# 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 $$A$$.

• The operators $T^\ast T$ and $TT^\ast$ do not always have the same spectrum, but $\sigma(T^\ast T)\setminus\{0\}=\sigma(T T^\ast)\setminus\{0\}$. Nov 26 '18 at 18:12
• @MaoWao Thank you. So the formula is only true for $\sigma(T^*T)$? Nov 26 '18 at 19:22
• No, it works either way. The maxima of $\sigma(T^\ast T)$ and $\sigma(T T^\ast)$ do coincide, as the formula $\sigma(T^\ast T)\setminus\{0\}=\sigma(TT^\ast)\setminus\{0\}$ shows. Nov 26 '18 at 20:28

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