How to show that $\|T\|^2=\|T^*T\|$ for a bounded linear operator $T$? I need to show that for a bounded linear operator, $T$, on a Hilbert space:
\begin{align*}
\|T\|^2=\|T^*T\|
\end{align*}
All I have so far:
\begin{align*}
\|T^*T\|&=\sup\{|\langle T^*Tf,g \rangle |:\|f\|\le1,\|g\|\le1\}
\\ &= \sup\{|\langle Tf,Tg \rangle|:\|f\|\le1,\|g\|\le1\}
\\ &\le \sup_{\|f\|\le1}\|Tf\|\sup_{\|g\|\le1}\|Tg\|
\end{align*}
Not too sure what to do from here (or even if this is right so far)
Any help is appreciated.
 A: $
\newcommand{\norm}[1]{\left\|{#1}\right\|}
\newcommand{\ip}[1]{\left\langle{#1}\right\rangle}
$You might find it easier to use an alternative but equivalent definition of the operator norm: if $S \in B(H)$, then
$$
 \norm{S} = \sup_{\norm{x}=1} \norm{Sx}.
$$
Now:


*

*On the one hand,
$$
 \norm{T} = \sup_{\norm{x}=1} \norm{Tx} = \sup_{\norm{x}=1} \sqrt{\ip{Tx,Tx}}.
$$
Since $\ip{Tx,Tx} = \ip{x, T^\ast T x} \leq \norm{T^\ast T}\norm{x}$, what can you conclude?

*On the other hand,
$$
 \norm{T^\ast T} = \sup_{\norm{x}=1} \norm{T^\ast Tx}.
$$ 
Since $\norm{T^\ast T x} \leq \norm{T^\ast} \norm{Tx}$, where $\norm{T^\ast}=\norm{T}$ (why?), what can you conclude?
In terms of wider context, the identity $\norm{T^\ast T} = \norm{T}^2$ is called the $C^\ast$-identity, and is the key fact that makes the theory of $C^\ast$-algebras (e.g., $B(H)$ for $H$ a Hilbert space) so much easier (for lack of a better word) than the theory of more general Banach ($\ast$-)algebras, just as the theory of Hilbert spaces is so much easier than the theory of more general Banach spaces.
A: The first notice that $\|T\|=\|T^*\|$:
From definition of operator norm
\begin{align*}
\|T^*\|&=\sup_{\|y\|=1}\|T^*y\|=\sup_{\|y\|=1}\sup_{\|x\|=1}|(x,T^*y)|=
\sup_{\|y\|=1}\sup_{\|x\|=1}|(Tx,y)|\\
&=\sup_{\|x\|=1}\sup_{\|y\|=1}|(Tx,y)|=\sup_{\|x\|=1}\|Tx\|=\|T\|
\end{align*}
Therefore $T^*\in\mathcal{L}(H,H)$ and $\|T\|=\|T^*\|$.
To conclude, let $x\in H$ with $\|x\|=1$. It follows from
\begin{align}
\|Tx\|^2=(Tx,Tx)=(x,T^*Tx)\leq\|x\|\|T^*Tx\|\leq \|T^*T\|\leq\|T^*\|\|T\|=\|T\|^2
\end{align}
that $\|T\|^2\leq\|T^*T\|=\|T\|^2$.
