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

Problem

[[1,−1,0],[−1,2,−1],[0,−1,1]]

Solution

1. Set up the characteristic equation by finding the determinant of A−λ*I where I is the identity matrix.

det(1−λ)=0

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

(1−λ)*((2−λ)*(1−λ)−1)−(−1)*(−1*(1−λ)−0)=0

(1−λ)*(λ2−3*λ+1)−(1−λ)=0

3. Factor the characteristic polynomial to find the eigenvalues.

(1−λ)*(λ2−3*λ+1−1)=0

(1−λ)*(λ2−3*λ)=0

−λ*(λ−1)*(λ−3)=0

(λ_1)=0,(λ_2)=1,(λ_3)=3

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

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

Row reduction leads to x=y and y=z

(v_1)=[[1],[1],[1]]

5. Find the eigenvector for (λ_2)=1 by solving (A−1*I)*v=0

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

From the first row, y=0 Substituting into the second row, −x−z=0 so x=−z

(v_2)=[[1],[0],[−1]]

6. Find the eigenvector for (λ_3)=3 by solving (A−3*I)*v=0

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

From the first row, y=−2*x From the third row, y=−2*z Thus x=z

(v_3)=[[1],[−2],[1]]

Final Answer

Eigenspaces: *(E_0)=span*{[1],[1],[1]},(E_1)=span*{[1],[0],[−1]},(E_3)=span*{[1],[−2],[1]}

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