# How to determine eigenvectors of symmetric circulant matrix {{A,B,B},{B,A,B},{B,B,A}}?

I'm trying to find the eigenvectors for the matrix $$\begin{bmatrix} A & B & B \\ B & A & B \\ B & B & A \end{bmatrix}$$ .

I determined the eigenvalues to be $$\lambda_1=\lambda_2=A-B$$ and $$\lambda_3=A+2B$$. I also detrrmined one eigenvector to be $$v_3=\frac{1}{\sqrt{3}}\begin{bmatrix} 1 \\1\\1 \end{bmatrix}$$ .

But I'm having issues with detrrmining the remaining ones. For the first eigenvalue, the matrix used to determine eigenvector coefficients reduces to $$\begin{bmatrix} 1 & 1 & 1 \\ 0& 0 & 0\\ 0& 0 & 0 \end{bmatrix}$$.

From this I managed to extract some conditions for the coefficients, but I don't know how to determine eigenvectors that actually satisfy the eigenvalue equation. Conditions:

$$v_1+v_2+v_3=0$$ $$v_1\neq v_2\neq v_3\neq 0$$ $$v_1^2+v_2^2+v_3^2=1$$

Any help very appreciated.

We can arbitrarily pick values for $$v_1,v_2,v_3$$ that satisfy $$v_1+v_2+v_3=0$$.
In particular we do not need them all to be different from $$0$$. We only need the vector to be different from the zero vector. And we don't need the vector to have length $$1$$ either, although we can normalize afterwards if we want to.
We can pick for instance $$v_1=1$$ and $$v_3=0$$, which means that $$v_2=-1$$. Just now we've found the eigenvector $$(1,-1,0)$$.