Question 6.1.5: Find the eigenvalues and a basis for each eigenspace of A =[...
Find the eigenvalues and a basis for each eigenspace of
A=\left[\begin{array}{rrr}1 & -3 & 3 \\2 & -2 & 2 \\2 & 0 & 0\end{array}\right]
We determine the eigenvalues of A by calculating the characteristic polynomial,
\operatorname{det}\left(A-\lambda I_3\right)=\left|\begin{array}{ccc}(1-\lambda) & -3 & 3 \\2 & (-2-\lambda) & 2 \\2 & 0 & (0-\lambda)\end{array}\right|=-\lambda^3-\lambda^2+2 \lambda=-\lambda(\lambda+2)(\lambda-1)
Hence the eigenvalues are \lambda_1=-2, \lambda_2=0, and \lambda_3=1.aking them in order, the eigenvectors associated with \lambda_1=-2 are found by solving the homogeneous system \left(A+2 I_3\right) u=0. The augmented matrix and echelon form are
\left[\begin{array}{rrrr}3 & -3 & 3 & 0 \\2 & 0 & 2 & 0 \\2 & 0 & 2 & 0\end{array}\right] \underset{\substack{\sim}}{\begin{aligned}R_1 \Leftrightarrow R_2 \\-\frac{3}{2}R_1+R_2 \Rightarrow R_2 \\-R_1+R_3 \Rightarrow R_3\end{aligned}}\left[\begin{array}{rrrr}2 & 0 & 2 & 0 \\0 & -3 & 0 & 0 \\0 & 0 & 0 & 0\end{array}\right]
Back substitution gives us
General Solution: u _1=s\left[\begin{array}{r}1 \\0 \\-1\end{array}\right] ⇒ Basis for Eigenspace of \lambda_1=-2:\left\{\left[\begin{array}{r}1 \\0 \\-1\end{array}\right]\right\}
For λ_2 = 0, the homogeneous system is (A−0· I_2)u = Au = 0. The augmented matrix and echelon form are
\left[\begin{array}{crrr}1 & -3 & 3 & 0 \\2 & -2 & 2 & 0 \\2 & 0 & 0 & 0\end{array}\right] \underset{\substack{\sim}}{\begin{aligned}-2 R_1+R_2 \Rightarrow R_2 \\-2 R_1+R_3 \Rightarrow R_3 \\-\frac{3}{2}R_2+R_3 \Rightarrow R_3\end{aligned}}\left[\begin{array}{rrrr}1 & -3 & 3 & 0 \\0 & 4 & -4 & 0 \\0 & 0 & 0 & 0\end{array}\right]
This time back substitution yields
General Solution: u _2=s\left[\begin{array}{l}0 \\1 \\1\end{array}\right] ⇒ Basis for Eigenspace of \lambda_2=0:\left\{\left[\begin{array}{l}0 \\1 \\1\end{array}\right]\right\}
The final eigenvalue is λ_3 = 1. The homogeneous system is (A − I_3)u = 0, and the augmented matrix and echelon form are
\left[\begin{array}{rrrr}0 & -3 & 3 & 0 \\2 & -3 & 2 & 0 \\2 & 0 & -1 & 0\end{array}\right] \underset{\substack{\sim}}{\begin{aligned}R_1 \Leftrightarrow R_3 \\-R_1+R_2 \Rightarrow R_2 \\-R_2+R_3 \Rightarrow R_3\end{aligned}}\left[\begin{array}{rrrr}2 & 0 & -1 & 0 \\0 & -3 & 3 & 0 \\0 & 0 & 0 & 0\end{array}\right]
With back substitution, we find
General Solution: u _3=s\left[\begin{array}{l}1 \\2 \\2\end{array}\right] ⇒ Basis for Eigenspace of \lambda_3=1:\left\{\left[\begin{array}{l}1 \\2 \\2\end{array}\right]\right\}