Question 6.1.6: 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 & -2 & 1 \\-1 & 0 & 1 \\-1 & -2 & 3\end{array}\right]
We start out by finding the eigenvalues for A by computing
\operatorname{det}\left(A-\lambda I_3\right)=\left|\begin{array}{rrr}(1-\lambda) & -2 & 1 \\-1 & -\lambda & 1 \\-1 & -2 & (3-\lambda)\end{array}\right|=-\lambda^3+4 \lambda^2-4 \lambda=-\lambda(\lambda-2)^2
From the factored form we see that our matrix has two distinct eigenvalues, λ_1 = 0 and λ_2 = 2.
Now we find the eigenvectors. Starting with those associated with λ_1 = 0, we solve the homogeneous system (A−0 · I_3)u = Au = 0. The augmented matrix and echelon form are
\left[\begin{array}{rrrr} \\1 & -2 & 1 & 0 \\-1 & 0 & 1 & 0 \\-1 & -2 & 3 & 0\end{array}\right] \stackrel{\substack{R_1+R_2 \Rightarrow R_2 \\R_1+R_3 \Rightarrow R_3 \\-2 R_2+R_3 \Rightarrow R_3}}{\sim}\left[\begin{array}{rrrr}1 & -2 & 1 & 0 \\0 & -2 & 2 & 0 \\0 & 0 & 0 & 0\end{array}\right]
Back substitution produces
General Solution: u _1=s\left[\begin{array}{l}1 \\1 \\1\end{array}\right] ⇒ Basis for Eigenspace of \lambda_1=0:\left\{\left[\begin{array}{l}1 \\1 \\1\end{array}\right]\right\}
Turning to the eigenvalue \lambda_2=2, we form the augmented matrix for the system \left(A-2 I_3\right) u=0 and reduce to echelon form,
\left[\begin{array}{llll}-1 & -2 & 1 & 0 \\-1 & -2 & 1 & 0 \\-1 & -2 & 1 & 0\end{array}\right] \stackrel{\substack{\sim}}{\begin{aligned}-R_1+R_2 \Rightarrow R_2 \\-R_1+R_3 \Rightarrow R_3\end{aligned}}\left[\begin{array}{rrrr}-1 & -2 & 1 & 0 \\0 & 0 & 0 & 0 \\0 & 0 & 0 & 0\end{array}\right]
This time back substitution produces
General Solution: u _2=s_1\left[\begin{array}{l}1 \\0 \\1\end{array}\right]+s_2\left[\begin{array}{r}-2 \\1 \\0\end{array}\right]
⇒ Basis for Eigenspace of \lambda_2=2:\left\{\left[\begin{array}{l}1 \\0 \\1\end{array}\right],\left[\begin{array}{r}-2 \\1 \\0\end{array}\right]\right\}