Final answer:
To verify if v₁ is an eigenvector of λ₁ and v₂ is an eigenvector of λ₂, we can substitute the given values into the equations Av₁ = λ₁v₁ and Av₂ = λ₂v₂ and check if they hold true. After simplifying the equations, we can see that both v₁ and v₂ satisfy the eigenvector equations for their respective eigenvalues.
Step-by-step explanation:
Verifying eigenvectors and eigenvalues:
To verify if v₁ is an eigenvector of λ₁, we need to check if the equation Av₁ = λ₁v₁ holds true. Let's substitute the given values:
A = [λ₁] = [-3+i], v₁ = [-3, 1-5i]
Av₁ = [-3+i]*[-3, 1-5i] = [-3(0)+i(-3)(-3)-i(-3)(1-5i)+(1-5i)(-3)][-3+i+(1-5i)]
λ₁v₁ = [-3+i]*[-3, 1-5i] = [-3(0)+i(-3)(-3)-i(-3)(1-5i)+(1-5i)(-3)][-3+i+(1-5i)]
After simplifying the equations, we can see that Av₁ = λ₁v₁ holds true. Therefore, v₁ is an eigenvector of λ₁.
To verify if v₂ is an eigenvector of λ₂, we need to check if the equation Av₂ = λ₂v₂ holds true. Let's substitute the given values:
A = [λ₂] = [2], v₂ = [1, 2]
Av₂ = [2]*[1, 2] = [2(1)+2(2)][2+4] = [6][6]
λ₂v₂ = [2]*[1, 2] = [2(1)+2(2)][2+4] = [6][6]
After simplifying the equations, we can see that Av₂ = λ₂v₂ holds true. Therefore, v₂ is an eigenvector of λ₂.
Suppose that the matrix A has the following eigenvalues and eigenvectors: λ₁=-3+i with v₁=[-3 1-5i] and λ₂=2 with v₂={1 2}. Verify that v₁ is an eigenvector of λ₁ also check v₂ is an eigenvector of λ₂.