### Instruction: Find all real numbers $k$ for which there exists a nonzero, 2-dimensional vector $\mathbf{v}$ such that
\[\begin{pmatrix} 1 & 8 \\ 2 & 1 \end{pmatrix} \mathbf{v} = k \mathbf{v}.\]Enter all the solutions, separated by commas.### Response: This problem is asking me to find the eigenvalues of the matrix $A = \begin{pmatrix} 1 & 8 \\ 2 & 1 \end{pmatrix}$.
The eigenvalues are the values of $k$ that make the equation $A \mathbf{v} = k \mathbf{v}$ have a nonzero solution for $\mathbf{v}$.
To find them, I need to rearrange the equation to get $(A - kI) \mathbf{v} = \mathbf{0}$, where $I$ is the identity matrix.
Then I need to find the values of $k$ that make the matrix $A - kI$ singular, meaning its determinant is zero.
The determinant of $A - kI$ is \[\begin{vmatrix} 1 - k & 8 \\ 2 & 1 - k \end{vmatrix} = (1 - k)^2 - 16.\]
To make this zero, I need to solve the quadratic equation $(1 - k)^2 - 16 = 0$.
By factoring, I get $(1 - k - 4)(1 - k + 4) = 0$.
So the solutions are $k = 1 - 4 = -3$ or $k = 1 + 4 = 5$.