Answer :
(a) The eigenvalues of matrix A are λ₁ = 1 and λ₂ = -1. The corresponding eigenvectors are [1, 0] and [0, 1], respectively.
(b) The eigenvalues of matrix B are λ₁ = 1 and λ₂ = 0. The corresponding eigenvectors are [3, 4] and [-4, 3], respectively.
(a) Matrix A represents a reflection in the x-axis, which means it flips points over the x-axis. The eigenvalues represent the scaling factors of the eigenvectors under this transformation. Since the x-axis reflection doesn't change the direction of vectors, the eigenvectors remain unchanged, resulting in eigenvalues of 1 and -1.
Calculation:
To find the eigenvalues of A, we solve the characteristic equation det(A - λI) = 0:
det([[1-λ, 0], [0, -1-λ]]) = (1-λ)(-1-λ) = λ² - 1 = 0
This yields λ₁ = 1 and λ₂ = -1.
Now, to find the eigenvectors, substitute each eigenvalue back into the equation (A - λI)v = 0 and solve for v:
For λ₁ = 1: [[1-1, 0], [0, -1-1]]v = 0 => [[0, 0], [0, -2]]v = 0
This gives the eigenvector [1, 0].
For λ₂ = -1: [[1+1, 0], [0, -1+1]]v = 0 => [[2, 0], [0, 0]]v = 0
This gives the eigenvector [0, 1].
(b) Matrix B represents a projection onto the line y=(4)/(3)x. The eigenvalues represent the scaling factors of the eigenvectors under this transformation. For any vector along the line, the projection keeps its direction intact, hence the eigenvalue of 1. The eigenvector [3, 4] lies on the line and is stretched by a factor of 1, while the eigenvector [-4, 3] is perpendicular to the line and gets projected onto the origin, resulting in an eigenvalue of 0.
Calculation:
To find the eigenvalues of B, we solve the characteristic equation det(B - λI) = 0:
det([[(9/25)-λ, (12/25)], [(12/25), (16/25)-λ]]) = (9/25-λ)(16/25-λ) - (12/25)(12/25) = λ² - (25/25)λ = λ(λ-1) = 0
This yields λ₁ = 1 and λ₂ = 0.
Now, to find the eigenvectors, substitute each eigenvalue back into the equation (B - λI)v = 0 and solve for v:
For λ₁ = 1: [[(9/25)-1, (12/25)], [(12/25), (16/25)-1]]v = 0 => [[-16/25, 12/25], [12/25, -9/25]]v = 0
This gives the eigenvector [3, 4].
For λ₂ = 0: [[(9/25)-0, (12/25)], [(12/25), (16/25)-0]]v = 0 => [[9/25, 12/25], [12/25, 16/25]]v = 0
This gives the eigenvector [-4, 3].