I know that operator norm of a self adjoint operator $T$ on $\Bbb C^n$ is given by$\|T\|=\mathrm {sup} \left \{|\left <Tx,x \right > |\ :\ \|x\|=1 \right \}$. But I just came to know that the result also holds for normal operators. But I couldn't find the proof of it after googling it for almost an hour.
Would anybody please provide me the proof of it? Then it will really be very helpful for me.
Thank you very much.
Really you should view the hypothesis of this theorem as "$T$ is unitarily diagonalizable", not "$T$ is self-adjoint", because this is the property that enters the proof. Once you unitarily diagonalize it is straightforward, because $\langle T\sum_{i=1}^n c_i v_i,\sum_{i=1}^n c_i v_i \rangle = \sum_{i=1}^n \lambda_i |c_i|^2$, which is a convex combination of the eigenvalues if $\sum_{i=1}^n c_i v_i$ was a unit vector. You need the unitarity of the eigenvector matrix to remove the cross-terms, whose indefinite sign would spoil this convex structure.
It is a standard theorem that an operator on a vector space over $\mathbb{C}$ is unitarily diagonalizable if and only if it is normal. The proof is slightly harder than the proof of the spectral theorem but not that hard.