### 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: I recognize this problem as asking for the eigenvalues of the matrix $\begin{pmatrix} 1 & 8 \\ 2 & 1 \end{pmatrix}$.
To find the eigenvalues, I need to solve the equation $\begin{pmatrix} 1 & 8 \\ 2 & 1 \end{pmatrix} \mathbf{v} = k \mathbf{v}$ for $k$ and $\mathbf{v}$.
I can rewrite this equation as $\begin{pmatrix} 1 & 8 \\ 2 & 1 \end{pmatrix} \mathbf{v} - k \mathbf{v} = \mathbf{0}$.
Then, I can factor out $\mathbf{v}$ and get $(\begin{pmatrix} 1 & 8 \\ 2 & 1 \end{pmatrix} - k \mathbf{I}) \mathbf{v} = \mathbf{0}$.
Here, $\mathbf{I}$ is the 2x2 identity matrix.
Now, for this equation to have a nonzero solution for $\mathbf{v}$, the matrix $\begin{pmatrix} 1 & 8 \\ 2 & 1 \end{pmatrix} - k \mathbf{I}$ must be singular, which means its determinant must be zero.
So, I need to solve the equation $\det (\begin{pmatrix} 1 & 8 \\ 2 & 1 \end{pmatrix} - k \mathbf{I}) = 0$.
To find the determinant of a 2x2 matrix, I multiply the diagonal entries and subtract the product of the off-diagonal entries.
So, I get $(1-k)(1-k) - 16 = 0$.
Expanding and simplifying, I get $k^2 - 2k - 15 = 0$.
To solve for $k$, I can factor this quadratic equation as $(k-5)(k+3) = 0$.
So, the solutions are $k=5$ and $k=-3$.