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In Serge Lang's Linear Algebra this problem was posed in Ch VIII, Section 6:

Let $V$ be a finite dimensional complex vector space equipped with a positive definite hermitian product.

An operator $A:V$$\rightarrow$$V$ is said to be normal if $AA^*=A^*A$.

(a) Let $A$,$B$ be normal operators such that $AB=BA$. Show that $AB$ is normal.

I tried at first to check whether or not $AB$ and $(AB)^*$ commute and thus would know by definition whether or not it is normal, but that didn't work out.

I tried to work on their matrix representations in an orthogonal basis but Spectral Theorem doesn't work out here as $A$ and $B$ are not necessarily Hermitian. When I looked for the solution, he said that by an exercise from the previous section we can find a basis of $V$ formed of eigenvectors of each operator.

The problem is that in that exercise it was given that $A$ and $B$ are real symmetric operators and thus one could use spectral theorem for the real case. That is not available here. This is the question he referred to:

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How would this imply the existence of a basis of $V$ formed of eigenvectors of $A$,$B$,$A^*$ and $B^*$? Any help is appreciated!

  • There is a spectral theorem for normal operators, perhaps this is what the solution refers to – Ben Grossmann Mar 23 '21 at 15:28
  • @BenGrossmann In fact, the author asks in the second part of the same problem to prove spectral theorem for normal operators. –  Mar 23 '21 at 15:34
  • If we could use Fuglede's theorem, then there would be a way forward – Ben Grossmann Mar 23 '21 at 15:43
  • @BenGrossmann Honestly I am not very familiar with notion of bounded operators on a complex Hilbert space and thus not familiar with such a theorem. I wonder if there's any other approach that uses more elementary methods. –  Mar 23 '21 at 15:47
  • The only point I'm making here is that we could technically show that $AB = BA \implies AB^* = B^*A$, and then it would be possible to show that $AB$ is normal. However, I don't know any "elementary" way of doing this; the nicest proof I could think of in this context uses simultaneous upper triangularization – Ben Grossmann Mar 23 '21 at 15:49
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    See this post for details on that problem – Ben Grossmann Mar 23 '21 at 15:50
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    as far as elementary methods go, to get Fuglede's theorem, I'd suggest computing the Frobenius norm of the (commutator) $AB^-B^A$ (i.e. of the associated matrices with respect to a basis that is orthonormal under the hermitian form). I did that here https://math.stackexchange.com/questions/3611846/fugledes-theorem-in-finite-dimensional-vector-space – user8675309 Mar 24 '21 at 19:42

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