How to evaluate $\lim_{\epsilon \rightarrow 0^+} \epsilon x(\epsilon)$ where $x(\epsilon)$ is the solution of $\mathbf{A}x + \epsilon x = v$

Question

This is an excercise for matrix computations. Let $\mathbf{A} \in \mathbb{R}^{n \times n}$ be a symmetric matrix with $0$ as a simple eigenvalue, and given $\mathbf{v} \in \mathbb{R}^n$. First prove that for sufficiently small positive $\epsilon$ ($\epsilon$ is smaller than all the absolute value of nonzero eigenvalues), the equation $\mathbf{A}\mathbf{x} + \epsilon \mathbf{x} = \mathbf{v}$ has a unique solution $\mathbf{x} = \mathbf{x}(\epsilon)$. Then evaluate $\lim_{\epsilon \rightarrow 0^+} \epsilon \mathbf{x}(\epsilon)$ in terms of $\mathbf{v}$ and the eigenvectors of $\mathbf{A}$.

I already showed that the solution of the equation $\mathbf{A}\mathbf{x} + \epsilon \mathbf{x} = \mathbf{v}$ is unique. Since $0$ is the simple eigenvalue of $\mathbf{A}$, we have $\det(\mathbf{A}) = 0$. Through the characteristic polynomial of $\mathbf{A}$ we can then derive $\det(\mathbf{A} + \epsilon \mathbf{I}) \neq 0$, which shows uniqueness of the solution of this equation.
But for the limit evaluation, I have no idea about how the eigenvectors are related to this. This is an exercise marked with diagonalization, but I don't know how they are related. Maybe there's some properties between the inner products of the eigenvectors?
Thanks!

Answer 1

I would do something like this. $\epsilon x = \epsilon(A+\epsilon I)^{-1}v\implies \epsilon x = \epsilon Q(\Lambda + \epsilon I)^{-1} Q^Tv = Q(\epsilon^{-1}\Lambda + I)^{-1}Q^Tv$ so your desired limit equals $QQ^Tv$, where $A = Q\Lambda Q^T$ is the eigen decomposition of your symmetric matrix $A.$

Note that you need to fill in the gaps here and there to make the proof rigorous.