Find the Eigenvalues

Problem

A=[[0,0,2,0],[6,0,2,0],[0,1,−7,0],[0,0,0,1]]

Solution

1. Set up the characteristic equation by finding the determinant of A−λ*I

det(A−λ*I)=|[−λ,0,2,0],[6,−λ,2,0],[0,1,−7−λ,0],[0,0,0,1−λ]|=0

2. Expand the determinant along the fourth column or fourth row to simplify the calculation.

det(A−λ*I)=(1−λ)*|[−λ,0,2],[6,−λ,2],[0,1,−7−λ]|=0

3. Calculate the 3×3 determinant using expansion along the first row.

|[−λ,0,2],[6,−λ,2],[0,1,−7−λ]|=−λ*|[−λ,2],[1,−7−λ]|+2*|[6,−λ],[0,1]|

4. Simplify the internal 2×2 determinants.

−λ*(λ2+7*λ−2)+2*(6)=0

−λ3−7*λ2+2*λ+12=0

5. Combine with the factor from step 2 to get the full characteristic polynomial.

(1−λ)*(−λ3−7*λ2+2*λ+12)=0

(λ−1)*(λ3+7*λ2−2*λ−12)=0

6. Find the roots of the cubic factor λ3+7*λ2−2*λ−12 By testing small integers, we find λ=−2 is a root because (−2)3+7*(−2)2−2*(−2)−12=−8+28+4−12=12

(λ−1)*(λ+2)*(λ2+5*λ−6)=0

7. Factor the remaining quadratic λ2+5*λ−6

(λ−1)*(λ+2)*(λ+6)*(λ−1)=0

(λ−1)2*(λ+2)*(λ+6)=0

8. Identify the eigenvalues from the roots of the polynomial.

(λ_1)=1,(λ_2)=1,(λ_3)=−2,(λ_4)=−6

Final Answer

λ=1,1,−2,−6

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