Question 6.1.2: Take as known that λ = 3 and λ = 2 are eigenvalues for A =[4...
Take as known that λ = 3 and λ = 2 are eigenvalues for
A=\left[\begin{array}{lll}4 & 4 & -2 \\1 & 4 & -1 \\3 & 6 & -1\end{array}\right]
Find the eigenvectors associated with each eigenvalue.
Starting with λ = 3,we need to find all nonzero vectors u such that Au = 3u. Since 3u = 3I_3u, our equation becomes
A u=3 I_3 u \quad \Rightarrow \quad A u -3 I_3 u=0 \quad \Rightarrow \quad\left(A-3 I_3\right) u=0
Thus we need to find the solutions to the homogeneous system with coefficient matrix
A-3 I_3=\left[\begin{array}{lll}4 & 4 & -2 \\1 & 4 & -1 \\3 & 6 & -1\end{array}\right]-\left[\begin{array}{lll}3 & 0 & 0 \\0 & 3 & 0 \\0 & 0 & 3\end{array}\right]=\left[\begin{array}{lll}1 & 4 & -2 \\1 & 1 & -1 \\3 & 6 & -4\end{array}\right]
Forming an augmented matrix and performing the indicated row operations, we have
\left[\begin{array}{cccc}1 & 4 & -2 & 0 \\1 & 1 & -1 & 0 \\3 & 6 & -4 & 0\end{array}\right] \underset{\substack{-R_1+R_2 \Rightarrow R_2 \\-3 R_1+R_3 \Rightarrow R_3 \\-2 R_2+R_3 \Rightarrow R_3}}{\sim}\left[\begin{array}{rrrr}1 & 4 & -2 & 0 \\0 & -3 & 1 & 0 \\0 & 0 & 0 & 0\end{array}\right]
After back substitution, we find that the system (A − 3I_3)u = 0 has general solution
u=s\left[\begin{array}{l}2 \\1 \\3\end{array}\right]
We can verify that u is an eigenvector associated with λ = 3 by computing
A u=\left[\begin{array}{lll}4 & 4 & -2 \\1 & 4 & -1 \\3 & 6 & -1\end{array}\right]\left(s\left[\begin{array}{l}2 \\1 \\3\end{array}\right]\right)=s\left[\begin{array}{lll}4 & 4 & -2 \\1 & 4 & -1 \\3 & 6 & -1\end{array}\right]\left[\begin{array}{l}2 \\1 \\3\end{array}\right]=s\left[\begin{array}{l}6 \\3 \\9\end{array}\right]=3 u
The procedure is the same for λ = 2. This time we need to find the general solution to the homogeneous system (A − 2I_3)u = 0, where
A-2 I_3=\left[\begin{array}{lll}4 & 4 & -2 \\1 & 4 & -1 \\3 & 6 & -1\end{array}\right]-\left[\begin{array}{lll}2 & 0 & 0 \\0 & 2 & 0 \\0 & 0 & 2\end{array}\right]=\left[\begin{array}{lll}2 & 4 & -2 \\1 & 2 & -1 \\3 & 6 & -3\end{array}\right]
The augmented matrix and corresponding echelon form are
\left[\begin{array}{cccc}2 & 4 & -2 & 0 \\1 & 2 & -1 & 0 \\3 & 6 & -3 & 0\end{array}\right] \underset{\sim }{\substack{R_1 \Leftrightarrow R_2 \\-2 R_1+R_2 \Rightarrow R_2 \\-3 R_1+R_3 \Rightarrow R_3}}\left[\begin{array}{rrrr}1 & 2 & -1 & 0 \\0 & 0 & 0 & 0 \\0 & 0 & 0 & 0\end{array}\right]
After back substitution, we find that this system has general solution
u=s_1\left[\begin{array}{r}-2 \\1 \\0\end{array}\right]+s_2\left[\begin{array}{l}1 \\0 \\1\end{array}\right] (2)
As long as at least one of s_1 and s_2 is nonzero, then u will be an eigenvector associated with λ = 2.We can check our answer by computing
\begin{aligned}A u &=A\left(s_1\left[\begin{array}{r}-2 \\1 \\0\end{array}\right]+s_2\left[\begin{array}{l}1 \\0 \\1\end{array}\right]\right)=s_1 A\left[\begin{array}{r}-2 \\1 \\0\end{array}\right]+s_2 A\left[\begin{array}{l}1 \\0 \\1\end{array}\right] \\&=s_1\left[\begin{array}{r}-4 \\2 \\0\end{array}\right]+s_2\left[\begin{array}{l}2 \\0 \\2\end{array}\right]=2\left(s_1\left[\begin{array}{r}-2 \\1 \\0\end{array}\right]+s_2\left[\begin{array}{l}1 \\0 \\1\end{array}\right]\right)=2 u\end{aligned}