Is every element in $SO(2)$ diagonalizable in $\mathfrak{so}(2)$?
I am trying to prove that if $A\in SO(2)$ then $A$ is normal, to do this I need to prove that $AA^*=A^*A$. I know that $AA^T=Id$ and so $AA^*=Id$, but how can I prove that $A^*A=Id$?
On the other hand I also have the following question:
Since $A\in SO(2)$ is normal, $A$ is diagonalizable, that is, $A=PDP^{-1}$ where $D$ is diagonal, can we say that $P\in \mathfrak{so}(2)$? Why?
$A$ being in $SO(2)$ means (among other things) that the determinant of $A$ is one, so that the product of the eigenvalues of $A$ (which form the non-zero entries of $D$) equals 1. $D$ being in $\mathfrak{so}(2)$ means that the trace of $D$ is zero, which means that the sum of the eigenvalues equals zero.
Intuitively it is clear that the condition that two numbers multiply to 1 does not imply that their sum is zero, so it remains to find a counterexample.
How about rotation over 45 degrees? The corresponding matrix $A$ satisfies $A^8 = I$ so both entries of $D$ must satisfy $\lambda^8 = 1$. This equation has 8 solutions, but the two solutions $\lambda = \frac{1+i}{\sqrt{2}}$ and $\lambda = \frac{1-i}{\sqrt{2}}$ are more interesting as they do not satisfy $\lambda^n = 1$ for any $n < 8$.
My intuition tells me that these two special solutions are the eigenvalues, but leave checking that to you. If this is correct their sum is $\sqrt{2}$ and not 0.