How to prove: "Every subspace of $V$ invariant under $T$ is also invariant under $T^*$ if and only if $T$ is normal."? 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.
 A: 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.


*

*Either by Primary Decomposition Theorem or as simple as the fundamental theorem of Algebra, one can show that there has to be an eigenvector and an eigenvalue. Denote that pair by $v_1$ and $\lambda_1$. 

*Let $W=(span\{v_1\})^\perp$. Show that $W$ is invariant under $T$. This can be done by noticing that since $span\{v_1\}$ is invariant under $T$, it is invariant under $T^*$. 

*Restrict $T$ to $W$, you get a new operator (this is possible because you proved in step 2 that $W$ is invariant under $T$). Repeat step 1 and step 2 until you have got yourself $n$ eigenvectors.

*By construction, these $n$ eigenvectors are orthogonal to each other. And I trust that you can derive that $T$ is normal from this on.


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.
