Answer on question #41199 – Math – Linear Algebra
We have
A = ( 2 2 − 2 1 3 1 1 2 2 ) A = \left( \begin{array}{ccc} 2 & 2 & -2 \\ 1 & 3 & 1 \\ 1 & 2 & 2 \end{array} \right) A = ⎝ ⎛ 2 1 1 2 3 2 − 2 1 2 ⎠ ⎞
To find the eigen values of this matrix we should find the following determinant and equate it to 0
∣ A − λ E ∣ = ∣ 2 − λ 2 − 2 1 3 − λ 1 1 2 2 − λ ∣ = ( 2 − λ ) 2 ( 3 − λ ) − 4 + 2 + 2 ( 3 − λ ) − 4 ( 2 − λ ) = 0 − λ 3 + 7 λ 2 − 14 λ + 8 = 0 ( λ − 1 ) ( λ − 2 ) ( λ − 4 ) = 0 λ = 1 ; 2 ; 4. \begin{aligned}
|A - \lambda E| &= \left| \begin{array}{ccc} 2 - \lambda & 2 & -2 \\ 1 & 3 - \lambda & 1 \\ 1 & 2 & 2 - \lambda \end{array} \right| = (2 - \lambda)^2 (3 - \lambda) - 4 + 2 + 2(3 - \lambda) - 4(2 - \lambda) = 0 \\
&\quad - \lambda^3 + 7\lambda^2 - 14\lambda + 8 = 0 \\
&\quad (\lambda - 1)(\lambda - 2)(\lambda - 4) = 0 \\
&\quad \lambda = 1; 2; 4.
\end{aligned} ∣ A − λ E ∣ = ∣ ∣ 2 − λ 1 1 2 3 − λ 2 − 2 1 2 − λ ∣ ∣ = ( 2 − λ ) 2 ( 3 − λ ) − 4 + 2 + 2 ( 3 − λ ) − 4 ( 2 − λ ) = 0 − λ 3 + 7 λ 2 − 14 λ + 8 = 0 ( λ − 1 ) ( λ − 2 ) ( λ − 4 ) = 0 λ = 1 ; 2 ; 4.
The corresponding eigen vectors are
For λ = 1 \lambda = 1 λ = 1
Ax − Ex = 0 ( 1 2 − 2 1 2 1 1 2 1 ) x = 0 { x 1 + 2 x 2 − 2 x 3 = 0 x 1 + 2 x 2 + x 3 = 0 x 1 + 2 x 2 + x 3 = 0 ⇒ { x 1 = − 2 x 2 + 2 x 3 − 2 x 2 + 2 x 3 + 2 x 2 + x 3 = 0 ⇒ { x 3 = 0 x 1 = − 2 x 2 \begin{aligned}
& \text{Ax} - \text{Ex} = 0 \\
& \left( \begin{array}{ccc} 1 & 2 & -2 \\ 1 & 2 & 1 \\ 1 & 2 & 1 \end{array} \right) x = 0 \\
&\left\{ \begin{array}{l} x_1 + 2x_2 - 2x_3 = 0 \\ x_1 + 2x_2 + x_3 = 0 \\ x_1 + 2x_2 + x_3 = 0 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x_1 = -2x_2 + 2x_3 \\ -2x_2 + 2x_3 + 2x_2 + x_3 = 0 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x_3 = 0 \\ x_1 = -2x_2 \end{array} \right.
\end{aligned} Ax − Ex = 0 ⎝ ⎛ 1 1 1 2 2 2 − 2 1 1 ⎠ ⎞ x = 0 ⎩ ⎨ ⎧ x 1 + 2 x 2 − 2 x 3 = 0 x 1 + 2 x 2 + x 3 = 0 x 1 + 2 x 2 + x 3 = 0 ⇒ { x 1 = − 2 x 2 + 2 x 3 − 2 x 2 + 2 x 3 + 2 x 2 + x 3 = 0 ⇒ { x 3 = 0 x 1 = − 2 x 2
The eigen vector is x = ( − 2 ; 1 ; 0 ) x = (-2; 1; 0) x = ( − 2 ; 1 ; 0 )
For λ = 2 \lambda = 2 λ = 2
Ax − 2 Ex = 0 ( 0 2 − 2 1 1 1 1 2 0 ) x = 0 { 2 x 2 − 2 x 3 = 0 x 1 + x 2 + x 3 = 0 x 1 + 2 x 2 = 0 ⇒ { x 1 = − 2 x 2 x 3 = x 2 − 2 x 2 + x 2 + x 2 = 0 ⇒ { x 1 = − 2 ; x 2 = 1 ; x 3 = 1 \begin{aligned}
& \text{Ax} - 2\text{Ex} = 0 \\
& \left( \begin{array}{ccc} 0 & 2 & -2 \\ 1 & 1 & 1 \\ 1 & 2 & 0 \end{array} \right) x = 0 \\
&\left\{ \begin{array}{l} 2x_2 - 2x_3 = 0 \\ x_1 + x_2 + x_3 = 0 \\ x_1 + 2x_2 = 0 \end{array} \right. \Rightarrow \left\{ \begin{array}{c} x_1 = -2x_2 \\ x_3 = x_2 \\ -2x_2 + x_2 + x_2 = 0 \end{array} \right. \Rightarrow \left\{ \begin{array}{c} x_1 = -2; \\ x_2 = 1; \\ x_3 = 1 \end{array} \right.
\end{aligned} Ax − 2 Ex = 0 ⎝ ⎛ 0 1 1 2 1 2 − 2 1 0 ⎠ ⎞ x = 0 ⎩ ⎨ ⎧ 2 x 2 − 2 x 3 = 0 x 1 + x 2 + x 3 = 0 x 1 + 2 x 2 = 0 ⇒ ⎩ ⎨ ⎧ x 1 = − 2 x 2 x 3 = x 2 − 2 x 2 + x 2 + x 2 = 0 ⇒ ⎩ ⎨ ⎧ x 1 = − 2 ; x 2 = 1 ; x 3 = 1
The eigen vector is x = ( − 2 ; 1 ; 1 ) x = (-2; 1; 1) x = ( − 2 ; 1 ; 1 )
For λ = 4 \lambda = 4 λ = 4
Ax − 4 Ex = 0 ( − 2 2 − 2 1 − 1 1 1 2 − 2 ) x = 0 { − 2 x 1 + 2 x 2 − 2 x 3 = 0 x 1 − x 2 + x 3 = 0 x 1 + 2 x 2 − 2 x 3 = 0 ⇒ { x 1 = 0 x 2 = 1 x 3 = 1 \begin{aligned}
& \text{Ax} - 4\text{Ex} = 0 \\
&\left( \begin{array}{ccc} -2 & 2 & -2 \\ 1 & -1 & 1 \\ 1 & 2 & -2 \end{array} \right) x = 0 \\
&\left\{ \begin{array}{l} -2x_1 + 2x_2 - 2x_3 = 0 \\ x_1 - x_2 + x_3 = 0 \\ x_1 + 2x_2 - 2x_3 = 0 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x_1 = 0 \\ x_2 = 1 \\ x_3 = 1 \end{array} \right.
\end{aligned} Ax − 4 Ex = 0 ⎝ ⎛ − 2 1 1 2 − 1 2 − 2 1 − 2 ⎠ ⎞ x = 0 ⎩ ⎨ ⎧ − 2 x 1 + 2 x 2 − 2 x 3 = 0 x 1 − x 2 + x 3 = 0 x 1 + 2 x 2 − 2 x 3 = 0 ⇒ ⎩ ⎨ ⎧ x 1 = 0 x 2 = 1 x 3 = 1
The eigen vector is x = ( 0 ; 1 ; 1 ) x = (0; 1; 1) x = ( 0 ; 1 ; 1 )
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