Home · A-Level Further Maths 9231 · Matrices: systems and eigenvalues · Eigenvalues and eigenvectors · Practice
AS & A Level · Practice questions · Eigenvalues and eigenvectors

Find them, check them, use them.

Six original Cambridge-style questions on eigenvalues and eigenvectors: characteristic equations of \(2 \times 2\) and \(3 \times 3\) matrices, the factor theorem on the cubic, vector products for eigenvectors, proofs from \(\mathbf{A}\mathbf{e} = \lambda\mathbf{e}\), and powers of a matrix acting on a vector.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
The method

Determinant, roots, vectors, check.

01
Recall
[4 marks]

Find the eigenvalues and corresponding eigenvectors of the matrix \(\begin{pmatrix}5&-2\\1&2\end{pmatrix}\).

  • \((5 - \lambda)(2 - \lambda) + 2 = \lambda^{2} - 7\lambda + 12 = (\lambda - 3)(\lambda - 4) = 0\), so \(\lambda = 3\) or \(4\). ✓
  • \(\lambda = 3\): \(\begin{pmatrix}2&-2\\1&-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \mathbf{0}\) gives \(x = y\): \(\begin{pmatrix}1\\1\end{pmatrix}\). ✓
  • \(\lambda = 4\): \(\begin{pmatrix}1&-2\\1&-2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix} = \mathbf{0}\) gives \(x = 2y\): \(\begin{pmatrix}2\\1\end{pmatrix}\). ✓
  • Check: \(\begin{pmatrix}5&-2\\1&2\end{pmatrix}\begin{pmatrix}2\\1\end{pmatrix} = \begin{pmatrix}8\\4\end{pmatrix} = 4\begin{pmatrix}2\\1\end{pmatrix}\). ✓
02
Calculation
[5 marks]

The matrix \(\mathbf{A} = \begin{pmatrix}a&4\\1&1\end{pmatrix}\) has an eigenvalue \(5\).

(a) Find the value of \(a\).

(b) Find the other eigenvalue and an eigenvector for each eigenvalue.

(c) Explain why \(\mathbf{A}\) has no inverse.

  • (a) \(\det(\mathbf{A} - 5\mathbf{I}) = (a - 5)(1 - 5) - 4 = -4(a - 5) - 4 = 0\), so \(a - 5 = -1\) and \(a = 4\). ✓
  • (b) \((4 - \lambda)(1 - \lambda) - 4 = \lambda^{2} - 5\lambda = \lambda(\lambda - 5) = 0\), so the other eigenvalue is \(0\). ✓
  • \(\lambda = 5\): \(-x + 4y = 0\), so \(\begin{pmatrix}4\\1\end{pmatrix}\). \(\lambda = 0\): \(4x + 4y = 0\), so \(\begin{pmatrix}1\\-1\end{pmatrix}\). ✓
  • Check: \(\mathbf{A}\begin{pmatrix}4\\1\end{pmatrix} = \begin{pmatrix}20\\5\end{pmatrix}\) and \(\mathbf{A}\begin{pmatrix}1\\-1\end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}\). ✓
  • (c) \(0\) is an eigenvalue, so putting \(\lambda = 0\) in the characteristic equation gives \(\det\mathbf{A} = 0\) (indeed \(4 \times 1 - 4 \times 1 = 0\)). A matrix with zero determinant has no inverse. ✓
03
Calculation
[8 marks]

The matrix \(\mathbf{A}\) is given by \[\mathbf{A} = \begin{pmatrix}0&1&2\\0&3&2\\2&0&2\end{pmatrix}.\]

(a) Show that the characteristic equation of \(\mathbf{A}\) is \(\lambda^{3} - 5\lambda^{2} + 2\lambda + 8 = 0\).

(b) Find the eigenvalues of \(\mathbf{A}\) and a corresponding eigenvector for each.

  • (a) Expand along the first row: \[\begin{aligned}&\det(\mathbf{A} - \lambda\mathbf{I})\\ &= -\lambda\big[(3 - \lambda)(2 - \lambda) - 0\big]\\ &\quad - 1\big[0 - 4\big] + 2\big[0 - 2(3 - \lambda)\big]\\ &= -\lambda(\lambda^{2} - 5\lambda + 6) + 4 - 12 + 4\lambda\\ &= -\lambda^{3} + 5\lambda^{2} - 2\lambda - 8\end{aligned}\] ✓
  • Setting this to zero and multiplying by \(-1\): \(\lambda^{3} - 5\lambda^{2} + 2\lambda + 8 = 0\), as required. ✓
  • (b) \(\lambda = -1\): \(-1 - 5 - 2 + 8 = 0\), so \(\lambda^{3} - 5\lambda^{2} + 2\lambda + 8 = (\lambda + 1)(\lambda^{2} - 6\lambda + 8) = (\lambda + 1)(\lambda - 2)(\lambda - 4)\). ✓
  • \(\lambda = -1\): \((1, 1, 2) \times (0, 4, 2) = (2 - 8, 0 - 2, 4 - 0) = (-6, -2, 4)\), so \(\begin{pmatrix}3\\1\\-2\end{pmatrix}\). ✓
  • \(\lambda = 2\): \((-2, 1, 2) \times (0, 1, 2) = (2 - 2, 0 + 4, -2 - 0) = (0, 4, -2)\), so \(\begin{pmatrix}0\\2\\-1\end{pmatrix}\). ✓
  • \(\lambda = 4\): \((-4, 1, 2) \times (0, -1, 2) = (2 + 2, 0 + 8, 4 - 0) = (4, 8, 4)\), so \(\begin{pmatrix}1\\2\\1\end{pmatrix}\). ✓
  • Check: \(\mathbf{A}\begin{pmatrix}3\\1\\-2\end{pmatrix} = \begin{pmatrix}-3\\-1\\2\end{pmatrix}\), \(\mathbf{A}\begin{pmatrix}0\\2\\-1\end{pmatrix} = \begin{pmatrix}0\\4\\-2\end{pmatrix}\), \(\mathbf{A}\begin{pmatrix}1\\2\\1\end{pmatrix} = \begin{pmatrix}4\\8\\4\end{pmatrix}\); also \(-1 + 2 + 4 = 5\), the diagonal sum. ✓
04
Proof
[4 marks]

The square matrices \(\mathbf{A}\) and \(\mathbf{B}\) have a common eigenvector \(\mathbf{e}\), with \(\mathbf{A}\mathbf{e} = \lambda\mathbf{e}\) and \(\mathbf{B}\mathbf{e} = \mu\mathbf{e}\).

(a) Prove that \(\mathbf{e}\) is an eigenvector of \(\mathbf{A}\mathbf{B}\) and state the corresponding eigenvalue.

(b) Prove that \(\mathbf{e}\) is an eigenvector of \(\mathbf{A} + \mathbf{B}\) and state the corresponding eigenvalue.

  • (a) \(\mathbf{A}\mathbf{B}\mathbf{e} = \mathbf{A}(\mu\mathbf{e}) = \mu\mathbf{A}\mathbf{e} = \mu\lambda\mathbf{e}\). ✓
  • \(\mathbf{e} \ne \mathbf{0}\) because it is an eigenvector, so \(\mathbf{e}\) is an eigenvector of \(\mathbf{A}\mathbf{B}\) with eigenvalue \(\lambda\mu\). ✓
  • (b) \((\mathbf{A} + \mathbf{B})\mathbf{e} = \mathbf{A}\mathbf{e} + \mathbf{B}\mathbf{e} = \lambda\mathbf{e} + \mu\mathbf{e} = (\lambda + \mu)\mathbf{e}\). ✓
  • So \(\mathbf{e}\) is an eigenvector of \(\mathbf{A} + \mathbf{B}\) with eigenvalue \(\lambda + \mu\). Both proofs need the SAME \(\mathbf{e}\) for \(\mathbf{A}\) and \(\mathbf{B}\). ✓
05
Analysis
[7 marks]

The matrix \(\mathbf{A}\) is given by \[\mathbf{A} = \begin{pmatrix}3&1&3\\1&3&1\\0&0&3\end{pmatrix}.\]

(a) Find the eigenvalues of \(\mathbf{A}\) and a corresponding eigenvector for each.

(b) Hence write down the eigenvalues of \(\mathbf{A}^{-1}\) and of \(\mathbf{A}^{3} - 5\mathbf{A}\), justifying the second.

  • (a) Expand along the third row: \[\begin{aligned}\det(\mathbf{A} - \lambda\mathbf{I}) &= (3 - \lambda)\big[(3 - \lambda)^{2} - 1\big]\\ &= (3 - \lambda)(2 - \lambda)(4 - \lambda)\end{aligned}\] using \(u^{2} - 1 = (u - 1)(u + 1)\). So \(\lambda = 2\), \(3\), \(4\). ✓
  • \(\lambda = 2\): \((1, 1, 3) \times (1, 1, 1) = (1 - 3, 3 - 1, 1 - 1) = (-2, 2, 0)\), so \(\begin{pmatrix}1\\-1\\0\end{pmatrix}\). ✓
  • \(\lambda = 3\): \((0, 1, 3) \times (1, 0, 1) = (1 - 0, 3 - 0, 0 - 1) = (1, 3, -1)\). ✓
  • \(\lambda = 4\): \((-1, 1, 3) \times (1, -1, 1) = (1 + 3, 3 + 1, 1 - 1) = (4, 4, 0)\), so \(\begin{pmatrix}1\\1\\0\end{pmatrix}\). ✓
  • Check: \(\mathbf{A}\begin{pmatrix}1\\3\\-1\end{pmatrix} = \begin{pmatrix}3 + 3 - 3\\1 + 9 - 1\\-3\end{pmatrix} = \begin{pmatrix}3\\9\\-3\end{pmatrix}\). ✓
  • (b) \(\mathbf{A}^{-1}\): \(\tfrac{1}{2}\), \(\tfrac{1}{3}\), \(\tfrac{1}{4}\). ✓
  • \((\mathbf{A}^{3} - 5\mathbf{A})\mathbf{e} = \lambda^{3}\mathbf{e} - 5\lambda\mathbf{e} = (\lambda^{3} - 5\lambda)\mathbf{e}\), giving \(8 - 10 = -2\), \(27 - 15 = 12\) and \(64 - 20 = 44\), with the same eigenvectors. ✓
06
Analysis
[7 marks]

The transformation T of the plane has matrix \(\mathbf{M} = \begin{pmatrix}1&2\\2&-2\end{pmatrix}\).

(a) Find the eigenvalues and corresponding eigenvectors of \(\mathbf{M}\).

(b) State the equations of the invariant lines of T through the origin, and show that they are perpendicular.

(c) Write \(\begin{pmatrix}4\\-3\end{pmatrix}\) as a combination of your eigenvectors, and hence find \(\mathbf{M}^{5}\begin{pmatrix}4\\-3\end{pmatrix}\) without working out \(\mathbf{M}^{5}\).

  • (a) \((1 - \lambda)(-2 - \lambda) - 4 = \lambda^{2} + \lambda - 6 = (\lambda + 3)(\lambda - 2) = 0\), so \(\lambda = 2\) or \(-3\). ✓
  • \(\lambda = 2\): \(-x + 2y = 0\), so \(\begin{pmatrix}2\\1\end{pmatrix}\). \(\lambda = -3\): \(4x + 2y = 0\), so \(\begin{pmatrix}1\\-2\end{pmatrix}\). ✓
  • (b) The invariant lines are \(y = \tfrac{1}{2}x\) and \(y = -2x\). The gradients multiply to \(\tfrac{1}{2} \times (-2) = -1\), so they are perpendicular. ✓
  • (c) \(a\begin{pmatrix}2\\1\end{pmatrix} + b\begin{pmatrix}1\\-2\end{pmatrix} = \begin{pmatrix}4\\-3\end{pmatrix}\): \(2a + b = 4\) and \(a - 2b = -3\) give \(a = 1\), \(b = 2\). ✓
  • \(\mathbf{M}^{5}\) multiplies each eigenvector by \(\lambda^{5}\): \(2^{5} = 32\) and \((-3)^{5} = -243\). ✓
  • \(\mathbf{M}^{5}\begin{pmatrix}4\\-3\end{pmatrix} = 32\begin{pmatrix}2\\1\end{pmatrix} + 2(-243)\begin{pmatrix}1\\-2\end{pmatrix} = \begin{pmatrix}64 - 486\\32 + 972\end{pmatrix} = \begin{pmatrix}-422\\1004\end{pmatrix}\). ✓

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