About proof of theorem (T is normal if and only if there exists an orthonormal basis for V consisting of eigenvectors of T) Definition: Let V be an inner product space, and let T be a linear operator on V. Say T is normal if $TT^{*} =T^{*}T$. 
Theorem: Let T be a linear operator on a finite-dimensional complex inner product space V Then T is normal if and only if there exists an orthonormal basis for V consisting of eigenvectors of T.
How to prove this? A sketch is appreciated.
 A: If $T : X\rightarrow X$ is a linear transformation on a complex inner product space $X$, then $T$ is normal iff $T^*T=TT^*$. This is equivalent to
$$
             \|Tx\|^2=\langle Tx,Tx\rangle = \langle T^*Tx,x\rangle=\langle TT^*x,x\rangle=\|T^*x\|^2,\;\;\; x\in X.
$$
Therefore, if $T^2x=0$, then $T^*Tx=0$ follows from the above, which implies that $Tx=0$ because
$$
         0 = \langle T^*Tx,x\rangle=\|Tx\|^2 \implies Tx=0.
$$
Consequently, the minimal polynomial for $T$ has no repeated roots, which proves that $T$ is diagonalizable.
To show that the eigenvectors corresponding to distinct eigenvalues are orthogonal, suppose that $Tx=\lambda x$ and $Ty=\mu y$ for some $\mu\ne\lambda$. Then, $x\perp y$ because
$$
    |\lambda-\mu|^2\langle x,y\rangle=(\lambda\overline{\lambda}-\lambda\overline{\mu}-\overline{\lambda}\mu+\mu\overline{\mu})\langle x,y\rangle \\
   =\langle T^*Tx,y\rangle-\langle Tx,Ty\rangle-\langle T^*x,T^*y\rangle+\langle x,TT^*y\rangle \\
   =\langle Tx,Ty\rangle-\langle Tx,Ty\rangle-\langle T^*x,T^*y\rangle+\langle T^*x,T^*y\rangle=0
$$
It follows that $T$ has an orthonormal basis of eigenvectors. The converse that $T$ is normal if it has an orthonormal basis of eigenvectors can be proved by directly showing $T^*T=TT^*$.
