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in class, we learned that over the complex numbers or the real numbers field, a normal operator is diagnosable, I got the idea why a normal operator is diagnosable over Complex field, but I didn't got the idea why a normal operator is diagonalizable over the real number field?

in the real number field, we can get that the characteristic polynomial is not a product of linear equations. why why normal operator is diagonalizable over the real number field ?

can we say that if a normal operator is diagonalizable over the real number field, then the operator is actually self-adjoint? Thank you

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I assume by "diagnosable" you mean "diagonalizable".

It isn't. For example, the $2 \times 2$ rotation matrix $$ \pmatrix{\cos(\theta) & \sin(\theta)\cr -\sin(\theta) & \cos(\theta)}$$ is normal. It has eigenvalues $\exp(\pm i \theta)$, and is thus not diagonalizable over the reals unless $\sin(\theta) = 0$.

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  • $\begingroup$ On the other hand, it's normal, as it's diagonalizable over $\Bbb C$. $\endgroup$
    – Berci
    Jun 21 '20 at 17:47
  • $\begingroup$ What does "normal" mean over an arbitrary field? $\endgroup$ Jun 21 '20 at 18:28
  • $\begingroup$ T : V ---> V is a linear operator over finite dimensional inner product space over Field F (some field ) and T is a normal operator . can I deduce that T is diagonalizable ? if yes why ? $\endgroup$
    – user14732
    Jun 21 '20 at 18:58
  • $\begingroup$ I asked you to define "normal operator" over an arbitrary field. $\endgroup$ Jun 22 '20 at 0:23
  • $\begingroup$ I mean by" normal" operator a linear operator such that $T T^* = T^*T $, and our vector field is finite dimensional and is over some artistry Field (not some specified field)? is that now more clear ? $\endgroup$
    – user14732
    Jun 22 '20 at 5:07

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