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From the definition of the normal matrix using the property $A^\dagger A=AA^\dagger$, where $A^\dagger$ is the conjugate transpose of matrix A, it is obvious that every hermitian and skew-hermitian matrix is also normal, since (skew-)hermitian matrices satisfy $A^\dagger=(-)A$. What I'm wondering is: is there a theorem which explicitly states why normal matrices are not necessarily hermitian, i.e. why doesn't it go the other way around? I'm aware of some counterexamples which prove that some normal matrices are not hermitian, and I don't question that they aren't the same, I just want to know why that's so in a general case.

Also, if for some regular matrices $A,B \in \mathcal{M}_{n\times n}$ we have $(AB)^\dagger=B^\dagger A^\dagger$, does that mean that for a normal matrix $A$ the product $C=A^\dagger A$ is neccesarily hermitian, since $C^\dagger=(A^\dagger A)^\dagger=A^\dagger A^{\dagger \dagger}=A^\dagger A=C$? Is there a way to relate that to the above question?

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  • $\begingroup$ Hello. It might help to clarify the notation. Does $(.)^\dagger$ mean conjugate transpose? $\endgroup$ – mathreadler Oct 11 '15 at 20:38
  • $\begingroup$ @mathreadler yes $\endgroup$ – user223391 Oct 11 '15 at 20:39
  • $\begingroup$ Yes, that's the conjugate transpose, I've edited the question accordingly. Thanks for the remark! $\endgroup$ – Soba noodles Oct 11 '15 at 20:40
  • $\begingroup$ I'm not really sure what you're asking for. If you understand examples of matrices that are normal but not hermitian, what more do you want? Why would you expect normal matrices to be hermitian in the first place? $\endgroup$ – Eric Wofsey Oct 11 '15 at 20:50
  • $\begingroup$ Well, I want to know if there is a property of normal matrices which prevents them from necessarily being hermitian. If we denote the set of hermitian matrices as $\mathcal{H}$ and the set of normal matrices as $\mathcal{N}$, what I want to know is why is it that $\mathcal{H}\subset \mathcal{N}$ instead of $\mathcal{H}\subseteq \mathcal{N}$, in other words, from what property are the counterexamples generated. $\endgroup$ – Soba noodles Oct 11 '15 at 20:59
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The basic example of a normal matrix is any diagonal matrix $A$; then $A^\dagger$ is the diagonal matrix with complex conjugate entries, which commutes with $A$ since any two diagonal matrices commute. Such an $A$ is Hermitian iff all the diagonal entries are real and skew-Hermitian iff all the diagonal entries are purely imaginary. This example demonstrates how Hermitian matrices are very unusual and special among the normal matrices, much as real numbers are very unusual and special among the complex numbers.

(In fact, the spectral theorem says that this is essentially the only example: a matrix is normal iff it is conjugate by a unitary to a diagonal matrix. Similarly, a matrix is Hermitian iff it is conjugate by a unitary to a diagonal matrix with real entries.)

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  • $\begingroup$ Yes, this way the type of answer I was looking for: elaborating why hermitian matrices stand out from normal matrices. Indeed, the spectral theorem provides an elegant way of discerning their nature. Thank you for the insight! $\endgroup$ – Soba noodles Oct 11 '15 at 21:24
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$A^\dagger A$ and $A A^\dagger$ are always hermitian, whether $A$ is normal or not.

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  • $\begingroup$ True, I see now that I haven't used the normal matrix property anywhere so the proof targets a more general class of matrices. Thanks! $\endgroup$ – Soba noodles Oct 11 '15 at 20:54

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