### 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 as an eigenvalue problem, where $k$ is an eigenvalue of the matrix and $\mathbf{v}$ is a corresponding eigenvector.
To find the eigenvalues of a matrix, I need to solve the equation $\det(A - kI) = 0$, where $A$ is the matrix and $I$ is the identity matrix of the same size.
In this case, $A = \begin{pmatrix} 1 & 8 \\ 2 & 1 \end{pmatrix}$ and $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$, so $A - kI = \begin{pmatrix} 1 - k & 8 \\ 2 & 1 - k \end{pmatrix}$.
To find the determinant of a 2 by 2 matrix, I multiply the diagonal entries and subtract the product of the off-diagonal entries.
So $\det(A - kI) = (1 - k)^2 - 16$.
Setting this equal to zero, I get $(1 - k)^2 - 16 = 0$, which I can rewrite as $(1 - k)^2 = 16$.
Taking the square root of both sides, I get $1 - k = \pm 4$.
Solving for $k$, I get $k = 1 \pm 4$, which means $k = 5$ or $k = -3$.
These are the only two possible values of $k$ for which there exists a nonzero vector $\mathbf{v}$ such that $A \mathbf{v} = k \mathbf{v}$.