I have been asked to verify whether $v = \begin{bmatrix}1\\4\end{bmatrix}$ is an eigenvector of $A = \begin{bmatrix}-3&1\\-3&8\end{bmatrix}$? If yes, find the eigenvalue.
The way that I approached this question is to find eigenvalues, then use eigenvalues to verify whether $v$ is an eigenvector of the matrix. Here is how I find the eigenvalues:
$$\begin{align*} &det(A - \lambda I)\\ &=det(\begin{bmatrix}-3&1\\-3&8\end{bmatrix} - \begin{bmatrix} \lambda & 0 \\ 0 & \lambda \end{bmatrix})\\ &=det(\begin{bmatrix} -3-\lambda & 1 \\ -3 & 8 - \lambda \end{bmatrix})\\ &=(-3-\lambda)(8-\lambda) - (-3)\\ \therefore \lambda &= \frac{1}{2} (5 \pm \sqrt{109}) \end{align*} $$ To verify: $$ \begin{align*} Av = \lambda v\\ \begin{bmatrix}-3&1\\-3&8\end{bmatrix} \begin{bmatrix}1\\4\end{bmatrix} = \begin{bmatrix}1\\4\end{bmatrix}\\ \begin{bmatrix}1\\29\end{bmatrix} = \frac{1}{2} (5 \pm \sqrt{109}) \begin{bmatrix}1\\ 4 \end{bmatrix} \end{align*} $$
I am not sure this is the right approach. Appreciate your comment, insight, etc.