A Question on Normal Operators Let $T$ be a compact operator on a Hilbert space $H$. I want to prove the following:
If there exists Orthonormal Basis from the eigen vectors of $T$, then $T$ is normal operator.
 A: I'll assume a complex Hilbert space $H$. If $\{ e_n \}_{n=1}^{\infty}$ is an orthnormal basis of $H$ with $Te_n =\lambda_n e_n$, then there is a constant $M$ such that $|\lambda_n| \le M$ for all $n$ because $T$ must be bounded, which ensures the norm converges of all vector sums in this post, such as
$$
         T x = \lim_{N} T\sum_{n=1}^{N}\langle x,e_n\rangle e_n = \lim_N\sum_{n=1}^{N}\lambda_n \langle x,e_n\rangle e_n =\sum_{n=1}^{\infty}\lambda_n\langle x,e_n\rangle e_n
$$
Therefore,
\begin{align}
     \langle Tx,y\rangle & =\lim_N \langle \sum_{n=1}^{N}\lambda_n\langle x,e_n\rangle e_n,y\rangle \\
    & = \lim_N\sum_{n=1}^{N}\lambda_n\langle x,e_n \rangle\langle e_n,y\rangle \\
    & = \lim_N\langle x,\sum_{n=1}^{N}\overline{\lambda_n}\langle y,e_n\rangle e_n\rangle \\
    & = \langle x,\sum_{n=1}^{\infty}\overline{\lambda_n}\langle y,e_n\rangle e_n\rangle
\end{align}
Hence $T^*$ is given by
$$
         T^*y = \sum_{n=1}^{\infty}\overline{\lambda_n}\langle y,e_n\rangle e_n
$$
Consequently,
$$
             \|Tx\|^2 = \sum_{n=1}^{\infty}|\lambda_n|^2|\langle x,e_n\rangle|^2 = \|T^*x\|^2.
$$
That's enough to imply $T^*T=TT^*$. (Alternatively, you can directly verify $T^*Tx = TT^*x$ for all $x$ using the known forms of $T$ and $T^*$.)
