Consider the following 4x4 real symmetric matrix:
$$ M = 2 \, \begin{pmatrix} 1 & c_{xy} & c_{xz} & c_{yz} \\ c_{xy} & 1 & c_{y\overline{z}} & c_{x\overline{z}} \\ c_{xz} & c_{y\overline{z}} & 1 & c_{x\overline{y}}\\ c_{yz} & c_{x\overline{z}} & c_{x\overline{y}} & 1 \end{pmatrix} $$ where shortcuts $c_{xy} = \cos(\frac{x+ y}{4})$ and $c_{x\overline{y}} = \cos(\frac{x - y}{4})$ etc. are introduced.
I know the eigenvalues $\lambda$ of $M$: $$ \lambda = 0, 0, 4\pm 2\sqrt{1+Q} $$ where $Q = (c_{xy}^2 + c_{x\overline{y}}^2 + c_{yz}^2 + c_{y\overline{z}}^2 + c_{xz}^2 + c_{x\overline{z}}^2 - 3)$, i.e. two zeros and two non-zeros.
Question
Now my question is, if I know for a fact that 2 of the eigenvalues are 0, how can I arrive simply at the other two non-zero eigenvalues that I shown above? Note I am interested in the analytical form (I know I can solve these easily with numerical methods).
I'm thinking maybe there is a way to reduce $M$ to a 2x2 matrix block (?)
My attempt
Not much of an attempt, but the usual trick of trace of matrix is the sum of eigenvalues and determinant is the product of eigenvalues doesn't really help here.
I solved for the eigenvalues symbolically using sympy in Python (I believe one can do the same in Mathematica), but this method doesn't scale well (see below context).
Context
The point of asking this is that I have another, more complicated, 4x4 real symmetric matrix filled with trigonometric expressions, that I know has 2 zero eigenvalues (like $M$).
I want to find the analytical form of the other two eigenvalues, but solving for all 4 of the eigenvalues symbolically is too computationally expensive. I was hoping I can make use of the fact that two of the eigenvalues are (known and) zero in some way.
Hence, a useful answer here would describe a method in which I can obtain the two non-zero $\lambda$'s from $M$ -- then hopefully I will be able to generalize it for my own, complicated matrix.
Possible idea. Since you know two eigenvalues are $0$ you know the $0$-eigenspace is $2$ dimensional. If you can (symbolically) see the null space, then its orthogonal complement will be $2$ dimensional and invariant. It will be spanned by the two missing eigenvectors. Finding the eigenvalues should be easier in just two dimensions.