Find the Eigenvalues [[x,y,z],[2x,y,z],[x,y,z]]

Problem

[[x,y,z],[2*x,y,z],[x,y,z]]

Solution

1. Set up the characteristic equation by subtracting λ from the diagonal elements and setting the determinant of the resulting matrix to zero.

det(x−λ)=0

2. Simplify the determinant by performing row operations. Subtract the third row from the first row ((R_1)→(R_1)−(R_3).

det(−λ)=0

3. Expand the determinant along the first row.

(−λ)*((y−λ)*(z−λ)−y*z)+λ*(2*x*y−x*(y−λ))=0

4. Simplify the terms inside the parentheses.

(−λ)*(y*z−y*λ−z*λ+λ2−y*z)+λ*(2*x*y−x*y+x*λ)=0

(−λ)*(λ2−y*λ−z*λ)+λ*(x*y+x*λ)=0

5. Factor out λ from the entire expression.

λ*(−λ2+y*λ+z*λ+x*y+x*λ)=0

λ*(−λ2+(x+y+z)*λ+x*y)=0

6. Identify the first eigenvalue from the factor λ

(λ_1)=0

7. Solve the remaining quadratic equation −λ2+(x+y+z)*λ+x*y=0 using the quadratic formula λ=(−b±√(,b2−4*a*c))/(2*a)

λ=(−(x+y+z)±√(,(x+y+z)2−4*(−1)*(x*y)))/(2*(−1))

λ=(x+y+z±√(,(x+y+z)2+4*x*y))/2

Final Answer

λ=0,(x+y+z±√(,(x+y+z)2+4*x*y))/2

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