A = [ 2 2 1 1 3 1 1 2 2 ] A=\begin{bmatrix}
2 & 2&1 \\
1 & 3&1\\
1&2&2
\end{bmatrix} A = ⎣ ⎡ 2 1 1 2 3 2 1 1 2 ⎦ ⎤
1.1. det ( A − λ I ) = ∣ 2 − λ 2 1 1 3 − λ 1 1 2 2 − λ ∣ = − λ 3 + 7 λ 2 − 11 λ + 5 = − ( λ − 5 ) ( λ − 1 ) 2 = 0 \det (A-\lambda I)=\begin{vmatrix}
2-\lambda & 2&1 \\
1 & 3-\lambda &1\\ 1&2&2-\lambda
\end{vmatrix}=-\lambda^3+7\lambda^2-11\lambda+5=-(\lambda-5)(\lambda-1)^2=0 det ( A − λ I ) = ∣ ∣ 2 − λ 1 1 2 3 − λ 2 1 1 2 − λ ∣ ∣ = − λ 3 + 7 λ 2 − 11 λ + 5 = − ( λ − 5 ) ( λ − 1 ) 2 = 0
Eigenvalues: λ 1 = 5 \lambda _1=5 λ 1 = 5 and λ 2 = 1 \lambda_2=1 λ 2 = 1
1.2.
λ 1 = 5 \lambda_1=5 λ 1 = 5 : ( A − 5 I ) x = [ − 3 2 1 1 − 2 1 1 2 − 3 ] [ x y z ] = [ 0 0 0 ] : \ \ (A-5I)\mathbf{x}= \begin{bmatrix}
-3 & 2&1 \\
1 & -2&1\\
1&2&-3
\end{bmatrix}
\begin{bmatrix}
x\\y\\z
\end{bmatrix}=\begin{bmatrix}
0\\0\\0
\end{bmatrix} : ( A − 5 I ) x = ⎣ ⎡ − 3 1 1 2 − 2 2 1 1 − 3 ⎦ ⎤ ⎣ ⎡ x y z ⎦ ⎤ = ⎣ ⎡ 0 0 0 ⎦ ⎤ gives the eigenvector x 1 = [ 1 1 1 ] \mathbf{x}_1=\begin{bmatrix}1\\1\\1\end{bmatrix} x 1 = ⎣ ⎡ 1 1 1 ⎦ ⎤ .
Answer: 1.1. λ 1 = 5 \lambda_1=5 λ 1 = 5 and λ 2 = 1 \lambda_2=1 λ 2 = 1 ; 1.2. x 1 = [ 1 1 1 ] \mathbf{x}_1=\begin{bmatrix}1\\1\\1\end{bmatrix} x 1 = ⎣ ⎡ 1 1 1 ⎦ ⎤
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