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

Problem

[[5,2],[−1,3]]

Solution

1. Identify the matrix A and the characteristic equation det(A−λ*I)=0 to find the eigenvalues.

A=[[5,2],[−1,3]]

det(5−λ)=0

2. Expand the determinant to form the characteristic polynomial.

(5−λ)*(3−λ)−(2)*(−1)=0

λ2−8*λ+15+2=0

λ2−8*λ+17=0

3. Solve for λ using the quadratic formula λ=(−b±√(,b2−4*a*c))/(2*a)

λ=(8±√(,(−8)2−4*(1)*(17)))/(2*(1))

λ=(8±√(,64−68))/2

λ=(8±√(,−4))/2

λ=4±i

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

[[5−(4+i),2],[−1,3−(4+i)]]*[[(x_1)],[(x_2)]]=[[0],[0]]

[[1−i,2],[−1,−1−i]]*[[(x_1)],[(x_2)]]=[[0],[0]]

5. Solve the system of equations. From the first row: (1−i)*(x_1)+2*(x_2)=0

2*(x_2)=−(1−i)*(x_1)

(x_2)=(−1+i)/2*(x_1)

Setting (x_1)=2 we get (x_2)=−1+i

(v_1)=[[2],[−1+i]]

6. Find the eigenvector for (λ_2)=4−i Since the matrix is real, the eigenvector is the complex conjugate of (v_1)

(v_2)=[[2],[−1−i]]

Final Answer

Eigenspace=span*{[2],[−1+i]}* for *λ=4+i, span*{[2],[−1−i]}* for *λ=4−i

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