Consider the following question:

Define the adjoint $\alpha^*$ of an endomorphism $\alpha$ of a complex inner-product space $V$. Show that if $W$ is a subspace of $V$, then $\alpha(W) \subseteq W$ if and only if $\alpha^*\left(W^{\perp}\right) > \subseteq W^{\perp}$.

An endomorphism of a complex inner-product space is said to be normal if it commutes with its adjoint. Prove the following facts about a normal endomorphism $\alpha$ of a finite-dimensional space $V$.

(i) $\alpha$ and $\alpha^*$ have the same kernel.

(ii) $\alpha$ and $\alpha^*$ have the same eigenvectors, with complex conjugate eigenvalues.

(iii) If $E_\lambda=\{x \in V: \alpha(x)=\lambda x\}$, then $\alpha\left(E_\lambda^{\perp}\right) \subseteq E_\lambda^{\perp}$.

(iv) There is an orthonormal basis of $V$ consisting of eigenvectors of $\alpha$.

Deduce that an endomorphism $\alpha$ is normal if and only if it can be written as a product $\beta \gamma$, where $\beta$ is Hermitian, $\gamma$ is unitary and $\beta$ and $\gamma$ commute with each other. [Hint: Given $\alpha$, define $\beta$ and $\gamma$ in terms of their effect on the basis constructed in (iv).]

I have already seen a proof of (iv) with the spectral theorem, however, in this course the theorem was not lectured. (I checked similar questions here but they all seem to be using it).

Moreover, I can't seem to think of another way to quickly get an answer as this question is not meant to be too long since the whole thing should take about 25 minutes. I also get the sense that this construction must be explicit because of the very last part.

Now I should say that this is from an exam that was 23 years ago so there is a chance the syllabus has changed and that I am asking for something that is unreasonable.

If anybody sees a way to do (iv) without the spectral theorem I would greatly appreciate it if they could share.

Note: The only version of the Spectral Theorem that is allowed in the current version of the course is that if an ednomorphism is self adjoint then there is an orthonormal basis of eigenvectors of that endomorphism. I am sure that is not the full version of the theorem which seems to be used by most answers associated with this question on the website.

Remark: I would only like to be able to answer (iv) of the question.

  • $\begingroup$ Please do not rely on pictures of text. $\endgroup$
    – Shaun
    Mar 12 at 18:19
  • $\begingroup$ This is from an exam. What would you like me to do? $\endgroup$ Mar 12 at 18:35
  • $\begingroup$ May I ask why is this being down voted? $\endgroup$ Mar 12 at 18:36
  • 1
    $\begingroup$ I have now typed the question for the convenience of the forum. How can I address the focus issue? I have done research as to how to approach it, and have found no similar questions that have been asked here. What would be the appropriate next step? $\endgroup$ Mar 12 at 18:43
  • 1
    $\begingroup$ No worries. I am glad I was able to clarify my intentions with the question. $\endgroup$ Mar 12 at 18:54

1 Answer 1


Part (iii) tells us that $\alpha(E_\lambda^\perp) \subseteq E_\lambda^\perp$, but it is also true that $\alpha^*(E_\lambda^\perp) \subseteq E_\lambda^\perp$. One way to see this is that by (ii), a different definition of $E_\lambda$ is $\{x \in V : \alpha^*(x) = \overline \lambda x\}$, so we can apply (iii) from $\alpha^*$'s point of view instead of $\alpha$'s. Therefore $\alpha$ and $\alpha^*$, restricted to $E_\lambda^\perp$, are also endomorphisms of $E_\lambda^\perp$.

This lets us inductively solve the problem:

  1. Pick an eigenvalue $\lambda$ of $\alpha$ (and $\alpha^*$);
  2. Find an arbitrary orthonormal basis of $E_\lambda$.
  3. If $E_\lambda = V$ we are done. (In particular, this must always happen if $V$ is $1$-dimensional, proving our base case.)
  4. Otherwise, by strong induction on the dimension of $V$, we can find an orthonormal basis of eigenvectors of $\alpha$ restricted to $E_\lambda^\perp$; together with the orthonormal basis found in step 2, this gives us an orthonormal basis of all of $V$ consisting of eigenvectors of $\alpha$.
  • $\begingroup$ Hi, thank you for the answer. Just to make sure I am understanding the inductive step correctly. In 4 we say that we can split $V$ into $E_\lambda$ and its orthogonal complement. If the complement is empty we are done, else if it is not, then its dimension must be between $1$ and $\dim V -1$ which are all covered by the strong induction. For the rest, that is $E_\lambda$ we know that orthogonal basis exists (of eigen values). So combining the two gives an orthogonal eigen basis for the whole of $V$. Am I interpreting this correctly? $\endgroup$ Mar 12 at 20:53
  • $\begingroup$ That's correct. (For $E_\lambda$, we know that an orthonormal basis exists because any subspace of $V$ has an orthonormal basis - we don't have any further constraints because $E_\lambda$ already consists only of eigenvectors.) $\endgroup$ Mar 12 at 21:00
  • $\begingroup$ Also, for the last part, does the inductive construction allow us to answer it "deductively"? Part (iv) clearly tells us that it is diagonalizable. So we can think of how the matrix acts on each eigen space. I would think that the unitary requirement is some sort of rotation for the eigen basis where as the other matrix is to introduce scaling in each direction, however, I don't know why it would be hermitian. For the backwards argument, I can verify it algebraically $\endgroup$ Mar 12 at 21:15
  • $\begingroup$ Think about the $1$-dimensional case first. Why can every complex number be written as the product of a "unitary complex number" and a "Hermitian complex number"? Then generalize by dealing with each eigenvector in the basis separately. $\endgroup$ Mar 12 at 21:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .