Questions on eigenvalues and eigenvectors

Question 1

M has rows (4, 1) and (2, 3). Find its eigenvalues and a corresponding eigenvector for each.

Accepted 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.

Question 2

Prove that if λ is an eigenvalue of M with eigenvector v, then λ² is an eigenvalue of M² with the same eigenvector.

Accepted 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 λ^k for M^k.

Question 3

Explain what it means to say that λ is an eigenvalue of a matrix M with eigenvector v, and state why v is required to be non-zero.

Accepted Answer

It means that Mv = λv: the matrix sends v to a multiple of itself, so v keeps its direction and is scaled by λ. B1 for Mv = λv. The zero vector is excluded because M0 = λ0 holds for every number λ, so allowing it would make every number an eigenvalue and the definition would say nothing. B1 for that reason. The eigenvalue λ may be zero; it is the vector that may not.

Question 4

The matrix M has rows (2, 0) and (0, −3). Write down its eigenvalues and an eigenvector for each, and describe the transformation M represents.

Accepted Answer

A diagonal matrix carries its eigenvalues on the diagonal, so they are 2 and −3. M(1, 0) = (2, 0) = 2(1, 0) and M(0, 1) = (0, −3) = −3(0, 1), so (1, 0) and (0, 1) serve. B1 for the eigenvalues, B1 for the two eigenvectors. The transformation stretches by a factor 2 parallel to the x-axis and by a factor 3 parallel to the y-axis, reversing that second direction because the eigenvalue is negative; both axes are invariant lines. B1 for the description.

Question 5

A has rows (2, 0, 1), (0, 3, 0) and (1, 0, 2). Find its eigenvalues and an eigenvector for each distinct one.

Accepted 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.

Question 6

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.

Accepted 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.

Question 7

Explain what the eigenvectors of a 2 by 2 matrix mean geometrically for the transformation it represents.

Accepted 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.

Question 8

The matrix M with rows (a, 2) and (3, b) has eigenvalue 5 with eigenvector (1, 1). Find a and b, then find the other eigenvalue and an eigenvector for it.

Accepted Answer

Mv = 5v reads a + 2 = 5 and 3 + b = 5, so a = 3 and b = 2. The trace is then 5, and the eigenvalues sum to the trace, so the other eigenvalue is 5 − 5 = 0; the determinant 6 − 6 = 0 says the same thing, since the product of the eigenvalues must vanish. For λ = 0, solve Mv = 0: the row 3x + 2y = 0 gives (2, −3), and M(2, −3) = (0, 0) confirms it. M1 for the two equations from Mv = 5v, A1 for a = 3 and b = 2, B1 for the other eigenvalue 0 from the trace or the determinant, M1 for solving Mv = 0, A1 for (2, −3) or any non-zero multiple. A zero eigenvalue always means a singular matrix: the whole line spanned by (2, −3) is squashed onto the origin.

Question 9

The matrix N has rows (3, 1) and (−1, 5). Show that N has a repeated eigenvalue, and find every eigenvector of N.

Accepted Answer

The trace is 8 and the determinant is 15 + 1 = 16, so the characteristic equation is λ² − 8λ + 16 = 0, that is (λ − 4)² = 0. The discriminant 64 − 64 is zero, so λ = 4 is a repeated eigenvalue. Solving (N − 4I)v = 0, both rows of N − 4I read (−1, 1), so the single equation is −x + y = 0 and the eigenvectors are the non-zero multiples of (1, 1). M1 for the characteristic equation, A1 for the repeated root 4, M1 for solving (N − 4I)v = 0, A1 for the multiples of (1, 1). A repeated eigenvalue does not have to come with two independent eigenvectors; here the two rows of N − 4I coincide, leaving one equation and one direction, so N cannot be diagonalised.

Question 10

Find the eigenvalues of the matrix with rows (1, −2) and (2, 1), and explain what their form says about the invariant lines of the transformation and about the field the calculation is made over.

Accepted Answer

The trace is 2 and the determinant is 1 + 4 = 5, so the characteristic equation is λ² − 2λ + 5 = 0. Its discriminant is 4 − 20 = −16, so λ = (2 ± 4i)/2 = 1 ± 2i, a conjugate pair. M1 for the characteristic equation, A1 for the two eigenvalues. Over the real numbers there is no eigenvalue and no eigenvector, so no line through the origin is mapped to itself: as a transformation of the real plane this matrix turns every direction off its own line. B1 for no real invariant line. The eigenvalues exist over the complex numbers, where the eigenvectors (1, −i) and (1, i) go with them, so how many eigenvalues the matrix has depends on the field the calculation is made over. B1 for naming the field.

Question 11

The matrix A has rows (2, 2, 1), (1, 3, 1) and (1, 2, 2). Verify that (1, 1, 1) is an eigenvector of A and state the eigenvalue. Show that 1 is also an eigenvalue of A, and find two independent eigenvectors for it.

Accepted Answer

A(1, 1, 1) = (2 + 2 + 1, 1 + 3 + 1, 1 + 2 + 2) = (5, 5, 5) = 5(1, 1, 1), so (1, 1, 1) is an eigenvector with eigenvalue 5. M1 for the multiplication, A1 for 5.
For the second part, A − I has rows (1, 2, 1), (1, 2, 1) and (1, 2, 1). Three identical rows make the determinant zero, so det(A − I) = 0 and 1 is an eigenvalue. M1 for forming A − I and observing that its rows coincide. The single equation left is x + 2y + z = 0, a plane, so any two independent solutions serve: (2, −1, 0) and (1, 0, −1) both satisfy it and neither is a multiple of the other. A1 for the plane, A1 for two independent eigenvectors. The trace 7 equals 5 + 1 + 1 and the determinant 5 equals 5 × 1 × 1, so 1 is repeated, and this time the repeated eigenvalue does carry two independent directions.

Question 12

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.

Accepted 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.

Question 13

R has rows (cos θ, −sin θ) and (sin θ, cos θ), a rotation through the angle θ. Show that the eigenvalues of R are cos θ ± i sin θ, and state for which θ they are real, interpreting that case geometrically.

Accepted Answer

det(R − λI) = 0 gives (cos θ − λ)² + sin²θ = 0, using (−sin θ)(sin θ) = −sin²θ. So (cos θ − λ)² = −sin²θ, and taking square roots gives λ = cos θ ± i sin θ. The eigenvalues are real only when sin θ = 0, that is θ = 0 or π (up to full turns): the identity, where every direction is invariant with λ = 1, and the half-turn, where every line through the origin maps to itself with λ = −1. Any other rotation turns every real direction off its own line. M1 for forming det(R − λI) = 0, A1 for (cos θ − λ)² + sin²θ = 0, A1 for λ = cos θ ± i sin θ, B1 for sin θ = 0, so θ = 0 or π, with the geometric reading.

Question 14

A has rows (2, 1, 1), (0, 1, 3) and (0, 2, 2). Show that A has three distinct integer eigenvalues, find an eigenvector for each, and give a normalised eigenvector for the largest eigenvalue.

Accepted Answer

The zeros in the first column invite expanding det(A − λI) down it: det(A − λI) = (2 − λ)[(1 − λ)(2 − λ) − 6] = (2 − λ)(λ² − 3λ − 4) = (2 − λ)(λ − 4)(λ + 1), so λ = 2, 4 or −1, three distinct integers.
For λ = 2 the remaining rows force y = z = 0 with x free, so (1, 0, 0). For λ = 4 the rows (0, −3, 3) and (0, 2, −2) give y = z, and −2x + y + z = 0 then gives x = y, so (1, 1, 1); indeed A(1, 1, 1) = (4, 4, 4). For λ = −1 the rows give 2y + 3z = 0, so take y = 9, z = −6, and 3x + y + z = 0 gives x = −1, so (−1, 9, −6), and A(−1, 9, −6) = (1, −9, 6), which is −1 times it.
The eigenvector for the largest eigenvalue 4 is (1, 1, 1), of length √3, so the normalised eigenvector is (1/√3)(1, 1, 1). M1 for expanding down the first column, A1 for the factorised cubic, A1 for 2, 4 and −1, M1 for solving (A − λI)v = 0, A1 for (1, 0, 0) and (1, 1, 1), A1 for (−1, 9, −6) or any non-zero multiple, B1 for (1/√3)(1, 1, 1). A check: the trace 2 + 1 + 2 = 5 equals 2 + 4 + (−1).

Question 15

M is a 2 × 2 matrix with real entries. Prove that its eigenvalues are either both real or a pair of complex conjugates, and prove that if λ is a non-real eigenvalue with eigenvector v then the conjugate of λ is an eigenvalue with the conjugate of v as an eigenvector.

Accepted Answer

The characteristic equation is det(M − λI) = 0, which for a 2 × 2 matrix is λ² − (trace)λ + (det) = 0, a quadratic whose coefficients are real because the entries of M are. M1 for the characteristic quadratic, A1 for its coefficients being real. Its roots are given by the quadratic formula: when the discriminant is positive or zero they are real, and when it is negative they are p ± iq with p and q real and q non-zero, a conjugate pair. A1 for the discriminant argument covering both cases.
For the second part, start from Mv = λv and take the conjugate of every entry on both sides. Conjugation respects sums and products, so the conjugate of a matrix product is the product of the conjugates, and M is real and equal to its own conjugate. Hence Mv* = λ*v*, where the star denotes the conjugate. M1 for conjugating the equation, A1 for Mv* = λ*v*. Since v is non-zero so is v*, so λ* is an eigenvalue with eigenvector v*. A1 for the conclusion.