Inequalities inside operator norm for a normal operator

Question

Suppose $T$ is normal, I am trying to show that

$\|T^{n}\|^2\|T\|^2 = \|T^{n+1}\|^2$

I know that

$\|T^{n+1}\|^2 = \|T^{n}T\|^2 \leq \|T^{n}\|^2\|T\|^2$.

Is there a way to show that

$ \|T^{n}\|^2\|T\|^2 \leq \|T^{n+1}\|^2 $?

I could be taking the wrong approach, although I am curious if it is possible to show if the second inequality is true.

Answer 1

We only need to prove $$\|T^n\|=\|T\|^n$$ when $T$ is a normal operator.

For any operator $R$, we have $$\|R^*R\|=\|R\|^2.$$ So for a self-ajoint operator $S$, we have $\|S^2\|=\|S\|^2$, thus $$\|S^{2^k}\|=\|S\|^{2^k} ~\forall k\in \mathbb{N}.$$ For any $n\in \mathbb{N}$, pick a positive integer $k$ such that $n<2^k$, we have $$\|S\|^{2^k}=\|S^{2^k}\|=\|S^nS^{2^k-n}\|\leq\|S^n\|\|S\|^{2^k-n}\leq \|S\|^{2^k},$$ hence $ \|S^n\|=\|S\|^n$. For a normal operator $T$, $\|T^n\|^2=\|(T^n)^*T^n\|=\|(T^*T)^n\|=\|T^*T\|^n=\|T\|^{2n}.$ Done!

Answer 2

Since $T$ is normal, you have $\|T\|=r(T)$, the spectral radius. So now your equality, without the squares is saying that $$ \max\{|\lambda|^n: \lambda\in\sigma(T)\}\,\max\{|\lambda|: \lambda\in\sigma(T)\}=\max\{|\lambda|^{n+1}: \lambda\in\sigma(T)\}. $$ This is trivial after you notice that $\max\{|\lambda|^n:\ \lambda\in\sigma(T)\}=(\max\{|\lambda|:\ \lambda\in\sigma(T)\})^n$.