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

Problem

[[−3,0,−1],[2,0,0],[2,0,0]]

Solution

1. Find the characteristic equation by calculating the determinant of A−λ*I=0

det(−3−λ)=0

2. Expand the determinant along the second column to simplify the calculation.

(−λ)⋅det(−3−λ)=0

(−λ)*[(−3−λ)*(−λ)−(−2)]=0

(−λ)*[λ2+3*λ+2]=0

3. Factor the polynomial to find the eigenvalues.

−λ*(λ+1)*(λ+2)=0

(λ_1)=0,(λ_2)=−1,(λ_3)=−2

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

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

2*x=0⇒x=0

−3*(0)+0*y−z=0⇒z=0

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

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

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

2*x+z=0⇒z=−2*x

2*x+y=0⇒y=−2*x

(v_2)=[[1],[−2],[−2]]

6. Find the eigenvector for (λ_3)=−2 by solving (A+2*I)*v=0

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

−x−z=0⇒z=−x

2*x+2*y=0⇒y=−x

(v_3)=[[1],[−1],[−1]]

Final Answer

(λ_1)=0:span*{[0],[1],[0]},(λ_2)=−1:span*{[1],[−2],[−2]},(λ_3)=−2:span*{[1],[−1],[−1]}

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