every diagonalizable operator is normal Suppose that $H $ is an infinite Hilbert space and $T$ is a diagonalizable operator on it i.e. for an orthonormal basis $\{e_n\}$ for $H$ an operator  $T$ is diagonalizable if $Te_n=\lambda_n e_n$ in which case $||T||=\sup\{|\lambda_n|:n\geq 1\}$ and $\{\lambda_n\}$ is eigenvalue of it. Could anyone show that $T$ is normal

$T^*$ is diagonal with respect to the same basis as $T$ since $(T^*e_n,e_m)=(e_n,Te_m)=\lambda_m\delta_{n,m}$. Diagonal operators obviously commute.
