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


*

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

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

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