\| \leq \|A\| \|Ax\| \leq \|A\| \left( \|A\| \|x\| \right) = \|A\|^2 \|x\|.$$
Therefore $$\|A^2\| = \sup_{x \neq 0}\dfrac{\|A^2x\|}{\|x\|} \leq \|A\|^2.$$
It is true. In fact, for all $x \in \mathbb{R}^n$ with $\|x\| = 1$. If $Ax \neq 0$, then by definition, $$\|A^2x\| = \|A(Ax)\| \leq \|Ax\| \|A\|\frac{\|x\|}{\|Ax\|} = \|A\|\|x\| = \|A\|.$$ The above inequality also holds for $x$ such that $Ax = 0$. Thus $$\|A^2\| = \sup_{\|x\| = 1} \|A^2 x \| \leq \|A\|.$$