Home / A-Level Further Maths 9231 / Matrices: systems and eigenvalues / Diagonalisation and the Cayley-Hamilton theorem
Cambridge International AS & A Level · Topic 2.2
A-Level Further Maths 9231 / Topic 2.2 / AS & A Level

Line up the eigenvectors, and powers become easy.

Put the eigenvectors of a matrix side by side and it splits as \(A = QDQ^{-1}\), with the eigenvalues on the diagonal of \(D\). Then \(A^{n} = QD^{n}Q^{-1}\) needs only powers of numbers, and the Cayley-Hamilton theorem turns the characteristic equation into a shortcut for high powers and for the inverse.

Topic 2.2All matrices: systems and eigenvalues lessons →
The key idea

If \(A\) has eigenvalues \(\lambda_{1}, \lambda_{2}, \ldots\) with eigenvectors \(\mathbf{v}_{1}, \mathbf{v}_{2}, \ldots\), put the eigenvectors as the columns of \(Q\) and the eigenvalues, in the same order, on the diagonal of \(D\). Then \(AQ = QD\), so \(A = QDQ^{-1}\) and \(A^{n} = QD^{n}Q^{-1}\), where \(D^{n}\) just raises each diagonal entry to the power \(n\). Cayley-Hamilton: every square matrix satisfies its own characteristic equation, so if \(\lambda^{2} - 7\lambda + 10 = 0\) then \(A^{2} - 7A + 10I = \mathbf{0}\).

x y 1 1 v₁ Av₁ = 5v₁ v₂ Av₂ = 2v₂ u Au A = (4 1 ; 2 3) v₁ = (1, 1), λ = 5 stretched along its own line v₂ = (1, −2), λ = 2 stretched along its own line u = (1, 0) → (4, 2) not an eigenvector: it turns Q = (v₁ v₂) and D = diag(5, 2) give A = QDQ⁻¹
Fig. 1: the matrix A = (4 1; 2 3) maps each eigenvector to a multiple of itself, (1, 1) to 5 times itself and (1, -2) to twice itself, while a vector such as (1, 0) changes direction. Using the two eigenvector directions as new axes, A is just a stretch by 5 and by 2, which is what A = QDQ⁻¹ says
Section 01

Eigenvectors side by side.

Suppose \(A\mathbf{v}_{1} = \lambda_{1}\mathbf{v}_{1}\) and \(A\mathbf{v}_{2} = \lambda_{2}\mathbf{v}_{2}\). Build \(Q\) with \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) as its columns. Multiplying \(A\) by \(Q\) acts on each column separately, so the columns of \(AQ\) are \(\lambda_{1}\mathbf{v}_{1}\) and \(\lambda_{2}\mathbf{v}_{2}\). Those are exactly the columns of \(QD\), where \(D\) is diagonal with \(\lambda_{1}, \lambda_{2}\) on the diagonal in that order. So \(AQ = QD\).

When the eigenvalues are real and distinct, the eigenvectors are independent, so \(\det Q \ne 0\) and \(Q^{-1}\) exists. Multiply \(AQ = QD\) on the right by \(Q^{-1}\):

\[A = QDQ^{-1}.\]

StepWhat you do
1Solve the characteristic equation \(\det(A - \lambda I) = 0\) for the eigenvalues.
2For each \(\lambda\), solve \((A - \lambda I)\mathbf{v} = \mathbf{0}\) for an eigenvector (any non-zero multiple will do).
3Write the eigenvectors as the COLUMNS of \(Q\).
4Write the eigenvalues on the diagonal of \(D\), in the SAME order as their eigenvectors in \(Q\).

The answer is not unique: you may reorder the columns (reordering \(D\) to match) or scale any eigenvector. A question that says 'find matrices \(Q\) and \(D\) such that \(A = QDQ^{-1}\)' accepts any consistent choice. Some questions write it as \(Q^{-1}AQ = D\); this is the same statement rearranged.

Worked example Diagonalise a 2 x 2 matrix
Find a matrix \(Q\) and a diagonal matrix \(D\) such that \(\begin{pmatrix}4&1\\2&3\end{pmatrix} = QDQ^{-1}\), and find \(Q^{-1}\).
  1. \(\det(A - \lambda I) = (4 - \lambda)(3 - \lambda) - 2 = \lambda^{2} - 7\lambda + 10 = (\lambda - 5)(\lambda - 2)\), so \(\lambda = 5\) or \(\lambda = 2\).
  2. \(\lambda = 5\): \(\begin{pmatrix}-1&1\\2&-2\end{pmatrix}\mathbf{v} = \mathbf{0}\) gives \(y = x\), so \(\mathbf{v}_{1} = \begin{pmatrix}1\\1\end{pmatrix}\).
  3. \(\lambda = 2\): \(\begin{pmatrix}2&1\\2&1\end{pmatrix}\mathbf{v} = \mathbf{0}\) gives \(y = -2x\), so \(\mathbf{v}_{2} = \begin{pmatrix}1\\-2\end{pmatrix}\).
  4. Eigenvectors as columns, eigenvalues in the same order: \(Q = \begin{pmatrix}1&1\\1&-2\end{pmatrix}\), \(D = \begin{pmatrix}5&0\\0&2\end{pmatrix}\).
  5. \(\det Q = -2 - 1 = -3\), so \(Q^{-1} = -\tfrac{1}{3}\begin{pmatrix}-2&-1\\-1&1\end{pmatrix} = \tfrac{1}{3}\begin{pmatrix}2&1\\1&-1\end{pmatrix}\).
  6. Check one column: \(A\begin{pmatrix}1\\-2\end{pmatrix} = \begin{pmatrix}4 - 2\\2 - 6\end{pmatrix} = \begin{pmatrix}2\\-4\end{pmatrix} = 2\begin{pmatrix}1\\-2\end{pmatrix}\).
Answer: \(Q = \begin{pmatrix}1&1\\1&-2\end{pmatrix}\), \(D = \begin{pmatrix}5&0\\0&2\end{pmatrix}\), \(Q^{-1} = \tfrac{1}{3}\begin{pmatrix}2&1\\1&-1\end{pmatrix}\)
Worked example A 3 x 3 matrix with given eigenvalues
The matrix \(B = \begin{pmatrix}3&2&-2\\1&3&-1\\1&2&0\end{pmatrix}\) has eigenvalues 1, 2 and 3. Find a matrix \(Q\) and a diagonal matrix \(D\) such that \(B = QDQ^{-1}\).
  1. \(\lambda = 1\): \(B - I = \begin{pmatrix}2&2&-2\\1&2&-1\\1&2&-1\end{pmatrix}\). Rows give \(x + y - z = 0\) and \(x + 2y - z = 0\); subtracting, \(y = 0\), then \(z = x\). So \(\begin{pmatrix}1\\0\\1\end{pmatrix}\).
  2. \(\lambda = 2\): \(B - 2I = \begin{pmatrix}1&2&-2\\1&1&-1\\1&2&-2\end{pmatrix}\). From \(x + 2y - 2z = 0\) and \(x + y - z = 0\): \(y = z\), then \(x = 0\). So \(\begin{pmatrix}0\\1\\1\end{pmatrix}\).
  3. \(\lambda = 3\): \(B - 3I = \begin{pmatrix}0&2&-2\\1&0&-1\\1&2&-3\end{pmatrix}\). Row 1 gives \(y = z\), row 2 gives \(x = z\). So \(\begin{pmatrix}1\\1\\1\end{pmatrix}\).
  4. (You can also take the vector product of two rows of \(B - \lambda I\): it is perpendicular to both, so it solves the system.)
  5. Check: \(B\begin{pmatrix}1\\1\\1\end{pmatrix} = \begin{pmatrix}3\\3\\3\end{pmatrix}\), three times the vector.
Answer: \(Q = \begin{pmatrix}1&0&1\\0&1&1\\1&1&1\end{pmatrix}\), \(D = \begin{pmatrix}1&0&0\\0&2&0\\0&0&3\end{pmatrix}\)
Your turn
For \(A = \begin{pmatrix}4&1\\2&3\end{pmatrix}\) above, write down a DIFFERENT valid pair \(Q\), \(D\) with \(A = QDQ^{-1}\).
Examiner trap

The eigenvectors go in as COLUMNS, not rows, and the order of the eigenvalues in \(D\) must match the order of the columns of \(Q\). With \(Q = \begin{pmatrix}1&1\\1&-2\end{pmatrix}\) the first column belongs to \(\lambda = 5\), so \(D = \operatorname{diag}(5, 2)\); writing \(\operatorname{diag}(2, 5)\) gives a matrix that is not \(A\).

Section 02

Powers made easy.

Squaring \(A = QDQ^{-1}\) gives \(A^{2} = QDQ^{-1}QDQ^{-1} = QD^{2}Q^{-1}\), because the \(Q^{-1}Q\) in the middle is \(I\). Every extra factor cancels the same way, so \(A^{n} = QD^{n}Q^{-1}\). The hard part, the power, falls on the diagonal matrix, and a diagonal matrix is raised to a power entry by entry.

\[D = \begin{pmatrix}\lambda_{1}&0\\0&\lambda_{2}\end{pmatrix} \;\Rightarrow\; D^{n} = \begin{pmatrix}\lambda_{1}^{\,n}&0\\0&\lambda_{2}^{\,n}\end{pmatrix}\]

To get a general formula for \(A^{n}\), multiply in two stages: first \(QD^{n}\) (which just multiplies each column of \(Q\) by its own \(\lambda^{n}\)), then \((QD^{n})Q^{-1}\). Keep any fraction from \(Q^{-1}\) outside the matrix until the end. Always check your formula with \(n = 1\): it must give \(A\) back.

Worked example A general formula for A to the n
Using \(A = \begin{pmatrix}4&1\\2&3\end{pmatrix} = QDQ^{-1}\) from the previous section, find \(A^{n}\) in terms of \(n\).
  1. \(QD^{n}\): multiply column 1 of \(Q\) by \(5^{n}\) and column 2 by \(2^{n}\): \[QD^{n} = \begin{pmatrix}5^{n}&2^{n}\\5^{n}&-2^{\,n+1}\end{pmatrix}\]
  2. Now multiply by \(Q^{-1} = \tfrac{1}{3}\begin{pmatrix}2&1\\1&-1\end{pmatrix}\): \[A^{n} = \tfrac{1}{3}\begin{pmatrix}5^{n}&2^{n}\\5^{n}&-2^{\,n+1}\end{pmatrix}\begin{pmatrix}2&1\\1&-1\end{pmatrix}\]
  3. Top row: \(2 \cdot 5^{n} + 2^{n}\) and \(5^{n} - 2^{n}\). Bottom row: \(2 \cdot 5^{n} - 2^{\,n+1}\) and \(5^{n} + 2^{\,n+1}\).
  4. Check \(n = 1\): \(\tfrac{1}{3}\begin{pmatrix}12&3\\6&9\end{pmatrix} = \begin{pmatrix}4&1\\2&3\end{pmatrix}\), which is \(A\).
Answer: \[A^{n} = \tfrac{1}{3}\begin{pmatrix}2 \cdot 5^{n} + 2^{n} & 5^{n} - 2^{n}\\ 2 \cdot 5^{n} - 2^{\,n+1} & 5^{n} + 2^{\,n+1}\end{pmatrix}\]
Worked example Powers of a 3 x 3 matrix
\(C = \begin{pmatrix}3&0&0\\1&2&0\\1&1&1\end{pmatrix}\). Find \(C^{n}\) and hence \(C^{4}\).
  1. \(C\) is lower triangular, so its eigenvalues are its diagonal entries: 3, 2 and 1.
  2. Solving \((C - \lambda I)\mathbf{v} = \mathbf{0}\): \(\lambda = 3\) gives \(\begin{pmatrix}1\\1\\1\end{pmatrix}\), \(\lambda = 2\) gives \(\begin{pmatrix}0\\1\\1\end{pmatrix}\), \(\lambda = 1\) gives \(\begin{pmatrix}0\\0\\1\end{pmatrix}\).
  3. \(Q = \begin{pmatrix}1&0&0\\1&1&0\\1&1&1\end{pmatrix}\), \(D = \begin{pmatrix}3&0&0\\0&2&0\\0&0&1\end{pmatrix}\), and \(Q^{-1} = \begin{pmatrix}1&0&0\\-1&1&0\\0&-1&1\end{pmatrix}\) (check \(QQ^{-1} = I\)).
  4. \[QD^{n} = \begin{pmatrix}3^{n}&0&0\\3^{n}&2^{n}&0\\3^{n}&2^{n}&1\end{pmatrix}\]
  5. Multiply by \(Q^{-1}\): \[C^{n} = \begin{pmatrix}3^{n}&0&0\\3^{n} - 2^{n}&2^{n}&0\\3^{n} - 2^{n}&2^{n} - 1&1\end{pmatrix}\]
  6. \(n = 1\) gives \(C\) back. For \(n = 4\): \(3^{4} - 2^{4} = 65\) and \(2^{4} - 1 = 15\).
Answer: \(C^{4} = \begin{pmatrix}81&0&0\\65&16&0\\65&15&1\end{pmatrix}\)
Your turn
Use the formula for \(A^{n}\) in the first example to find \(A^{3}\), and check one entry by multiplying.
Examiner trap

\(A^{n}\) is \(QD^{n}Q^{-1}\), NOT \(Q^{n}D^{n}(Q^{-1})^{n}\): only \(D\) is raised to the power, and \(Q\) and \(Q^{-1}\) appear once each, in that order. Also keep the order of multiplication: matrix products do not commute, so \(Q^{-1}D^{n}Q\) is a different matrix.

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Quick check+10 XP

The columns of \(Q\) are eigenvectors of \(A\), and \(D\) is diagonal with the matching eigenvalues in the same order. Which statement is true?

Quick check+10 XP

\(A\) has eigenvector \(\begin{pmatrix}1\\1\end{pmatrix}\) for \(\lambda = 5\) and \(\begin{pmatrix}1\\-2\end{pmatrix}\) for \(\lambda = 2\). If \(Q = \begin{pmatrix}1&1\\-2&1\end{pmatrix}\), which \(D\) gives \(A = QDQ^{-1}\)?

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If \(A = QDQ^{-1}\), then \(A^{n}\) equals

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A 2 x 2 matrix \(A\) has characteristic equation \(\lambda^{2} - 5\lambda + 6 = 0\). By the Cayley-Hamilton theorem,

Section 03

A matrix obeys its own equation.

The Cayley-Hamilton theorem says that a square matrix satisfies its own characteristic equation. Write the characteristic equation as a polynomial in \(\lambda\) equal to zero, then replace \(\lambda\) by \(A\), every power \(\lambda^{k}\) by \(A^{k}\), and every constant term \(c\) by \(cI\). The right-hand side becomes the zero matrix.

Characteristic equationCayley-Hamilton
\(\lambda^{2} - 7\lambda + 10 = 0\)\(A^{2} - 7A + 10I = \mathbf{0}\)
\(\lambda^{3} - 6\lambda^{2} + 11\lambda - 6 = 0\)\(A^{3} - 6A^{2} + 11A - 6I = \mathbf{0}\)

For a 2 x 2 matrix the characteristic equation is \(\lambda^{2} - (\text{trace})\lambda + \det A = 0\). For a 3 x 3 matrix, \(\det(A - \lambda I)\) starts with \(-\lambda^{3}\); multiply through by \(-1\) first, which does not change the equation. Two uses:

Worked example Inverse and cube of a 2 x 2 matrix
For \(A = \begin{pmatrix}4&1\\2&3\end{pmatrix}\), use the Cayley-Hamilton theorem to find \(A^{-1}\), and to express \(A^{3}\) in the form \(pA + qI\).
  1. Characteristic equation: \(\lambda^{2} - 7\lambda + 10 = 0\), so \(A^{2} - 7A + 10I = \mathbf{0}\).
  2. Multiply by \(A^{-1}\): \(A - 7I + 10A^{-1} = \mathbf{0}\), so \(A^{-1} = \tfrac{1}{10}(7I - A)\).
  3. \(7I - A = \begin{pmatrix}3&-1\\-2&4\end{pmatrix}\), so \(A^{-1} = \tfrac{1}{10}\begin{pmatrix}3&-1\\-2&4\end{pmatrix}\) (the usual 2 x 2 inverse, as it should be).
  4. \(A^{2} = 7A - 10I\). Multiply by \(A\): \(A^{3} = 7A^{2} - 10A\).
  5. Substitute again: \(A^{3} = 7(7A - 10I) - 10A = 39A - 70I\).
  6. Check: \(39 \times 4 - 70 = 86\), the top-left entry of \(A^{3}\) found in the previous section.
Answer: \(A^{-1} = \tfrac{1}{10}\begin{pmatrix}3&-1\\-2&4\end{pmatrix}\), \(A^{3} = 39A - 70I\)
Worked example Inverse of a 3 x 3 matrix
\(B = \begin{pmatrix}3&2&-2\\1&3&-1\\1&2&0\end{pmatrix}\) has eigenvalues 1, 2 and 3, and \(B^{2} = \begin{pmatrix}9&8&-8\\5&9&-5\\5&8&-4\end{pmatrix}\). Use the Cayley-Hamilton theorem to find \(B^{-1}\).
  1. The eigenvalues are the roots, so the characteristic equation is \((\lambda - 1)(\lambda - 2)(\lambda - 3) = 0\), that is \(\lambda^{3} - 6\lambda^{2} + 11\lambda - 6 = 0\).
  2. Cayley-Hamilton: \(B^{3} - 6B^{2} + 11B - 6I = \mathbf{0}\).
  3. Multiply by \(B^{-1}\): \(B^{2} - 6B + 11I - 6B^{-1} = \mathbf{0}\), so \(B^{-1} = \tfrac{1}{6}(B^{2} - 6B + 11I)\).
  4. \[B^{2} - 6B + 11I = \begin{pmatrix}9 - 18 + 11 & 8 - 12 & -8 + 12\\ 5 - 6 & 9 - 18 + 11 & -5 + 6\\ 5 - 6 & 8 - 12 & -4 + 11\end{pmatrix}\]
  5. That is \(\begin{pmatrix}2&-4&4\\-1&2&1\\-1&-4&7\end{pmatrix}\). Check: row 1 of \(B\) times column 1 is \(3 \times 2 + 2 \times (-1) + (-2)(-1) = 6\), and \(6 \div 6 = 1\).
Answer: \(B^{-1} = \tfrac{1}{6}\begin{pmatrix}2&-4&4\\-1&2&1\\-1&-4&7\end{pmatrix}\)
Your turn
For \(A = \begin{pmatrix}4&1\\2&3\end{pmatrix}\), use \(A^{3} = 39A - 70I\) to express \(A^{4}\) in the form \(pA + qI\).
Examiner trap

The constant term becomes a multiple of the identity: \(A^{2} - 7A + 10I = \mathbf{0}\), never \(A^{2} - 7A + 10 = 0\) (you cannot add a number to a matrix). And when you multiply by \(A^{-1}\), every term loses one power of \(A\): \(10I\) becomes \(10A^{-1}\), not \(10I\).

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\(E = \begin{pmatrix}3&2\\1&4\end{pmatrix}\) has eigenvalues 5 and 2. Which pair satisfies \(E = QDQ^{-1}\)?

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For \(E = \begin{pmatrix}3&2\\1&4\end{pmatrix} = QDQ^{-1}\) with \(Q = \begin{pmatrix}1&2\\1&-1\end{pmatrix}\) and \(D = \begin{pmatrix}5&0\\0&2\end{pmatrix}\), what is the top-right entry of \(E^{n}\)?

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Use the Cayley-Hamilton theorem to find the inverse of \(F = \begin{pmatrix}3&1\\2&2\end{pmatrix}\).

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A 3 x 3 matrix \(A\) has characteristic equation \(\lambda^{3} - 2\lambda^{2} - 5\lambda + 6 = 0\). Express \(A^{4}\) in the form \(pA^{2} + qA + rI\).

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Six original Cambridge-style questions
Checking a diagonalisation, a general \(A^{n}\) formula with a negative eigenvalue, Cayley-Hamilton for \(A^{4}\) and \(A^{-1}\), \(A^{n}\) for a 3 x 3 matrix, a 3 x 3 inverse from the characteristic equation, and a long-run limit of \(A^{n}\).