# Normal Operator $\| T^2\| = \|T\|^2$

Given a complex inner product space X, and an operator $$T: X \rightarrow X$$ is normal i.e. $$T^*T=TT^*$$ How can we show $$\| T^2\| = \|T\|^2$$?

By the definition of operator norm, it follows that ||T|| = sup $$\frac{||Tx||}{||x||}$$ and ||T$$^2$$|| = sup $$\frac{||T^2x||}{||x||}$$. Then I can express the numerator as a form of inner product. But I still am not able to make these two equal. Any good ideas?

If $$T$$ is normal, then $$\|Tx\|^2=\left=\left =\left=\|T^*x\|^2$$, so $$\|Tx\|=\|T^*x\|$$ (and therefore $$\|T\|=\|T^*\|$$). Then (replacing $$x$$ by $$Tx$$) $$\|T^2x\|=\|T^*Tx\|$$ so that $$\|T^2\|=\|T^*T\|$$. But also $$\|Tx\|^2=\left\le\|T^*T\|\|x\|^2$$ so that $$\|T\|^2\le\|T^*T\| =\|T^2\|$$. But $$\|T^2\|\le\|T\|^2$$. We conclude that $$\|T^2\|=\|T\|^2$$ whenever $$T$$ is normal.
• (ii) why can we have <x, T$^*$Tx> $\leq$ ||T$^*$T|| ||x||$^2$? – Jonny Apr 23 at 4:56
• By the definition: $\|T\|=\sup_{\|x\|=1}\|Tx\|$. @Eric – Lord Shark the Unknown Apr 23 at 4:56
• @eric $|\left<x,Ax\right>|\le\|x\|\|Ax\|\le\|A\|\|x\|^2$. – Lord Shark the Unknown Apr 23 at 4:58