Find the Eigenvectors/Eigenspace A=[[1,0,0],[1,2,0],[-3,5,2]]

Problem

A=[[1,0,0],[1,2,0],[−3,5,2]]

Solution

1. Find the eigenvalues by solving the characteristic equation det(A−λ*I)=0 Since A is a lower triangular matrix, the eigenvalues are the diagonal entries.

(λ_1)=1

(λ_2)=2

2. Set up the system (A−λ*I)*v=0 for the eigenvalue (λ_1)=1

[[1−1,0,0],[1,2−1,0],[−3,5,2−1]]*[[x],[y],[z]]=[[0],[0],[0]]

[[0,0,0],[1,1,0],[−3,5,1]]*[[x],[y],[z]]=[[0],[0],[0]]

3. Solve for the eigenvector (v_1) corresponding to (λ_1)=1 From the second row, x+y=0⇒y=−x From the third row, −3*x+5*y+z=0 Substituting y=−x gives −3*x−5*x+z=0⇒z=8*x

(v_1)=[[1],[−1],[8]]

4. Set up the system (A−λ*I)*v=0 for the eigenvalue (λ_2)=2

[[1−2,0,0],[1,2−2,0],[−3,5,2−2]]*[[x],[y],[z]]=[[0],[0],[0]]

[[−1,0,0],[1,0,0],[−3,5,0]]*[[x],[y],[z]]=[[0],[0],[0]]

5. Solve for the eigenvectors corresponding to (λ_2)=2 From the first row, −x=0⇒x=0 From the third row, −3*(0)+5*y+0*z=0⇒y=0 The variable z is free.

(v_2)=[[0],[0],[1]]

6. Identify the eigenspaces by expressing the sets of all linear combinations of the basis vectors found for each eigenvalue.

(E_1)=span*{[1],[−1],[8]}

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

Final Answer

(E_1)=span*{[1],[−1],[8]},(E_2)=span*{[0],[0],[1]}

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