Find the Eigenvectors/Eigenspace A=[[1,1,4],[0,-4,0],[-5,-1,-8]]

Problem

A=[[1,1,4],[0,−4,0],[−5,−1,−8]]

Solution

1. Find the characteristic equation by calculating the determinant of A−λ*I and setting it to zero.

det(A−λ*I)=|[1−λ,1,4],[0,−4−λ,0],[−5,−1,−8−λ]|=0

2. Expand the determinant along the second row since it contains two zeros.

(−4−λ)*|[1−λ,4],[−5,−8−λ]|=0

3. Solve for the eigenvalues by factoring the quadratic expression inside the determinant.

(−4−λ)*((1−λ)*(−8−λ)−(−20))=0

(−4−λ)*(λ2+7*λ+12)=0

−(λ+4)*(λ+4)*(λ+3)=0

(λ_1)=−4,(λ_2)=−3

4. Find the eigenspace for λ=−4 by solving (A+4*I)*v=0

[[5,1,4],[0,0,0],[−5,−1,−4]]*[[(x_1)],[(x_2)],[(x_3)]]=[[0],[0],[0]]

5*(x_1)+(x_2)+4*(x_3)=0

(x_2)=−5*(x_1)−4*(x_3)

(E_−4)=span*{[[1],[−5],[0]],[[0],[−4],[1]]}

5. Find the eigenspace for λ=−3 by solving (A+3*I)*v=0

[[4,1,4],[0,−1,0],[−5,−1,−5]]*[[(x_1)],[(x_2)],[(x_3)]]=[[0],[0],[0]]

−(x_2)=0⇒(x_2)=0

4*(x_1)+4*(x_3)=0⇒(x_1)=−(x_3)

(E_−3)=span*{[−1],[0],[1]}

Final Answer

(E_−4)=span*{[[1],[−5],[0]],[[0],[−4],[1]]},(E_−3)=span*{[−1],[0],[1]}

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