Let $A\in\mathbb{R}^{n\times n}$ be a real $n\times n $ matrix, and let $\lambda$ be an eigenvalue of $A$. For an arbitrary $b\in\mathbb{R}^n$, does the equation \begin{align} Ax = \lambda x + b \end{align} have a solution $x$ (not necessarily unique)? Is it possible that it has no solutions?
My attempt: We can re-write the equation as $(A - \lambda I) x = b$, which is a linear system. Since $\lambda$ is an eigenvalue of $A$, we know $A - \lambda I$ is singular. On the other hand, when a linear system of equations have a singular matrix coefficient, then it will have either no solution, or an infinite number of solutions. However, I wonder if some other properties of $A - \lambda I$ (i.e., the fact that $\lambda$ is an eigenvalue of $A$) will rule out the no-solution possibility? My goal is to prove that this system always has a solution.