Anti-Commuting $2\times2$ Hermitian Matrices 
Suppose $C$ and $D$ are $n \times n$ Hermitian matrices that anti-commute, $DC = -CD$, with $C^2 = D^2 = 1$.

*

*What are the allowed eigenvalues?

*Show that the traces of $C$ and $D$ must vanish and that $n$ must be even.


I'm trying to do it in the $2\times2$ case. I took a general matrix
$$ \begin{bmatrix} x & a + b i \\ a - b i & d \end{bmatrix} $$
I squared this matrix and set it equal to the identity, and I found that $c = \pm 1$, $d = \mp 1$, and $ a = \pm b i $. I also did it swapping the $a + bi$ and $a - bi$ in the matrix. Again I found the same numbers, just the signs and positions were swapped in the final multiplied out matrix.
However, I cannot seem to find a combination that comes out to $DC = -CD$. Is there a different way I should go about it to find what the general forms of $C$ and $D$ are? I believe once I find this matrix, part 2 will follow trivially since the $c$ and $d$ elements are $\pm 1$, and $\mp 1$, so they will always add to $0$ making the trace $0$. But I need a way to finish part 1. Any tips?
 A: In general
(a) $C^2=I$ and $D^2=I$ are involutions so for any eigenvector
$C^2\mathbf x = \lambda ^2 \mathbf x = I \mathbf x = \mathbf x\implies \lambda\in\{-1,1\}  $
and the same holds with $D$
(b) $DC = -CD$
$\implies C = -DCD$
$\implies \text{trace}\big(C\big)  =-\text{trace}\big(DCD\big)=-\text{trace}\big(C\big)$
$  \implies \text{trace}\big(C\big)=0$
thus the number of eigs = 1 must be the same as the number = -1 and the dimension is even.  (The dimension being even is also implied by the fact that $CD$ is skew hermitian and invertible.)
for the 2 x 2 case
the $2\times 2$ case may be is too small to be useful.  You can assume WLOG that $D$ is diagonal (by effecting a unitary similarity transform if needed).
So the options are (i) $D=I$, (ii) $D=-I$ or (iii) $D$ has a +1 and -1 on the diagonal (and WLOG that $d_{1,1}=-1$ which holds up to permutation similarity).  The first two are out as they commute, so it must be (iii).
$DC =-CD$ implies $c_{i,i}=-c_{i,i}\implies c_{i,i}=0$
so you have
$C =\begin{bmatrix} 
0 & \eta\\
\bar{\eta} & 0 
\end{bmatrix}$
Now consider the determinant:
$\vert \eta\vert^4 = (0-\vert \eta\vert^2)^2 = \det\big(C\big)^2  = \det\big(I\big)=1\implies \eta$ is on the unit circle.
