Question 6.1.4: Find the eigenvalues and a basis for each eigenspace for the...
Find the eigenvalues and a basis for each eigenspace for the matrix
A=\left[\begin{array}{rr}2 & -1 \\-1 & 2\end{array}\right]
We start by finding the eigenvalues of A by computing
\begin{aligned}\operatorname{det}\left(A-\lambda I_2\right) &=\left|\begin{array}{rr}(2-\lambda) & -1 \\-1 & (2-\lambda)\end{array}\right| \\&=(2-\lambda)^2-1 \\&=\lambda^2-4 \lambda+3=(\lambda-1)(\lambda-3)\end{aligned}
Therefore the eigenvalues are λ_1 = 1 and λ_2 = 3.Next,we find the eigenvectors, starting with those associated with λ_1 = 1. We find the associated eigenvectors by solving the homogeneous linear system (A − 1 · I_2)u = (A − I_2)u = 0. Since
A-I_2=\left[\begin{array}{rr}1 & -1 \\-1 & 1\end{array}\right]
the augmented matrix and corresponding echelon form are
\left[\begin{array}{rrr}1 & -1 & 0 \\-1 & 1 & 0\end{array}\right] \stackrel{R_1+R_2 \Rightarrow R_2}{\sim}\left[\begin{array}{rrr}1 & -1 & 0 \\0 & 0 & 0\end{array}\right]
Back substitution gives us
General Solution: u _1=s\left[\begin{array}{l}1 \\1\end{array}\right] ⇒ Basis for Eigenspace of \lambda_1=1:\left\{\left[\begin{array}{l}1 \\1\end{array}\right]\right\}
The eigenvectors associated with λ_2 = 3 are found by solving the homogeneous linear system (A − 3I_2)u = 0. We have
A-3 I_2=\left[\begin{array}{ll}-1 & -1 \\-1 & -1\end{array}\right]
so the augmented matrix and corresponding echelon form are
\left[\begin{array}{lll}-1 & -1 & 0 \\-1 & -1 & 0\end{array}\right]\stackrel{{-R_1+R_2 \Rightarrow R_2}}\sim\left[\begin{array}{rrr}-1 & -1 & 0 \\0 & 0 & 0\end{array}\right]
Back substitution gives us
General Solution: u _2=s\left[\begin{array}{r}-1 \\1\end{array}\right] ⇒ Basis for Eigenspace of \lambda_2=3:\left\{\left[\begin{array}{r}-1 \\1\end{array}\right]\right\}