Find the Eigenvectors/Eigenspace

Problem

[[6,−3,0,12],[0,4,1,−2],[0,0,3,0],[0,0,0,3]]

Solution

1. Identify the eigenvalues by looking at the diagonal entries of the upper triangular matrix. The eigenvalues are (λ_1)=6 (λ_2)=4 and (λ_3)=3 (with algebraic multiplicity 2).

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

[[0,−3,0,12],[0,−2,1,−2],[0,0,−3,0],[0,0,0,−3]]*[[(x_1)],[(x_2)],[(x_3)],[(x_4)]]=[[0],[0],[0],[0]]

From the last two rows, (x_4)=0 and (x_3)=0 From the first row, −3*(x_2)=0⇒(x_2)=0 (x_1) is free.

(v_1)=[[1],[0],[0],[0]]

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

[[2,−3,0,12],[0,0,1,−2],[0,0,−1,0],[0,0,0,−1]]*[[(x_1)],[(x_2)],[(x_3)],[(x_4)]]=[[0],[0],[0],[0]]

From the last two rows, (x_4)=0 and (x_3)=0 From the first row, 2*(x_1)−3*(x_2)=0⇒(x_1)=3/2*(x_2)

(v_2)=[[3],[2],[0],[0]]

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

[[3,−3,0,12],[0,1,1,−2],[0,0,0,0],[0,0,0,0]]*[[(x_1)],[(x_2)],[(x_3)],[(x_4)]]=[[0],[0],[0],[0]]

Row 2 gives (x_2)=−(x_3)+2*(x_4) Row 1 gives 3*(x_1)−3*(x_2)+12*(x_4)=0⇒(x_1)=(x_2)−4*(x_4)
Substitute (x_2) (x_1)=(−(x_3)+2*(x_4))−4*(x_4)=−(x_3)−2*(x_4)
Setting (x_3)=1,(x_4)=0 gives (v_3)=[[−1],[−1],[1],[0]]
Setting (x_3)=0,(x_4)=1 gives (v_4)=[[−2],[2],[0],[1]]

Final Answer

(E_6)=span*{[1],[0],[0],[0]},(E_4)=span*{[3],[2],[0],[0]},(E_3)=span*{[[−1],[−1],[1],[0]],[[−2],[2],[0],[1]]}

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