Show that every diagonalizable operator is normal

Question

Could anyone show that a diagonalizable operator is normal in Hilbert spaces? Every hint is appreciated.

Answer 1

If a (continuous) linear operator $A$ is normal (i.e. it commutes with its adjoint $[A,A^+]=0$), then it is diagonalizable. The converse however in general is not true. It is not true even for finite dimensional spaces. Consider the following operator $A$ in a 2-D H.S. given is some orthonormal basis by the following matrix:

$$A= \begin{pmatrix} 1&2\\ 3&4\\ \end{pmatrix} $$

$A$ is diagonalizable (it has two different eigenvalues) and has $[A,A^+] \ne0$:

$$ [A,A^+] = \begin{pmatrix} 1&2\\ 3&4\\ \end{pmatrix} \begin{pmatrix} 1&3\\ 2&4\\ \end{pmatrix} - \begin{pmatrix} 1&3\\ 2&4\\ \end{pmatrix} \begin{pmatrix} 1&2\\ 3&4\\ \end{pmatrix} = \begin{pmatrix} -5&-3\\ -3&+5\\ \end{pmatrix} \ne \begin{pmatrix} 0&0\\ 0&0\\ \end{pmatrix} $$

Therefore $A$ is not normal. In fact, most operator are diagonalizable, being non-diagonalizable is an oddity (in some sense it is a set of measure zero). At the same time most operators are not normal, even if they are diagonalizable.