Questions on diagonalisation and the Cayley-Hamilton theorem

Question 1

The matrix M has rows (4, 1) and (2, 3), with eigenvalues 5 and 2 and corresponding eigenvectors (1, 1) and (1, −2). Write down matrices P and D such that M = PDP⁻¹.

Accepted Answer

Put the eigenvectors in as the columns of P, so P has rows (1, 1) and (1, −2). Then D is diagonal with the matching eigenvalues in the same order, 5 then 2. B1 for P, B1 for D in the matching order. Swapping the columns of P works provided the entries of D are swapped with them.

Question 2

State the Cayley-Hamilton theorem, and verify it for the matrix M with rows (4, 1) and (2, 3).

Accepted Answer

Every square matrix satisfies its own characteristic equation. Here det(M − λI) = (4 − λ)(3 − λ) − 2 = λ² − 7λ + 10, so the claim is M² − 7M + 10I = 0.
M² has rows (18, 7) and (14, 11), 7M has rows (28, 7) and (14, 21), and 10I has 10 on the diagonal. Then M² − 7M + 10I has entries 18 − 28 + 10, 7 − 7, 14 − 14 and 11 − 21 + 10, all zero, as required. B1 for the statement of the theorem, M1 for forming M² − 7M + 10I, A1 for the zero matrix. Verifying means reaching the zero matrix; quoting the characteristic equation alone does not verify it.

Question 3

State the condition an n × n matrix must satisfy to be diagonalisable, and state what the columns of P and the entries of D are when it holds.

Accepted Answer

The matrix must have n linearly independent eigenvectors. B1 for that condition. The columns of P are those n eigenvectors and the diagonal entries of D are the matching eigenvalues, taken in the same order, so that P⁻¹AP = D. B1 for the columns and the matching order. Independence is what makes P invertible, which is why it is the condition rather than the eigenvalues being distinct: distinct eigenvalues are sufficient but not necessary.

Question 4

A matrix A satisfies A = PDP⁻¹, where D is the diagonal matrix with entries 2 and 5. Write down the eigenvalues of A, the eigenvalues of A³, and the value of det A.

Accepted Answer

The diagonal entries of D are the eigenvalues of A, so they are 2 and 5. Raising to a power raises each eigenvalue to that power, since A³ = PD³P⁻¹, so the eigenvalues of A³ are 2³ = 8 and 5³ = 125. The determinant is the product of the eigenvalues, 2 × 5 = 10. B1 for 2 and 5, B1 for 8 and 125, B1 for 10.

Question 5

The matrix M with rows (4, 1) and (2, 3) has eigenvalues 5 and 2 with eigenvectors (1, 1) and (1, −2). Use a diagonalisation to find M⁵.

Accepted Answer

Take P with rows (1, 1) and (1, −2) and D diagonal with 5 and 2, so that M = PDP⁻¹ and M⁵ = PD⁵P⁻¹. D⁵ is diagonal with entries 3125 and 32; only the diagonal entries are ever raised to a power.
Since det P = −3, P⁻¹ is (−1/3) times the matrix with rows (−2, −1) and (−1, 1).
Multiplying the three matrices in the order P, then D⁵, then P⁻¹ gives M⁵ with rows (2094, 1031) and (2062, 1063). M1 for using M raised to a power as P times D to that power times P⁻¹, A1 for 3125 and 32, M1 for P⁻¹, M1 for the triple product, A1 for the four entries. Matrix multiplication does not commute, so reversing P and P⁻¹ gives a different matrix.

Question 6

The matrix M has rows (4, 1) and (2, 3). Use the Cayley-Hamilton theorem to express M⁻¹ in terms of M and I, and hence evaluate M⁻¹.

Accepted Answer

The characteristic equation is λ² − 7λ + 10 = 0, so M² − 7M + 10I = 0. Multiply throughout by M⁻¹: M − 7I + 10M⁻¹ = 0, so M⁻¹ = (7I − M)/10. That is one tenth of the matrix with rows (3, −1) and (−2, 4). M1 for multiplying the relation through by M⁻¹, A1 for M⁻¹ = (7I − M)/10, M1 for evaluating 7I − M, A1 for the entries. Checking, M times that matrix is 10I, so dividing by 10 gives the identity.

Question 7

Give a 2 by 2 matrix that cannot be diagonalised, and explain what goes wrong.

Accepted Answer

Take the matrix with rows (1, 1) and (0, 1). Its characteristic equation is (1 − λ)² = 0, so 1 is a repeated eigenvalue, but solving (M − I)v = 0 gives only multiples of (1, 0). With one independent eigenvector there is no invertible P of eigenvectors, so no diagonalisation exists. B1 for a suitable matrix, M1 for solving for the eigenvectors, A1 for only one independent eigenvector, so no invertible P. A repeated eigenvalue is a warning rather than a verdict, since some repeated cases still supply two independent eigenvectors.

Question 8

The matrix N has rows (5, −2) and (6, −2). Show that N² = 3N − 2I, and hence express N⁴ in the form aN + bI, giving the entries of N⁴.

Accepted Answer

det(N − λI) = (5 − λ)(−2 − λ) + 12 = λ² − 3λ + 2, so by the Cayley-Hamilton theorem N² − 3N + 2I = 0, that is N² = 3N − 2I. Squaring that, N⁴ = (3N − 2I)² = 9N² − 12N + 4I, and replacing N² again gives 9(3N − 2I) − 12N + 4I = 15N − 14I. Then 15N has rows (75, −30) and (90, −30), and taking 14 off each diagonal entry leaves N⁴ with rows (61, −30) and (90, −44). M1 for the characteristic equation, A1 for N² = 3N − 2I, M1 for squaring and reducing again, A1 for 15N − 14I with its entries.

Question 9

The symmetric matrix S has rows (5, 2) and (2, 2). Find an orthogonal matrix P and a diagonal matrix D such that P^TSP = D.

Accepted Answer

The trace is 7 and the determinant is 10 − 4 = 6, so the characteristic equation is λ² − 7λ + 6 = 0 and λ = 6 or λ = 1. For λ = 6 the first row gives −x + 2y = 0, so (2, 1) serves; for λ = 1 it gives 4x + 2y = 0, so (1, −2) serves. Their scalar product is 2 − 2 = 0, as a symmetric matrix with distinct eigenvalues guarantees, and each has length √5. So take P = (1/√5) times the matrix with rows (2, 1) and (1, −2), and D diagonal with 6 then 1. M1 for the characteristic equation, A1 for both eigenvalues, A1 for both eigenvectors, B1 for P normalised with D in the matching order.

Question 10

A 3 × 3 matrix A has characteristic equation λ³ − 2λ² − λ + 2 = 0. Use the Cayley-Hamilton theorem to express A⁴ in the form aA² + bA + cI.

Accepted Answer

Cayley-Hamilton gives A³ − 2A² − A + 2I = 0, so A³ = 2A² + A − 2I. Multiplying through by A gives A⁴ = 2A³ + A² − 2A, and replacing A³ by the line above gives A⁴ = 2(2A² + A − 2I) + A² − 2A = 5A² − 4I, so a = 5, b = 0 and c = −4. M1 for the Cayley-Hamilton relation, M1 for multiplying through by A, A1 for substituting for A³, A1 for 5A² − 4I. A check costs one line: the equation factorises as (λ − 1)(λ + 1)(λ − 2) = 0, and each of λ = 1, −1 and 2 satisfies λ⁴ = 5λ² − 4.

Question 11

The matrix M with rows (4, 1) and (2, 3) has eigenvalues 5 and 2. Show that for every positive integer n, Mⁿ = a_nM + b_nI where a_n = (5ⁿ − 2ⁿ)/3, and find b_n. Hence evaluate M⁶.

Accepted Answer

Cayley-Hamilton gives M² = 7M − 10I, so every power collapses onto M and I, and Mⁿ = a_nM + b_nI for some numbers a_n and b_n.
To find them, use the eigenvalues. If Mv = λv then Mⁿv = λⁿv, and applying the relation to that same eigenvector gives λⁿ = a_nλ + b_n. Both eigenvalues obey it, so 5a_n + b_n = 5ⁿ and 2a_n + b_n = 2ⁿ.
Subtracting, 3a_n = 5ⁿ − 2ⁿ, which is the given a_n, and back-substituting gives b_n = (5·2ⁿ − 2·5ⁿ)/3.
At n = 6: a = (15625 − 64)/3 = 5187 and b = (320 − 31250)/3 = −10310, so M⁶ = 5187M − 10310I, with rows (10438, 5187) and (10374, 5251). B1 for M² = 7M − 10I, M1 for applying the relation to an eigenvector, A1 for the pair of simultaneous equations, A1 for b_n, M1 for substituting n = 6, A1 for the entries of M⁶.
A check at n = 2 gives a = 7 and b = −10, which is the Cayley-Hamilton relation itself.

Question 12

The symmetric matrix S has rows (3, 1) and (1, 3), with eigenvalues 4 and 2 and eigenvectors (1, 1) and (1, −1). Find an orthogonal matrix P and a diagonal matrix D such that P^TSP = D, and verify that P^TP = I.

Accepted Answer

The eigenvectors are already perpendicular, as a symmetric matrix with distinct eigenvalues guarantees, so they only need normalising: each has length √2. Take P = (1/√2) times the matrix with rows (1, 1) and (1, −1), and D diagonal with 4 then 2, matching the column order. P^TP has entries (1/2)(1 + 1), (1/2)(1 − 1) and their mirror images, which is I, so P is orthogonal and P⁻¹ = P^T with no inversion to do. Then P^TSP = D. M1 for normalising the eigenvectors by √2, A1 for P, B1 for D with the entries in the matching order, M1 for computing P^TP, A1 for reaching I and concluding P⁻¹ = P^T. The 1/√2 is required: the columns of an orthogonal matrix must be unit vectors, not merely perpendicular ones.

Question 13

Hence show that Sⁿ is one half of the matrix with rows (4ⁿ + 2ⁿ, 4ⁿ − 2ⁿ) and (4ⁿ − 2ⁿ, 4ⁿ + 2ⁿ), and check the case n = 2 directly.

Accepted Answer

S = PDP^T, so Sⁿ = PDⁿP^T, with Dⁿ diagonal with entries 4ⁿ and 2ⁿ. The two factors of 1/√2 combine into 1/2. Multiplying out, PDⁿ is (1/√2) times the matrix with rows (4ⁿ, 2ⁿ) and (4ⁿ, −2ⁿ), and applying P^T gives Sⁿ = (1/2) times the matrix with rows (4ⁿ + 2ⁿ, 4ⁿ − 2ⁿ) and (4ⁿ − 2ⁿ, 4ⁿ + 2ⁿ), as required. At n = 2 the formula gives (1/2) times rows (20, 12) and (12, 20), that is rows (10, 6) and (6, 10), and squaring S directly gives the same. M1 for Sⁿ = PDⁿP^T, A1 for Dⁿ, M1 for the double product, A1 for the stated form, B1 for the n = 2 check both ways. As n grows the 4ⁿ term swamps the 2ⁿ term, so Sⁿ approaches 4ⁿ/2 times the matrix of ones: the eigenvalue of largest modulus takes over.

Question 14

The matrix K has rows (2, 0, 1), (1, 1, 0) and (0, 1, 2). Show that its characteristic equation is λ³ − 5λ² + 8λ − 5 = 0, and use the Cayley-Hamilton theorem to find K⁻¹.

Accepted Answer

Expanding det(K − λI) along the top row: (2 − λ)(1 − λ)(2 − λ) + 1 × 1 = (1 − λ)(2 − λ)² + 1 = −λ³ + 5λ² − 8λ + 5, so the characteristic equation is λ³ − 5λ² + 8λ − 5 = 0. Cayley-Hamilton applies to 3 by 3 matrices exactly as to 2 by 2, so K³ − 5K² + 8K − 5I = 0. Multiply through by K⁻¹: K² − 5K + 8I − 5K⁻¹ = 0, so K⁻¹ = (K² − 5K + 8I)/5.
K² has rows (4, 1, 4), (3, 1, 1) and (1, 3, 4), so K² − 5K + 8I has rows (2, 1, −1), (−2, 4, 1) and (1, −2, 2), and K⁻¹ is one fifth of that. Multiplying K by it returns I, which is the check. M1 for expanding det(K − λI), A1 for λ³ − 5λ² + 8λ − 5 = 0, B1 for K³ − 5K² + 8K − 5I = 0, M1 for multiplying through by K⁻¹ and rearranging, A1 for K⁻¹ = (K² − 5K + 8I)/5, M1 for computing K², A1 for the nine entries. The constant term −5 is det K up to sign, which is why it is safe to divide by it: a zero constant term would mean a singular matrix and no inverse at all.

Question 15

A colony of 500 birds is spread over two islands. Each year 20% of the birds on island X move to island Y and 30% of those on island Y move to island X, and none are lost. At the start all 500 are on island X. Taking M to be the matrix with rows (0.8, 0.3) and (0.2, 0.7), find how the colony is divided between the islands in the long run.

Accepted Answer

Writing x_n and y_n for the numbers on the two islands after n years, x_n+1 = 0.8x_n + 0.3y_n and y_n+1 = 0.2x_n + 0.7y_n, so the state vector satisfies v_n+1 = Mv_n and v_n = Mⁿv₀. B1 for setting the model up as repeated multiplication by M.
The trace is 1.5 and the determinant is 0.56 − 0.06 = 0.5, so the characteristic equation is λ² − 1.5λ + 0.5 = 0 and λ = 1 or λ = 0.5. M1 for the characteristic equation, A1 for the two eigenvalues. For λ = 1 the first row gives −0.2x + 0.3y = 0, that is 2x = 3y, so (3, 2) serves; for λ = 0.5 it gives 0.3x + 0.3y = 0, so (1, −1) serves. A1 for both eigenvectors.
Now write the starting state in terms of them: a(3, 2) + b(1, −1) = (500, 0) gives 3a + b = 500 and 2a − b = 0, so a = 100 and b = 200. M1 for expressing v₀ in the eigenvector basis. Then v_n = 100(3, 2) + 200(0.5)ⁿ(1, −1), and 0.5ⁿ → 0, so the colony settles at 300 on island X and 200 on island Y. A1 for the limit.
The eigenvalue 1 is what makes a steady division exist at all, and its eigenvector fixes the ratio 3 : 2 whatever the starting split; the eigenvalue 0.5 measures how fast the memory of that split fades. After four years the model already reads 312.5 and 187.5.