Find the Eigenvectors/Eigenspace

Problem

A=[[2,0,0,1],[1,2,0,−1],[1,5,3,−1],[0,0,0,2]]

Solution

1. Find the characteristic equation by calculating det(A−λ*I)=0

det(2−λ)=0

2. Expand the determinant along the last row.

(2−λ)⋅det(2−λ)=0

3. Solve for eigenvalues by noting the remaining matrix is lower triangular, so its determinant is the product of the diagonal elements.

(2−λ)*(2−λ)*(2−λ)*(3−λ)=0

(λ_1)=3,(λ_2)=2

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

[[−1,0,0,1],[1,−1,0,−1],[1,5,0,−1],[0,0,0,−1]]*[[(x_1)],[(x_2)],[(x_3)],[(x_4)]]=[[0],[0],[0],[0]]

Row reduction leads to *(x_1)=0,(x_2)=0,(x_4)=0,(x_3)=free

(E_3)=span*{[0],[0],[1],[0]}

5. Find the eigenspace for λ=2 by solving (A−2*I)*v=0

[[0,0,0,1],[1,0,0,−1],[1,5,1,−1],[0,0,0,0]]*[[(x_1)],[(x_2)],[(x_3)],[(x_4)]]=[[0],[0],[0],[0]]

From row 1: *(x_4)=0

From row 2: *(x_1)−(x_4)=0⇒(x_1)=0

From row 3: *0+5*(x_2)+(x_3)−0=0⇒(x_3)=−5*(x_2)

(E_2)=span*{[0],[1],[−5],[0]}

Final Answer

(E_3)=span*{[0],[0],[1],[0]},(E_2)=span*{[0],[1],[−5],[0]}

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