Final answer:
To find the eigenvalues and eigenvectors of matrix A=[[ -15 25; -5 5 ]], we need to solve the characteristic equation det(A - λI) = 0, and then substitute each eigenvalue back into (A - λI)x = 0 to solve for the corresponding eigenvectors. The eigenvalue a + bi has an eigenvector [i, -i], and the eigenvalue a - bi has an eigenvector [-i, -i].
Step-by-step explanation:
To find the eigenvalues and eigenvectors of matrix A=[[ -15 25; -5 5 ]], we need to solve the characteristic equation det(A - λI) = 0, where det represents the determinant, A is the given matrix, λ is the eigenvalue and I is the identity matrix.
First, we subtract λ from the diagonal elements of A and set the determinant equal to zero:
| -15 - λ 25 |
| -5 5 - λ |
Simplifying, we get (-15 - λ)(5 - λ) - (25)(-5) = 0.
Expanding and rearranging, we have λ² - 10λ + 25 = 0.
Factoring, we get (λ - 5)(λ - 5) = 0, which means that λ = 5 is the only eigenvalue.
To find the eigenvectors, we substitute λ = 5 back into (A - λI)x = 0 and solve for x:
For the eigenvector corresponding to a + bi, we have:
| -20 + 25i 25 || x₁ | | 0 |
| -5 0 + 5i || x₂ | = | 0 |
Rearranging, we get (-20 + 25i)x₁ + 25x₂ = 0 and -5x₁ + (5i)x₂ = 0.
From the second equation, we get x₂ = -ix₁.
Substituting x₂ = -ix₁ into the first equation, we have (-20 + 25i)x₁ - 25(ix₁) = 0.
Simplifying, we get x₁ = i, and since x₂ = -ix₁, x₂ = -i.
Therefore, the eigenvector corresponding to a + bi is [i, -i].
For the eigenvector corresponding to a - bi, we have:
| -20 - 25i 25 || x₃ | | 0 |
| -5 0 - 5i || x₄ | = | 0 |
Rearranging, we get (-20 - 25i)x₃ + 25x₄ = 0 and -5x₃ - (5i)x₄ = 0.
From the second equation, we get x₄ = ix₃.
Substituting x₄ = ix₃ into the first equation, we have (-20 - 25i)x₃ + 25(ix₃) = 0.
Simplifying, we get x₃ = -i, and since x₄ = ix₃, x₄ = -i.
Therefore, the eigenvector corresponding to a - bi is [-i, -i].