Can square root of a normal operator be normal too ??

Question

Suppose $T$ is normal on a complex inner product space, I have proved that $T$ has a square root $S$. Can it be normal too ?

Answer 1

Certainly. In the event that $T$ is the identity operator, this is trivially true since $T$ is its own square root. Perhaps less trivially, if $T$ is a scaling operator, then its square root is well-defined and normal.

However, this is not necessarily the case. Consider, as a counter-example $T = 0\in \mathbb{C}^{2\times 2}$, with square root $$ S = \begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix} $$ We may note, on the other hand, that any finite-dimensional normal operator has a normal square root. Indeed, if $T$ is normal, then $$ T = UDU^* $$ For some unitary $U$ and diagonal $D$. $S = UD^{1/2}U^*$ is a normal matrix satisfying $S^2 = T$.