I got a homework question from linear algebra course. To prove "Suppose that $T$ is a linear operator and $T^*$ is its adjoint. Prove that every subspace of $V$ invariant under $T$ is also invariant under $T^*$ if and only if $T$ is normal. "
If $T$ is normal, then I know how to prove it.
But about the other direction: If every subspace of $V$ invariant under $T$ is also invariant under $T^*$,then $T$ is normal. I don't know how to show it.
Best Answer
This statement is in general not correct if the underlying field for $V$ is the real numbers. For an example, look at the case when $V=\mathbb{R}^2$ and the operator $T=(0,-1;1,0)$ (matlab notation as I don't know how to type matrix in LaTex unfortunately). In this case, not a single proper subspace is invariant under $T$, so the assumption is automatically correct. It's down to simple calculation to show that $T$ is not normal.
However, when the underlying field for $V$ is the complex field, then it's correct. I will sketch a proof.
A final remark, this kind of proof will not work in infinite dimensional space because the construction provided above is not guaranteed to end in a finite number of steps.