Find the Eigenvectors/Eigenspace

Problem

A=[[2,2,6,−8],[3,1,7,−4],[1,−1,1,4],[0,0,0,4]]

Solution

1. Find the eigenvalues by solving the characteristic equation det(A−λ*I)=0 Expanding along the bottom row gives (4−λ)*det(A′−λ*I)=0 where A′ is the upper-left 3×3 submatrix.

2. Calculate the determinant of the 3×3 submatrix. The characteristic polynomial is (4−λ)*(−λ3+4*λ2+4*λ−16)=0

3. Factor the polynomial to find the roots. The eigenvalues are (λ_1)=4 (with algebraic multiplicity 2), (λ_2)=2 and (λ_3)=−2

4. Find the eigenspace for λ=4 by solving (A−4*I)*v=0 This leads to the system:

[[−2,2,6,−8],[3,−3,7,−4],[1,−1,−3,4],[0,0,0,0]]*[[(x_1)],[(x_2)],[(x_3)],[(x_4)]]=[[0],[0],[0],[0]]

Row reduction yields (x_1)−(x_2)=0 and (x_3)=0 and (x_4)=0 The basis is (v_1)=[[1,1,0,0]]T
5. Find the eigenspace for λ=2 by solving (A−2*I)*v=0

[[0,2,6,−8],[3,−1,7,−4],[1,−1,−1,4],[0,0,0,2]]*[[(x_1)],[(x_2)],[(x_3)],[(x_4)]]=[[0],[0],[0],[0]]

Row reduction yields (x_4)=0 (x_2)+3*(x_3)=0 and (x_1)+2*(x_3)=0 The basis is (v_2)=[[−2,−3,1,0]]T
6. Find the eigenspace for λ=−2 by solving (A+2*I)*v=0

[[4,2,6,−8],[3,3,7,−4],[1,−1,3,4],[0,0,0,6]]*[[(x_1)],[(x_2)],[(x_3)],[(x_4)]]=[[0],[0],[0],[0]]

Row reduction yields (x_4)=0 (x_1)+2/3*(x_3)=0 and (x_2)+5/3*(x_3)=0 The basis is (v_3)=[[−2,−5,3,0]]T

Final Answer

(E_4)=span*{[1],[1],[0],[0]},(E_2)=span*{[−2],[−3],[1],[0]},(E_−2)=span*{[−2],[−5],[3],[0]}

Want more problems? Check here!