Given one eigenvalue and eigenvector pair, how to find the possible other one

Question

I have an assignment problem:

If $A$ is a $2\times 2$ matrix $$A = \begin{bmatrix}a & 2a \\ 2b & -b\end{bmatrix}$$ with an eigenvalue $\lambda = 2$ and corresponding eigenvector $\begin{bmatrix}2 \\ 3\end{bmatrix}$, which of the following vector(s), if any, could be an eigenvector for $A$?

1. $\begin{bmatrix}-8 \\ 1\end{bmatrix}$

2. $\begin{bmatrix}8 \\ -1\end{bmatrix}$

3. $\begin{bmatrix}1 \\ -8\end{bmatrix}$

4. $\begin{bmatrix}1\\ 8\end{bmatrix}$

Here is what I have tried so far:

I know that I need to solve $\det (A – \lambda I) = 0$ to find the eigenvalues $\lambda$. When I do it for $A$, I get $$(a-\lambda) (-b -\lambda) – 4ab = 0$$

$$(a-\lambda) (b +\lambda) + 4ab = 0$$
$$-\lambda^2 +a\lambda – b\lambda+5ab = 0$$

And I know that $\lambda = 2$ is one solution, so $(\lambda – 2)$ must be a factor, and also $$2a -2b + 5ab = 4$$

I don't know what to do next.

Thanks for help.

Answer 1

We have:

$$\begin{bmatrix} a-2 & 2a \\ 2b & -b-2 \end{bmatrix} \begin{bmatrix} 2 \\ 3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$

which gives:

$$2(a-2)+6a = 0$$ $$4b-3(b+2) = 0$$