# 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.

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 $A$.

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.

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 $A$.