Could any one tell me how this $\|x\|^2=\|x*x\|$ and the rest of it?
I know $\|x\|=\|x^*\|$, I also understand $x^*x$ is hermitian and so diagonalizale but then did not understand the norm square equal to max of eigen valuess etc thank you for helping
From your first expression $$\|x\|^{2}=\sup_{\|\xi\|=1}\overline{\xi}^{t}x^{\star}x\xi.$$ If $\lambda$ is an eigenvalue of $x^{\star}x$ then $\lambda$ is real and non-negative, and there exist a unit eigenvector $\xi$ of $x^{\star}x$ with eigenvalue $\lambda$, from which it follows that $\|x\|^{2} \ge \overline{\xi}^{t}x^{\star}x\xi=\lambda\|\xi\|^{2}=\lambda$. If $\lambda_{0} > \lambda_{1} > \cdots > \lambda_{k} \ge 0$ are the distinct eigenvalues of $x^{\star}x$, then, because $x^{\star}x$ is Hermitian, every unit vector $\xi$ can be written as an orthogonal sum $$ \xi = \xi_{0}+\xi_{1}+\cdots +\xi_{k},\;\;\; \overline{\xi_{j}}^{t}\xi_{k}=0 \mbox{ for } j\ne k, $$ where $$ x^{\star}x\xi_{k}=\lambda_{k}\xi_{k},\;\;\;1=\|\xi_{0}\|^{2}+\|\xi_{1}\|^{2}+\cdots+\|x_{i}\|^{2}. $$ (We allow any $\xi_{k}$ to be $0$ in order to have a general such representation.) Then $$ \begin{align} \overline{\xi^{t}}x^{\star}x\xi & = \lambda_{0}\|\xi_{0}\|^{2}+\lambda_{1}\|\xi_{1}\|^{2}+\cdots+\lambda_{k}\|\xi_{k}\|^{2} \\ & \le \lambda_{0}(\xi_{0}\|^{2}+\|\xi_{1}\|^{2}+\cdots+\|\xi_{k}\|^{2}) = \lambda_{0}. \end{align} $$ Therefore, $\|x\|^{2} \le \lambda_{0}$ also holds. Thus, $\|x\|^{2}$ is the maximum eigenvalue of $x^{\star}x$.