Every eigenvector of $T$ is an eigenvector of $T^*$. Prove that $T$ is a normal operator.

eigenvalues-eigenvectorslinear algebralinear-transformations

The question is as in the title:

$T:V \to V$ is a linear transformation, $V$ is a finite dimensional vector-space over the complex field.
Every eigenvector of T is an eigenvector of $T^*$. Prove that T is a normal operator.

I have found the following questions already asked previously, however the answers to them left me kind of stumbled, as they are not fully explained.

For example in this thread which attempted to provide a clearer proof than this thread.

The answer with the highest votes states in the middle of the proof:

By the induction assumption, the matrix $\tilde T$ is diagonal since it has
the same eigenvectors as $\tilde T^∗$ .

Specifically, I couldn't figure out why the following statement is true:

$\tilde T$ has the same eigenvectors as $\tilde T^*$.

If somebody could explain to me I will be very thankful 🙂

Thanks in advance

EDIT

I've posted an answer below which further breaks down Ben Grossman's answer, and also continues the proof for the original problem.

Best Answer

I'll pick up from the point in the proof in which the statement arises, in which we have already shown that $$ T = \pmatrix{\tau & 0\\0 & \tilde T}, \quad T^* = \pmatrix{\bar \tau & 0\\0 & \tilde T^*}. $$ Note that for any vector $v \in \Bbb C^{n-1}$ and $\alpha \in \Bbb C$, we have $$ T \pmatrix{\alpha \\ v}= \pmatrix{\tau & 0\\0 & \tilde T} \pmatrix{\alpha \\ v} = \pmatrix{\tau \alpha\\ \tilde Tv}. $$ Use this to observe that the following statements are equivalent:

  • $v$ is an eigenvector of $\tilde T$
  • the block-vector $(0,v)$ is an eigenvector of $T$
  • the block-vector $(0,v)$ is an eigenvector of $T^*$
  • $v$ is an eigenvector of $\tilde T^*$

Thus, $\tilde T$ and $\tilde T^*$ indeed have the same eigenvectors.

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