Here is a linear algebra problem from an interview and I have no clue. Matrix C of size n$\times$n is symmetric . Zero is a simple eigenvalue of C. The associated eigenvector is q. For  $\epsilon$>0, the equation $Cx+\epsilon x=d$ in x, where x and d are n-dimensional Column vectors and d is known, has a solution that depends on  $\epsilon$. Call this solution  $x(\epsilon)$. Express $ \lim\limits_{\epsilon \to 0^+} \epsilon x(\epsilon)$ in terms of vectors q and d.
I have some inspirations:

*

*When $\epsilon \to 0$, $\epsilon q$ also comes to $0$. So we have: $C(x+q)+\epsilon (x+q)=d$

*Since C is a symmetrix matrix, we use diagonalization like: $U\Lambda U^Hx+\epsilon x=d$
But to be honest, I have no clue on this and it bothers me a lot. I can't figure out how to use these conditions we have.
Thank you for reading my question!
Could you give me some clues on this problem?

I assume here that $C$ is a real symmetric matrix.
HINT: We assume $q$ is a unit vector. Apply the Spectral Theorem to write
$$C+\epsilon I = \epsilon qq^\top + \sum_{i=1}^{n-1} (\lambda_i + \epsilon) v_iv_i^\top,$$
where $q,v_1,\dots,v_{n-1}$ form an orthonormal basis. Then
$$(C+\epsilon I)^{-1} = \frac1{\epsilon} qq^\top + \sum_{i=1}^{n-1} \frac1{\lambda_i + \epsilon} v_iv_i^\top.$$
Of course, we assume $\epsilon>0$ small enough so that no $\lambda_i+\epsilon$ becomes $0$.
Can you finish now?
