Question 6.1.7: 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}0 & 2 & -1 \\1 & -1 & 0 \\1 & -2 & 0\end{array}\right]
The characteristic polynomial of A is
\operatorname{det}\left(A-\lambda I_3\right)=\left|\begin{array}{rrr}-\lambda & 2 & -1 \\1 & (-1-\lambda) & 0 \\1 & -2 & -\lambda\end{array}\right|=-\lambda^3-\lambda^2+\lambda+1=-(\lambda-1)(\lambda+1)^2
Thus A has two distinct eigenvalues,\lambda_1=-1 (multiplicity 2) and \lambda_2=1 (multiplicity 1).
To find the eigenvectors associated with \lambda_1=-1, we solve the homogeneous system (A + I_3)u = 0. The augmented matrix and echelon form are
\left[\begin{array}{rrrr}1 & 2 & -1 & 0 \\1 & 0 & 0 & 0 \\1 & -2 & 1 & 0\end{array}\right] \underset{\substack{\sim}}{\begin{aligned}-R_1+R_2 \Rightarrow R_2 \\-R_1+R_3 \Rightarrow R_3 \\-2 R_2+R_3 \Rightarrow R_3\end{aligned}}\left[\begin{array}{rrrr}1 & 2 & -1 & 0 \\0 & -2 & 1 & 0 \\0 & 0 & 0 & 0\end{array}\right]
Back substitution produces
General Solution: u _1=s\left[\begin{array}{l}0 \\1 \\2\end{array}\right] ⇒ Basis for Eigenspace of \lambda_1=-1:\left\{\left[\begin{array}{l}0 \\1 \\2\end{array}\right]\right\}
Note that although the eigenvalue λ_1 = −1 has multiplicity 2, the eigenspace has dimension 1.
For λ_2 = 1, the augmented matrix for the system (A− I_3)u = 0 and echelon form are
\left[\begin{array}{rrrr}-1 & 2 & -1 & 0 \\1 & -2 & 0 & 0 \\1 & -2 & -1 & 0\end{array}\right] \underset{\substack{\sim}}{\begin{aligned}R_1+R_2 \Rightarrow R_2 \\R_1+R_3 \Rightarrow R_3 \\-2 R_2+R_3 \Rightarrow R_3\end{aligned}}\left[\begin{array}{rrrr}-1 & 2 & -1 & 0 \\0 & 0 & -1 & 0 \\0 & 0 & 0 & 0\end{array}\right]
Back substitution gives us
General Solution: u _2=s\left[\begin{array}{l}2 \\1 \\0\end{array}\right] ⇒ Basis for Eigenspace of \lambda_2=1:\left\{\left[\begin{array}{l}2 \\1 \\0\end{array}\right]\right\}