What can be said about the solution to $Ax=b$ when $b$ is an eigenvector of $A$?

Question

If I have some eigenvector $b$ of some matrix $A$, which entails $Ab=\lambda b$, what can I say about the solution, $x$, to $Ax=b$?

My first solution, which seems very weak, is

$$ Ab = \lambda b, Ax = b $$ $$ AAx = \lambda b $$ $$ A^2 x = \lambda b $$

Since we know that eigenvectors of $A$ are also eigenvectors of $A^2, A^{-1}$ (or any function of $A$), and $b$ in the last line above is still an eigenvector of $A$, then $x = b$?

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But I am pretty sure this fails because we don't know if $\lambda$ is an eigenvalue of $A^2$? Though, $\lambda ^2$ is an eigenvalue of $A^2$.. But I don't see how I can tie this in to fix the proposed solution above. I really feel like I need to invoke $det(A - \lambda I) x = 0$ for this to ever work out..

Answer 1

If $\lambda \neq 0$, then rearranging gives $$A \cdot \frac{1}{\lambda} b = \frac{1}{\lambda} A b = \frac{1}{\lambda} (\lambda b) = b,$$ so $x = \frac{1}{\lambda} b$ is a solution of $A x = b$. We can recover the full set of solutions as usual by adding solutions of the corresponding homogeneous equation, $A x = 0$. In particular, if $A$ is nondegenerate, the above solution is unique.

If $\lambda = 0$, there may or may not be a solution. If $0$ is not a defective eigenvalue (i.e., if its geometric and algebraic multiplicities agree), $A x = b$ has a solution only for $b = 0$. If $0$ is defective, then there will be a solution for some nonzero $b$ but not others. For example, take $A = \pmatrix{0&1\\0&0}$ and analyze separately the cases $b = \pmatrix{1\\0}$ and $b = \pmatrix{0\\1}$.

Answer 2

Since $Ab = \lambda b$ one solution to $Ax = b$ is $x = \frac 1 \lambda b$.

If $x$ is any other solution then $A(x- \frac 1 \lambda b ) = 0$ so that $x- \frac 1 \lambda b$ belongs to the null space of $A$.

Observe $x = \frac 1 \lambda b + (x- \frac 1 \lambda b)$.

Thus any solution $x$ to $Ax = b$ has the form $x = \frac 1 \lambda y + n$, where $n \in N(A)$.