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

Problem

[[−2,3,0],[3,0,1],[1,1,2]]

Solution

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

det(−2−λ)=0

2. Expand the determinant along the first row.

(−2−λ)*((−λ)*(2−λ)−1)−3*(3*(2−λ)−1)=0

3. Simplify the polynomial to find the eigenvalues.

(−2−λ)*(λ2−2*λ−1)−3*(5−3*λ)=0

(−λ3+5*λ+2)−(15−9*λ)=0

−λ3+14*λ−13=0

λ3−14*λ+13=0

4. Solve for eigenvalues by testing roots. Since 1−14*(1)+13=0 (λ_1)=1 is a root. Factoring out (λ−1) gives (λ−1)*(λ2+λ−13)=0

(λ_1)=1

(λ_2)=(−1+√(,53))/2

(λ_3)=(−1−√(,53))/2

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

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

6. Row reduce the matrix to find the relationship between variables.

[[1,−1,0],[0,2,1],[0,2,1]]⇒[[1,−1,0],[0,1,0.5],[0,0,0]]

x=y

y=−0.5*z

(v_1)=[[−0.5],[−0.5],[1]]⇒[[1],[1],[−2]]

7. Find the eigenvectors for (λ_2,3) by solving (A−λ*I)*v=0 using the quadratic roots.

x=(3*y)/(λ+2)

z=(λ−2)*y−x

(v_λ)=[[3/(λ+2)],[1],[λ−2−3/(λ+2)]]

Final Answer

(E_1)=span*{[1],[1],[−2]},(E_(−1±√(,53))/2)=span*{[6/(3±√(,53))],[1],[(−5±√(,53))/2−6/(3±√(,53))]}

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