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

Problem

A=[[1,2,1],[0,3,1],[0,5,−1]]

Solution

1. Find the characteristic equation by calculating det(A−λ*I)=0

det(1−λ)=0

2. Expand the determinant along the first column.

(1−λ)*((3−λ)*(−1−λ)−5)=0

3. Solve for the eigenvalues λ by factoring the quadratic expression.

(1−λ)*(λ2−2*λ−3−5)=0

(1−λ)*(λ2−2*λ−8)=0

(1−λ)*(λ−4)*(λ+2)=0

(λ_1)=1,(λ_2)=4,(λ_3)=−2

4. Find the eigenvector for (λ_1)=1 by solving (A−1*I)*v=0

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

⇒2*y+z=0

⇒5*y−2*z=0

⇒y=0,z=0,x=free

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

5. Find the eigenvector for (λ_2)=4 by solving (A−4*I)*v=0

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

⇒−y+z=0⇒y=z

⇒−3*x+2*y+z=0⇒−3*x+3*z=0⇒x=z

(E_4)=span*{[1],[1],[1]}

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

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

⇒5*y+z=0⇒z=−5*y

⇒3*x+2*y+(−5*y)=0⇒3*x−3*y=0⇒x=y

(E_−2)=span*{[1],[1],[−5]}

Final Answer

(E_1)=span*{[1],[0],[0]},(E_4)=span*{[1],[1],[1]},(E_−2)=span*{[1],[1],[−5]}

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