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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 $A$.

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  • $\begingroup$ The operators $T^\ast T$ and $TT^\ast$ do not always have the same spectrum, but $\sigma(T^\ast T)\setminus\{0\}=\sigma(T T^\ast)\setminus\{0\}$. $\endgroup$ – MaoWao Nov 26 '18 at 18:12
  • $\begingroup$ @MaoWao Thank you. So the formula is only true for $\sigma(T^*T)$? $\endgroup$ – Schüler Nov 26 '18 at 19:22
  • $\begingroup$ 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. $\endgroup$ – MaoWao Nov 26 '18 at 20:28
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  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 $\|T\|=\max\{\sqrt{\lambda};\;\lambda\in \sigma(T^*T)\}$.

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