Verify a vector is an eigenvector of a matrix 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.
 A: This is the long way!
Recall that $v$ is an eigenvector of $A$ if $Av=\lambda v$ for some $\lambda$.
Here we have
$$
\begin{bmatrix}
-3 & 1 \\ -3 & 8
\end{bmatrix}
\begin{bmatrix}
1\\ 4
\end{bmatrix}=
\begin{bmatrix}
1\\ 29
\end{bmatrix}
$$
But is $\begin{bmatrix}
1\\ 29
\end{bmatrix}$ a scalar multiple of $\begin{bmatrix}
1\\ 4
\end{bmatrix}$?

 No! So $\begin{bmatrix}1\\ 4\end{bmatrix}$ is not an eigenvector of $A$.

A: You should just multiply the matrix with the vector and then see if the result is a multiple of the original vector. The definition of eigenvector is
$AX = \lambda X$.
In this case, the vector is not an eigenvector, as the product is $\; \binom 1{29}\; $ which is not a multiple of the original vector. If you multiply and find that you get a multiple of the original vector, then the eigenvalue is the multiple.
A: if $v$ is an eigenvector of $A$, then we have $Av=\lambda v$. In your case,
$$Av=\begin{bmatrix}-3&1\\-3&8\end{bmatrix}\begin{bmatrix}1\\4\end{bmatrix}\\=\begin{bmatrix}1\\29\end{bmatrix}$$
there is not a $\lambda$ that satisfies $Av=\lambda v$, so it is not an eigenvector of $A$.
