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Eigenvalues and eigenvectors questions
Most vectors change direction under a matrix transformation. Eigenvectors do not: they are only scaled, by a factor called the eigenvalue. Solving det(M − λI) = 0 finds those factors, and the directions they belong to describe what the transformation does independently of the coordinates used.
6 original questions · 27 marks · the eigenvalues and eigenvectors notes · Further Pure 2
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M has rows (4, 1) and (2, 3). Find its eigenvalues and a corresponding eigenvector for each.
Worked answer
The characteristic equation is (4 − λ)(3 − λ) − 2 = 0, that is λ² − 7λ + 10 = 0, so λ = 5 or λ = 2. For λ = 5 the first row gives −x + y = 0, so (1, 1) works. For λ = 2 it gives 2x + y = 0, so (1, −2) works. M1 for det(M − λI) = 0, A1 for λ² − 7λ + 10 = 0, A1 for both eigenvalues, M1 for solving (M − λI)v = 0, A1 for both eigenvectors. Any non-zero multiple of either is equally valid.Prove that if λ is an eigenvalue of M with eigenvector v, then λ² is an eigenvalue of M² with the same eigenvector.
Worked answer
Start from Mv = λv. Then M²v = M(Mv) = M(λv) = λ(Mv) = λ(λv) = λ²v. Since v is non-zero, λ² is an eigenvalue of M² and v is an eigenvector for it. M1 for starting from Mv = λv, M1 for forming M²v = M(λv), A1 for the conclusion, using that v is non-zero. The same argument repeated gives λk for Mk.A has rows (2, 0, 1), (0, 3, 0) and (1, 0, 2). Find its eigenvalues and an eigenvector for each distinct one.
Worked answer
Expanding along the middle row, the characteristic equation factorises as (3 − λ)[(2 − λ)² − 1] = 0. The bracket gives 2 − λ = ±1, so λ = 1 or λ = 3. The eigenvalues are 1 and 3, with 3 repeated. For λ = 1, (1, 0, −1) works; for λ = 3, both (1, 0, 1) and (0, 1, 0) do, so that eigenvalue has two independent eigenvectors. M1 for expanding det(A − λI), A1 for the factorised form, A1 for the eigenvalue 1, A1 for the repeated eigenvalue 3, M1 for solving for an eigenvector, A1 for (1, 0, −1), A1 for two independent eigenvectors for the repeated eigenvalue.State how the trace and determinant of a 2 by 2 matrix relate to its eigenvalues, and verify both for the matrix in question 1.
Worked answer
The sum of the eigenvalues equals the trace, and their product equals the determinant. Both follow from comparing λ² − (trace)λ + (det) with the factorised form. For rows (4, 1) and (2, 3) the trace is 7 and 5 + 2 = 7; the determinant is 12 − 2 = 10 and 5 × 2 = 10. B1 for the two statements, B1 for the trace check, B1 for the determinant check. It is the fastest check on a pair of eigenvalues you have just found.Explain what the eigenvectors of a 2 by 2 matrix mean geometrically for the transformation it represents.
Worked answer
Each eigenvector spans a line through the origin that the transformation maps onto itself, an invariant line. Points on it are stretched by the factor λ without changing direction, and a negative λ reverses them along the same line. B1 for the invariant line through the origin, B1 for the stretch by the factor λ, B1 for a negative λ reversing direction. A matrix with no real eigenvalues, such as a rotation through 90°, has no invariant line at all.S has rows (3, 1) and (1, 3). Find its eigenvalues and eigenvectors, and explain why a symmetric matrix with distinct eigenvalues always has perpendicular ones.
Worked answer
λ² − 6λ + 8 = 0 gives λ = 4 with eigenvector (1, 1) and λ = 2 with eigenvector (1, −1), whose scalar product is 0. In general, if Su = λu and Sv = μv with λ ≠ μ, then u·(Sv) = (Su)·v because S is symmetric, so μ(u·v) = λ(u·v). Since λ ≠ μ this forces u·v = 0. M1 for the characteristic equation, A1 for both eigenvalues, A1 for both eigenvectors, M1 for using symmetry to write u·(Sv) = (Su)·v, A1 for μ(u·v) = λ(u·v), A1 for concluding u·v = 0.
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