Find the Eigenvectors/Eigenspace [[1,2,-1],[1,0,1],[-2,-2,0]]

Problem

[[1,2,−1],[1,0,1],[−2,−2,0]]

Solution

1. Set up the characteristic equation by calculating det(A−λ*I)=0 to find the eigenvalues.

det(1−λ)=0

2. Calculate the determinant using cofactor expansion along the first row.

(1−λ)*(λ^2+2)−2*(−λ+2)−1*(−2−2*λ)=0

3. Simplify the polynomial to find the roots.

λ^2+2−λ^3−2*λ+2*λ−4+2+2*λ=0

−λ^3+λ^2+2*λ=0

−λ*(λ−2)*(λ+1)=0

4. Identify the eigenvalues from the roots of the characteristic equation.

(λ_1)=0,(λ_2)=2,(λ_3)=−1

5. Find the eigenvector for (λ_1)=0 by solving (A−0*I)*v=0

[[1,2,−1],[1,0,1],[−2,−2,0]]*[[x],[y],[z]]=[[0],[0],[0]]

Row reduction yields *x+z=0* and *y−z=0

(v_1)=[[−1],[1],[1]]

6. Find the eigenvector for (λ_2)=2 by solving (A−2*I)*v=0

[[−1,2,−1],[1,−2,1],[−2,−2,−2]]*[[x],[y],[z]]=[[0],[0],[0]]

Row reduction yields *x+z=0* and *y=0

(v_2)=[[−1],[0],[1]]

7. Find the eigenvector for (λ_3)=−1 by solving (A+1*I)*v=0

[[2,2,−1],[1,1,1],[−2,−2,1]]*[[x],[y],[z]]=[[0],[0],[0]]

Row reduction yields *x+y=0* and *z=0

(v_3)=[[−1],[1],[0]]

Final Answer

(E_λ=0)=span*{[−1],[1],[1]},(E_λ=2)=span*{[−1],[0],[1]},(E_λ=−1)=span*{[−1],[1],[0]}

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