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
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 $\sin(\theta) = 0$.