Activity 10.3.1.
Calculate the eigenvalue and eigenvectors of this matrix. (Give the solutions in exact values).
\begin{equation*}
\left( \begin{matrix}
-2 \amp 0 \amp 1 \\
0 \amp -2 \amp 1 \\
1 \amp 1 \amp 0
\end{matrix} \right)
\end{equation*}
Solution.
I follow the algorithm for eigenvalue and eigenvectors. The first step is writing the matrix \(A - \lambda \Id\text{.}\)
\begin{equation*}
\left( \begin{matrix}
-2 - \lambda \amp 0 \amp 1 \\
0 \amp -2 - \lambda \amp 1 \\
1 \amp 1 \amp - \lambda
\end{matrix} \right)
\end{equation*}
Then I calculate the determinant of this matrix. I use cofactor expansion on the first row.
\begin{align*}
\amp \left| \begin{matrix}
-2 - \lambda \amp 0 \amp 1 \\
0 \amp -2 - \lambda \amp 1 \\
1 \amp 1 \amp - \lambda
\end{matrix} \right| \\
\amp =
(1)(-2 - \lambda) \left| \begin{matrix}
-2 - \lambda \amp 1 \\
1 \amp -\lambda
\end{matrix} \right| +
(-1)(0) \left| \begin{matrix}
0 \amp 1 \\
1 \amp -\lambda
\end{matrix} \right| +
(1)(1) \left| \begin{matrix}
0 \amp -2 - \lambda \\
1 \amp 1
\end{matrix} \right| \\
\amp = (-2 - \lambda) \left[ (-2 - \lambda)(-\lambda) -
1 \right] + 0 + \left[ (0)(1) - (-2 - \lambda)(1)
\right] \\
\amp = (-2 - \lambda)(\lambda^2 + 2 \lambda - 1) + 2 +
\lambda \\
\amp = -\lambda^3 - 2\lambda^2 - 2\lambda^2 - 4\lambda +
\lambda + 2 + 2 + \lambda \\
\amp = -\lambda^3 - 4\lambda^2 - 2\lambda + 4
\end{align*}
By inspection or by computer, I can see that \(\lambda =
-2\) is a root, so this polynomial factors as follows.
\begin{equation*}
-\lambda^3 - 4\lambda^2 - 2\lambda + 4 = -(\lambda +
2)(\lambda^2 + 2 \lambda - 2)
\end{equation*}
The quadratic piece does not factor into integer roots. The quadratic formula gives the other roots as \(\lambda = -1
\pm \sqrt{3}\text{.}\) Now I calculate the kernel of each \(A -
\lambda Id\) to find the matching eigenvectors. I’ll start with \(\lambda = -2\text{.}\)
\begin{align*}
\amp \left( \begin{array}{ccc|c}
0 \amp 0 \amp 1 \amp 0 \\
0 \amp 0 \amp 1 \amp 0 \\
1 \amp 1 \amp 2 \amp 0
\end{array} \right)
\end{align*}
I row reduce to find the kernel.
\begin{align*}
\amp \left( \begin{array}{ccc|c}
1 \amp 1 \amp 0 \amp 0 \\
0 \amp 0 \amp 1 \amp 0 \\
0 \amp 0 \amp 0 \amp 0
\end{array} \right)
\end{align*}
To show the kernel, I need to express the solution indicated by this matrix in terms of parameters. \(y\) is free and I can interpret the row of the matrix as expressions for \(x\) and \(z\) in terms of \(y\text{.}\) (Though, in this matrix, the second line simply indicatres that \(z =
0\text{.}\)) By choosing a specific value of \(y\text{,}\) I can get specific values of \(x\text{,}\) and thus a specific eigenvector. I choose \(y = -1\) for this particular system.
\begin{align*}
\amp \Ker \left( \begin{matrix}
0 \amp 0 \amp 1 \\
0 \amp 0 \amp 1 \\
1 \amp 1 \amp 2
\end{matrix} \right) = \Span \left\{ \left( \begin{matrix}
1 \\ -1 \\ 0
\end{matrix} \right) \right\}
\end{align*}
That gives me an eigenvector for \(\lambda = -2\text{.}\) Next, I’ll look at \(\lambda = -1 + \sqrt{3}\text{.}\)
\begin{align*}
\amp \left( \begin{array}{ccc|c}
-1 - \sqrt{3} \amp 0 \amp 1 \amp 0 \\
0 \amp -1 - \sqrt{3} \amp 1 \amp 0 \\
1 \amp 1 \amp 1 - \sqrt{3} \amp 0
\end{array} \right)
\end{align*}
I row reduce to find the kernel.
\begin{align*}
\amp \left( \begin{array}{ccc|c}
1 \amp 0 \amp \frac{-1}{1 + \sqrt{3}} \amp 0 \\
0 \amp 1 \amp \frac{-1}{1 + \sqrt{3}} \amp 0 \\
0 \amp 0 \amp 0 \amp 0
\end{array} \right)
\end{align*}
To show the kernel, I need to express the solution indicated by this matrix in terms of parameters. \(z\) is free and I can interpret the row of the matrix as expressions for \(x\) and \(y\) in terms of \(z\text{.}\) By choosing a specific value of \(z\text{,}\) I can get specific values of \(x\) and \(y\text{,}\) and thus a specific eigenvector. I choose \(z = 1 + \sqrt{3}\) for this particular system, to clear the denominator.
\begin{align*}
\amp \Ker \left( \begin{matrix}
-1 - \sqrt{3} \amp 0 \amp 1 \\
0 \amp -1 - \sqrt{3} \amp 1 \\
1 \amp 1 \amp 1 - \sqrt{3}
\end{matrix} \right) = \Span \left\{ \left( \begin{matrix}
1 \\ 1 \\ 1 + \sqrt{3}
\end{matrix} \right) \right\}
\end{align*}
That gives me an eigenvector for \(\lambda = -1 + \sqrt{3}\text{.}\) Next, I’ll look at \(\lambda = -1 - \sqrt{3}\text{.}\)
\begin{align*}
\amp \left( \begin{array}{ccc|c}
-1 + \sqrt{3} \amp 0 \amp 1 \amp 0 \\
0 \amp -1 + \sqrt{3} \amp 1 \amp 0 \\
1 \amp 1 \amp 1 + \sqrt{3} \amp 0
\end{array} \right)
\end{align*}
I row reduce to find the kernel.
\begin{align*}
\amp \left( \begin{array}{ccc|c}
1 \amp 0 \amp \frac{-1}{1 - \sqrt{3}} \amp 0 \\
0 \amp 1 \amp \frac{-1}{1 - \sqrt{3}} \amp 0 \\
0 \amp 0 \amp 0 \amp 0
\end{array} \right)
\end{align*}
To show the kernel, I need to express the solution indicated by this matrix in terms of parameters. \(z\) is free and I can interpret the row of the matrix as expressions for \(x\) and \(y\) in terms of \(z\text{.}\) By choosing a specific value of \(z\text{,}\) I can get specific values of \(x\) and \(y\text{,}\) and thus a specific eigenvector. I choose \(z = -1 - \sqrt{3}\) for this particular system, to clear the denominator.
\begin{align*}
\amp \Ker \left( \begin{matrix}
-1 + \sqrt{3} \amp 0 \amp 1 \\
0 \amp -1 + \sqrt{3} \amp 1 \\
1 \amp 1 \amp 1 + \sqrt{3}
\end{matrix} \right) = \Span \left\{ \left( \begin{matrix}
1 \\ 1 \\ 1 - \sqrt{3}
\end{matrix} \right) \right\}
\end{align*}
The spanning vector is an eigenvector for \(\lambda = -1 -
\sqrt{3}\text{.}\)
