Show that $\| T^{\ast}T \| = \| T \|^2$ Suppose that $T$ is any bounded operator. I want to show that $\| T^{\ast} T \| = \| T \|^2$. Can you please check my proof. 

Consider that $\| T \|^2 = \sup_{\| x \| \leq 1} \| Tx \|^2 = \sup_{\| x \| \leq 1} \left| \langle x, T^{\ast}Tx \rangle \right| \leq \| T^{\ast}T \|$ and $\| T^{\ast} T \| = \sup_{\| x \| \leq 1} \| T^{\ast}Tx\| = \sup_{\| x \| \leq 1} \left| \langle x, T^2 x \rangle \right| \leq \| T \|^2$.
You make a similar gap in both inequalities you are trying to show.  Specifically, you claim that $\sup_{\|x\|\leq 1}|\left&lt; x, Ax\right&gt;| \leq \|A\|$. This claim requires justification.
You make a similar gap in both inequalities you are trying to show.  Specifically, you claim that $\sup_{\|x\|\leq 1}|\left&lt; x, Ax\right&gt;| \leq \|A\|$. This claim requires justification.