Activity 4.2.2
The eigenvalues of a square matrix are defined by the condition that there be a nonzero solution to the homogeneous equation \((A\lambda I)\vvec=\zerovec\text{.}\)
If there is a nonzero solution to the homogeneous equation \((A\lambda I)\vvec = \zerovec\text{,}\) what can we conclude about the invertibility of the matrix \(A\lambda I\text{?}\)
If there is a nonzero solution to the homogeneous equation \((A\lambda I)\vvec = \zerovec\text{,}\) what can we conclude about the determinant \(\det(A\lambda I)\text{?}\)

Let's consider the matrix
\begin{equation*} A = \left[\begin{array}{rr} 1 \amp 2 \\ 2 \amp 1 \\ \end{array}\right] \end{equation*}from which we construct
\begin{equation*} A\lambda I = \left[\begin{array}{rr} 1 \amp 2 \\ 2 \amp 1 \\ \end{array}\right]  \lambda \left[\begin{array}{rr} 1 \amp 0 \\ 0 \amp 1 \\ \end{array}\right] = \left[\begin{array}{rr} 1\lambda \amp 2 \\ 2 \amp 1\lambda \\ \end{array}\right]\text{.} \end{equation*}Find the determinant \(\det(A\lambda I)\text{.}\) What kind of equation do you obtain when we set this determinant to zero to obtain \(\det(A\lambda I) = 0\text{?}\)
Use the determinant you found in the previous part to find the eigenvalues \(\lambda\) by solving \(\det(A\lambda I) = 0\text{.}\) We considered this matrix in the previous section so we should find the same eigenvalues for \(A\) that we found by reasoning geometrically there.
Consider the matrix \(A = \left[\begin{array}{rr} 2 \amp 1 \\ 0 \amp 2 \\ \end{array}\right]\) and find its eigenvalues by solving the equation \(\det(A\lambda I) = 0\text{.}\)
Consider the matrix \(A = \left[\begin{array}{rr} 0 \amp 1 \\ 1 \amp 0 \\ \end{array}\right]\) and find its eigenvalues by solving the equation \(\det(A\lambda I) = 0\text{.}\)
Find the eigenvalues of the triangular matrix \(\left[\begin{array}{rrr} 3 \amp 1 \amp 4 \\ 0 \amp 2 \amp 3 \\ 0 \amp 0 \amp 1 \\ \end{array}\right] \text{.}\) What is generally true about the eigenvalues of a triangular matrix?