We need to solve the equation
\(A\mathbf{v} = \lambda \mathbf{v}\text{,}\) where both
\(\lambda\) and
\(\mathbf{v} \neq 0\) are unknown.
The equation is equivalent to the equation
\((\lambda I - A) \mathbf{v} = \mathbf{0}\text{.}\) For each
\(\lambda\text{,}\) in order for this equation to have a nontrivial solution
\(\mathbf{v}\text{,}\) the matrix
\(\lambda I - A\) must be singular, i.e.,
\(\det(\lambda I - A) = 0\text{.}\) This allows us to temporarily eliminate
\(\mathbf{v}\) from the problem, determining the values of
\(\lambda\) that have solutions
\(\mathbf{v}\text{.}\)
We calculate that
\begin{align*}
\det(\lambda I - A) \amp = \det\begin{bmatrix} \lambda - 3 \amp 0 \\ 1 \amp 4 + \lambda \end{bmatrix}\\
\amp = (\lambda - 3)(\lambda + 4)
\end{align*}
Thus \(\det(\lambda I - A) = 0\) holds precisely when \((\lambda - 3)(\lambda + 4) = 0\text{,}\) i.e., when \(\lambda = 3\) or \(\lambda = -4\text{,}\) and so these two values are the two eigenvalues of \(A\text{.}\) It remains to find the eigenvectors for each of the eigenvalues.
We now find the nontrivial solutions to the equation
\((A - \lambda I) \mathbf{v} = \mathbf{0}\) for
\(\lambda = 3\) and
\(\lambda = -4\text{.}\) Each of these is a linear system that can be solved, either using the methods of
Unit 2, or more ad hoc methods, which may be faster since the matrices we are dealing with are small.
For \(\lambda = 3\text{,}\) we calculate that
\begin{equation*}
A - 3I = \begin{bmatrix} 0 \amp 0 \\ -1 \amp -7 \end{bmatrix}\text{.}
\end{equation*}
Thus the solutions to \((A - 3I) \mathbf{v} = 0\) are given by \(-v_1 - 7v_2 = 0\text{,}\) or \(v_1 = -7v_2\text{.}\) Thus the solutions to this equation are given by nonzero multiples of the basic solution
\begin{equation*}
\begin{bmatrix} -7 \\ 1 \end{bmatrix}\text{.}
\end{equation*}
These are all eigenvectors of \(A\) with eigenvalue \(3\text{.}\)
For \(\lambda = -4\text{,}\) we calculate that
\begin{equation*}
A + 4 I = \begin{bmatrix} 7 \amp 0 \\ -1 \amp 0 \end{bmatrix}\text{.}
\end{equation*}
Thus the solutions to \((A + 4I) \mathbf{v} = 0\) are those vectors \(\mathbf{v}\) such that \(v_1 = 0\text{,}\) and so the eigenvectors for \(A\) with eigenvalue \(-4\) are given by nonzero multiples of the basic solution
\begin{equation*}
\begin{bmatrix} 0 \\ 1 \end{bmatrix}\text{.}
\end{equation*}
To conclude, the two eigenvalues of the matrix \(A\) are \(\lambda = 3\) and \(\lambda = -4\text{.}\) The set of all eigenvectors of \(A\) associated with the eigenvalue \(\lambda = 3\) is given by the expression
\begin{equation*}
\left\{ \begin{bmatrix} -7t \\ t \end{bmatrix} : t \neq 0 \right\}\text{,}
\end{equation*}
and the set of eigenvectors of \(A\) associated with the eigenvalue \(\lambda = -4\) is given by
\begin{equation*}
\left\{ \begin{bmatrix} 0 \\ t \end{bmatrix} : t \neq 0 \right\}\text{.}
\end{equation*}
