### 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 that this problem is asking me to find the eigenvalues of the given matrix, which are the scalars $k$ that satisfy this equation for some nonzero vector $\mathbf{v}$.
To find the eigenvalues, I need to solve the characteristic equation of the matrix, which is
\[\det \begin{pmatrix} 1-k & 8 \\ 2 & 1-k \end{pmatrix} = 0.\]
Expanding the determinant, I get
\[(1-k)(1-k) - 16 = 0.\]
Simplifying and rearranging, I get
\[k^2 - 2k - 15 = 0.\]
To solve this quadratic equation, I can either use the quadratic formula or factor it.
I notice that it can be factored as
\[(k-5)(k+3) = 0.\]
Therefore, the eigenvalues are $k=5$ and $k=-3$.
These are the only real numbers that satisfy the original equation.