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Diagonalisation and the Cayley-Hamilton theorem questions

Change coordinates so the eigenvectors become the axes and the matrix turns diagonal, making high powers trivial. And every matrix, it turns out, satisfies its own characteristic equation.

6 original questions · 23 marks · the diagonalisation and the cayley-hamilton theorem notes · Further Pure 2

Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.

• 2 marks · warm-up

  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.

  Worked 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. Mismatching the order is the standard error, and it costs both marks.

• 3 marks · state · warm-up

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

  Worked 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 is one mark.

• 5 marks

  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.

  Worked 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. Reversing P and P⁻¹ gives a different matrix and loses the accuracy marks, since matrix multiplication does not commute.

• 4 marks

  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.

  Worked 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. The method needs no cofactors.

• 3 marks · give

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

  Worked 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.

• 6 marks · stretch

  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.

  Worked 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. Setting up the pair of equations from the two eigenvalues is where the method marks sit; grinding out M³, M⁴, M⁵ one at a time reaches the same place far more slowly.
  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.

The same practice on paper: the printable workbook for this topic, questions and a worked answer book.

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