Proving that $\|T\|=\max\{\sqrt{\lambda};\;\lambda\in \sigma(T^*T)\}$.

Question

Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on an infinite-dimensional complex Hilbert space $F$.

It is well known that if $T\in \mathcal{B}(F)$, then $$\|T\|=\displaystyle\sup_{\|x\|=1}\|Tx\|.$$

I want to prove that for $T\in \mathcal{B}(F)$, we have $$\|T\|=\max\{\sqrt{\lambda};\;\lambda\in \sigma(T^*T)=\sigma(TT^*)\},$$ where $\sigma(A)$ denotes the spectrum of an operator $A$.

Answer 1

1. for each $A \in \mathcal{B}(F)$ we have $||A^*A||=||A||^2$.

2. if $A \in \mathcal{B}(F)$ is self-adjoint, then $||A|| =\max \{| \mu|: \mu \in \sigma(A)\}$.

3. if $T \in \mathcal{B}(F)$, then $T^*T$ is self-adjoint and $ \sigma(T^*T) \subseteq [0, \infty)$.

Now you should be in a position to prove $\|T\|=\max\{\sqrt{\lambda};\;\lambda\in \sigma(T^*T)\}$.