I need to construct a 4x4 hermitian matrix $A={\overline A}^T=A^H$ with the 4 Eigenvalues $\lambda_0, \lambda_0, -\lambda_0, -\lambda_0$. For this, the characteristic polynome is $p(\lambda)=(\lambda-\lambda_0)^2 (\lambda+\lambda_0)^2$ and the trace is $Tr A = 0$
The Eigendecomposition of $A$ is
$A = W \Lambda W^H = \lambda_0 W \left( \matrix{ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 } \right) W^H = \lambda_0 \left( \pmb{w}_0 \cdot \pmb{w}_0^H + \pmb{w}_1 \cdot \pmb{w}_1^H - \pmb{w}_2 \cdot \pmb{w}_2^H - \pmb{w}_3 \cdot \pmb{w}_3^H \right) $
with $W^H W = \hat 1$ and where matrix $W$ contains the Eigenvectors of $A$
We can see, that $W$ contains also the eigenvectors for $A^2 = W \Lambda W^H W \Lambda W^H = \lambda_0^2 \hat 1$ and for higher exponents.
Let us assume, we have to construct another 4x4 hermitian matrix $B$ with the same 4 Eigenvalues $\lambda_0, \lambda_0, -\lambda_0, -\lambda_0$ with
$B = V \Lambda V^H = \lambda_0 V \left( \matrix{ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 } \right) V^H = \lambda_0 \left( \pmb{v}_0 \cdot \pmb{v}_0^H + \pmb{v}_1 \cdot \pmb{v}_1^H - \pmb{v}_2 \cdot \pmb{v}_2^H - \pmb{v}_3 \cdot \pmb{v}_3^H \right) $
with the following requirement
$A B + B A = \hat 0$
My questions:
How do I determine the orthogonal eigenvectors $\pmb{w}_i$, perhaps under consideration of the repeated eigenvalues?
How do I find a set of $\pmb{v}_i$ (another set of $\pmb{w}_i$) to fulfill $A B + B A = \hat 0$