# normal operator over Field R

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

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 $$\sin(\theta) = 0$$.
• On the other hand, it's normal, as it's diagonalizable over $\Bbb C$. Jun 21 '20 at 17:47
• 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 ? Jun 22 '20 at 5:07