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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.

Emanuel L
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2 Answers2

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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.

Ben Grossmann
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  • Thank you for the answer! Even after reading it, I don't understand why $v$ is necessarily an eigenvector of $\tilde T$. I don't understand why $\pmatrix{0 \ v}$ is an eigenvector of $T$ for any $v \in \Bbb C^{n-1}$.

    Following the fact $v \in \Bbb C^{n-1}$ and the final statements, wouldn't it mean that any such $v$ is an eigenvector of $\tilde T$?

     – Emanuel L Jun 16 '23 at 18:21
  • @EmanuelL Regarding "I don't understand why $v$ is necessarily an eigenvector of $\tilde T$": this is a question that I'm not sure how to answer directly because it indicates a fundamental miscommunication, but this might help: the statement "$\tilde T$ has the same eigenvectors as $\tilde T^$" can be rephrased as "$v$ is an eigenvector of $\tilde T$ if and only if $v$ is an eigenvector of $\tilde T^$". If you can understand that the four statements I list are equivalent, then you can show that if $v$ is an eigenvector of $T$ then it is an eigenvector of $\tilde T^*$ and vice versa – Ben Grossmann Jun 16 '23 at 20:06
  • @EmanuelL Perhaps it is more helpful to start here. Do you understand what I meant when I wrote "the following statements are equivalent"? – Ben Grossmann Jun 16 '23 at 20:12
  • Actually, after fiddling around with these in my notebook I think I now understand your proof. What confused me is that I thought the $v$ in the statement at the end is the same $v$ as above, (an arbitrary vector in $V$). After understanding that in the statements, $v$ is already an eigenvector of $\tilde T$, everything fell into it's place. Thank you for the patient explanation! – Emanuel L Jun 16 '23 at 20:56
  • I have posted an answer which basically breaks this down completely to the bones (for anyone who might have trouble understanding it). Thank you so much for taking your time answer :) – Emanuel L Jun 16 '23 at 21:12
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Ben's answer above basically solves the problem. However I want to post an answer to completely break down why this happens (especially if anyone has trouble understanding, like I did).

Continuing Ben's answer, we can choose any $v$ such that $v$ is an eigenvector of $\tilde T$. Constructing a new vector $\pmatrix{0 \\ v}$ leads us to:

$$ T \pmatrix{0 \\ v}= \pmatrix{\tau & 0\\0 & \tilde T} \pmatrix{0 \\ v} = \pmatrix{0 \\ \tilde Tv} $$

From this we can conclude that if $v$ is an eigenvector of $\tilde T$ then $\pmatrix{0 \\ v}$ is guaranteed to be an eigenvector of $T$.

However, it is given that any eigenvector of $T$ is also an eigenvector of $T^*$. In this case the following applies:

$$ T^* \pmatrix{0 \\ v}= \pmatrix{\overline \tau & 0\\0 & \tilde T^*} \pmatrix{0 \\ v} = \pmatrix{0 \\ \tilde T^*v} $$

From this we conclude that $\pmatrix{0 \\ v}$ is an eigenvector of $T^*$, only if $v$ is an eigenvector of $\tilde T^*$.

Hence this shows that any eigenvector of $T$ is an eigenvector of $T^*$.

Back to the original problem.

Since now $\tilde T$ is a matrix of $(n-1) \times (n-1)$, and every eigenvector of $\tilde T$ is an eigenvector of $\tilde T^*$, it follows by the induction's assumption that $\tilde T$ is also a diagonal matrix. In this case:

$$ T = \pmatrix{\tau & 0\\0 & \tilde T} $$

And since both matrix blocks are diagonal, it follows that $T$ is also a diagonal matrix.

Since we have shown that $T$ can be represented in some basis by a diagonal matrix, it follows that $T^*$ can also a represented by a diagonal matrix. Since diagonal matrices commute, we then conclude that $TT^*$=$T^*T$, hence $T$ is a normal operator.

Emanuel L
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