Find the Eigenvectors/Eigenspace [[13,3,1],[-56,-13,-4],[-14,-3,-2]]

Problem

A=[[13,3,1],[−56,−13,−4],[−14,−3,−2]]

Solution

1. Find the characteristic equation by calculating the determinant of A−λ*I

det(A−λ*I)=det(13−λ)=0

2. Expand the determinant to find the characteristic polynomial.

(13−λ)*((−13−λ)*(−2−λ)−12)−3*((−56)*(−2−λ)−56)+1*(168−(−14)*(−13−λ))=0

(13−λ)*(λ2+15*λ+14)−3*(56*λ+56)+(168−182−14*λ)=0

−λ3−2*λ2−λ=0

3. Solve for the eigenvalues by factoring the polynomial.

−λ*(λ2+2*λ+1)=0

−λ*(λ+1)2=0

(λ_1)=0,(λ_2)=−1

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

[[13,3,1],[−56,−13,−4],[−14,−3,−2]]*[[x],[y],[z]]=[[0],[0],[0]]

Row reduction leads to x=1/2*z and y=−5/2*z
Setting z=2 gives (v_1)=[[1],[−5],[2]]

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

[[14,3,1],[−56,−12,−4],[−14,−3,−1]]*[[x],[y],[z]]=[[0],[0],[0]]

The rows are multiples of each other, resulting in the single equation 14*x+3*y+z=0
Solving for z gives z=−14*x−3*y

6. Express the eigenspace for (λ_2)=−1 using two basis vectors.

v=[[x],[y],[−14*x−3*y]]=x*[[1],[0],[−14]]+y*[[0],[1],[−3]]

Final Answer

(E_0)=span*{[1],[−5],[2]},(E_−1)=span*{[[1],[0],[−14]],[[0],[1],[−3]]}

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