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International Baccalaureate · DP 2: Applications and Interpretation HL

Eigenvalues and Eigenvectors

⌚ ~83 min ▢ 10 blocks ✓ 42 exercises ➣ Prerequisites : Matrices
I Powers of a Matrix
I.1 Power of a Matrix
Recappreviously studied — click to reread
Definition — Power of a Matrix
Let \(\mathbf{A}\) be a square matrix and \(n\) an integer with \(n \geq 1\). The \(n\)-th power of \(\mathbf{A}\) is the product of \(n\) copies of \(\mathbf{A}\): $$\mathbf{A}^n = \underbrace{\mathbf{A}\times\mathbf{A}\times\dots\times\mathbf{A}}_{n \text{ factors}}.$$ In particular \(\mathbf{A}^1 = \mathbf{A}\) and \(\mathbf{A}^{n+1} = \mathbf{A}^n\times\mathbf{A}\); by convention \(\mathbf{A}^0 = \mathbf{I}\), the identity matrix. As for numbers, all the factors being equal to \(\mathbf{A}\), \(\mathbf{A}^{m+n} = \mathbf{A}^m\mathbf{A}^n\) and \((\mathbf{A}^m)^n = \mathbf{A}^{mn}\).
Skills to practice
  • Power of a Matrix
I.2 Powers and Diagonal Matrices
Example
Let \(\mathbf{A} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\).
  1. Show that \(\mathbf{A}^2 = 2\mathbf{A} - \mathbf{I_2}\).
  2. Deduce \(\mathbf{A}^3\) without computing any further matrix product.

  1. On the one hand \(\mathbf{A}^2 = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix}\); on the other hand \(2\mathbf{A} - \mathbf{I_2} = \begin{pmatrix} 2 & 4 \\ 0 & 2 \end{pmatrix} - \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 4 \\ 0 & 1 \end{pmatrix}\). The two results are equal, so \(\mathbf{A}^2 = 2\mathbf{A} - \mathbf{I_2}\).
  2. Multiplying by \(\mathbf{A}\): $$ \begin{aligned} \mathbf{A}^3 &= \mathbf{A}\,\mathbf{A}^2 = \mathbf{A}(2\mathbf{A} - \mathbf{I_2}) &&\scriptstyle(\text{previous result})\\ &= 2\mathbf{A}^2 - \mathbf{A} &&\scriptstyle(\text{distributivity})\\ &= 2(2\mathbf{A} - \mathbf{I_2}) - \mathbf{A} = 3\mathbf{A} - 2\mathbf{I_2} = \begin{pmatrix} 1 & 6 \\ 0 & 1 \end{pmatrix}. \end{aligned} $$

Definition — Diagonal Matrix
A square matrix is diagonal when all its entries outside the main diagonal are zero: $$\mathbf{D} = \begin{pmatrix} d_1 & 0 & \cdots & 0 \\ 0 & d_2 & \ddots & \vdots \\ \vdots & \ddots & \ddots & 0 \\ 0 & \cdots & 0 & d_p \end{pmatrix}.$$ The identity matrix \(\mathbf{I_p}\) and the zero matrix of order \(p\) are diagonal; so is \(\begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}\), but \(\begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}\) is not.
Proposition — Powers of a Diagonal Matrix
Let \(\mathbf{D}\) be the diagonal matrix of order \(p\) with diagonal entries \(d_1, d_2, \dots, d_p\). For every integer \(n \geq 1\), \(\mathbf{D}^n\) is the diagonal matrix with diagonal entries \(d_1^{\,n}, d_2^{\,n}, \dots, d_p^{\,n}\): $$\mathbf{D}^n = \begin{pmatrix} d_1^{\,n} & 0 & \cdots & 0 \\ 0 & d_2^{\,n} & \ddots & \vdots \\ \vdots & \ddots & \ddots & 0 \\ 0 & \cdots & 0 & d_p^{\,n} \end{pmatrix}.$$
Skills to practice
  • Computing Powers of a Matrix
II Eigenvalues and Eigenvectors
II.1 Eigenvalues and Eigenvectors
The vectors of interest are those \(\mathbf{x}\) that a matrix \(\mathbf{A}\) maps onto a multiple of themselves: the line they span is invariant. After multiplying by \(\mathbf{A}\), the new vector \(\mathbf{A}\mathbf{x}\) lies on the same line through the origin as the original one (it may be longer, shorter, or even point in the opposite direction). Such a vector is called an eigenvector, and the number \(\lambda\) that tells how the length (and possibly the direction) changes is called the eigenvalue.
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Definition — Eigenvalues and Eigenvectors
Let \(\mathbf{A}\) be a square matrix. An eigenvector of \(\mathbf{A}\) is a non-zero vector \(\mathbf{x}\) such that:$$ \mathbf{A}\mathbf{x} = \lambda \mathbf{x} $$ where \(\lambda\) is a scalar called the eigenvalue associated with the eigenvector \(\mathbf{x}\).
Where does the equation \(\det(\mathbf{A}-\lambda\mathbf{I}) = 0\) come from? The eigenvector condition \(\mathbf{A}\mathbf{x} = \lambda\mathbf{x}\) with \(\mathbf{x} \neq \mathbf{0}\) can be rewritten \((\mathbf{A}-\lambda\mathbf{I})\mathbf{x} = \mathbf{0}\). If the matrix \(\mathbf{A}-\lambda\mathbf{I}\) were invertible, multiplying by its inverse would force \(\mathbf{x} = \mathbf{0}\) — no eigenvector. A non-zero solution can therefore exist only when \(\mathbf{A}-\lambda\mathbf{I}\) is not invertible, that is, exactly when \(\det(\mathbf{A}-\lambda\mathbf{I}) = 0\). Conversely, a singular \(2\times 2\) matrix always sends some non-zero vector to \(\mathbf{0}\) (its columns are proportional), so every root of this equation really is an eigenvalue.
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Method — Finding Eigenvalues and Eigenvectors
To find the eigenvalues and eigenvectors of a matrix \(\mathbf{A}\):
  1. Find the eigenvalues (\(\lambda\)): Solve the characteristic equation $$ \det(\mathbf{A} - \lambda \mathbf{I}) = 0, $$ where \(\mathbf{I}\) is the identity matrix. The solutions are the eigenvalues.
  2. Find the eigenvectors (\(\mathbf{x}\)): For each eigenvalue \(\lambda\), substitute it into $$ \mathbf{A}\mathbf{x} = \lambda\mathbf{x}\text{ or } (\mathbf{A} - \lambda \mathbf{I})\mathbf{x} = \mathbf{0} $$ and solve for the non-zero vector \(\mathbf{x}\).
Example
Find the eigenvalues and corresponding eigenvectors of the matrix$$ \mathbf{A} = \begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix}. $$

  • Find eigenvalues: $$\begin{aligned} \det(\mathbf{A} - \lambda \mathbf{I}) &= 0\\ \det\left( \begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix} - \lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right) &= 0\\ \det\left( \begin{pmatrix} 4-\lambda & 2 \\ 1 & 3-\lambda \end{pmatrix} \right) &= 0\\ (4-\lambda)(3-\lambda) - (2)(1) &= 0\\ \lambda^2 - 7\lambda + 12 - 2 &= 0\\ \lambda^2 - 7\lambda + 10 &= 0\\ (\lambda - 5)(\lambda - 2) &= 0 \end{aligned}$$ The eigenvalues are \(\lambda_1 = 5\) and \(\lambda_2 = 2\).
  • Find an eigenvector for \(\lambda_1 = 5\): $$\begin{aligned} (\mathbf{A} - 5 \mathbf{I})\mathbf{x} &= \mathbf{0}\\ \left( \begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix} - 5 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right)\begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix}\\ \begin{pmatrix} 4-5 & 2 \\ 1 & 3-5 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix}\\ \begin{pmatrix} -1 & 2 \\ 1 & -2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix}\\ \begin{pmatrix} -x + 2y \\ x - 2y \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \end{aligned}$$ This gives \(-x + 2y = 0\) and \(x - 2y = 0\), so \(x = 2y\).
    Letting \(y = t\), we have \(x = 2t\), so $$ \mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2t \\ t \end{pmatrix} = t\begin{pmatrix} 2 \\ 1 \end{pmatrix}, \quad t \neq 0. $$ Any vector of the form \(t\begin{pmatrix} 2 \\ 1 \end{pmatrix}\), \(t\neq 0\), is an eigenvector corresponding to the eigenvalue \(5\).
  • Find an eigenvector for \(\lambda_2 = 2\): $$\begin{aligned} (\mathbf{A} - 2 \mathbf{I})\mathbf{x} &= \mathbf{0}\\ \left( \begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix} - 2 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \right)\begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix}\\ \begin{pmatrix} 4-2 & 2 \\ 1 & 3-2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix}\\ \begin{pmatrix} 2 & 2 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix}\\ \begin{pmatrix} 2x + 2y \\ x + y \end{pmatrix} &= \begin{pmatrix} 0 \\ 0 \end{pmatrix} \end{aligned}$$ This gives \(2x + 2y = 0\) and \(x + y = 0\), so \(y = -x\).
    Letting \(x = t\), we have \(y = -t\), so $$ \mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} t \\ -t \end{pmatrix} = t\begin{pmatrix} 1 \\ -1 \end{pmatrix}, \quad t \neq 0. $$ Any vector of the form \(t\begin{pmatrix} 1 \\ -1 \end{pmatrix}\), \(t\neq 0\), is an eigenvector corresponding to the eigenvalue \(2\).

Skills to practice
  • Recognising Eigenvectors and Eigenvalues
  • Finding Eigenvalues from the Characteristic Equation
  • Calculating Eigenvalues
  • Finding Eigenvectors
III Matrix Diagonalisation
III.1 Diagonalising a 2x2 Matrix
Definition — Diagonalisable Matrix
A square matrix \(\mathbf{A}\) is diagonalisable if there exists an invertible matrix \(\mathbf{P}\) such that$$ \mathbf{D} = \mathbf{P}^{-1}\mathbf{A}\mathbf{P} $$ is a diagonal matrix. We say that \(\mathbf{P}\) diagonalises \(\mathbf{A}\).
Equivalently, we can write$$ \mathbf{A} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}. $$
From the definition of diagonalisation, we start with the relationship:$$ \mathbf{D} = \mathbf{P}^{-1}\mathbf{A}\mathbf{P} $$ We can rearrange this equation to make \(\mathbf{A}\) the subject. This specific form is particularly useful for calculating powers of matrices, since a diagonal matrix is raised to a power entry by entry.
To isolate \(\mathbf{A}\), we multiply by \(\mathbf{P}\) on the left and by \(\mathbf{P}^{-1}\) on the right:$$\begin{aligned}\mathbf{D} &= \mathbf{P}^{-1}\mathbf{A}\mathbf{P} \\ \mathbf{P}\mathbf{D} &= \underbrace{\mathbf{P}\mathbf{P}^{-1}}_{\mathbf{I}}\mathbf{A}\mathbf{P} && \text{(multiply by } \mathbf{P} \text{ on the left)} \\ \mathbf{P}\mathbf{D} &= \mathbf{A}\mathbf{P} \\ \mathbf{P}\mathbf{D}\mathbf{P}^{-1} &= \mathbf{A}\underbrace{\mathbf{P}\mathbf{P}^{-1}}_{\mathbf{I}} && \text{(multiply by } \mathbf{P}^{-1} \text{ on the right)} \\ \mathbf{A} &= \mathbf{P}\mathbf{D}\mathbf{P}^{-1}\end{aligned}$$
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Proposition — Diagonalising a \(2 \times 2\) Matrix
If \(\mathbf{A}\) is a \(2 \times 2\) matrix with distinct real eigenvalues \(\lambda_1, \lambda_2\) and corresponding eigenvectors \(\mathbf{x}_1, \mathbf{x}_2\), then the matrix \(\mathbf{P}\) formed by these eigenvectors diagonalises \(\mathbf{A}\):$$\text{If } \mathbf{P} = (\mathbf{x}_1 \mid \mathbf{x}_2)\text{ then }\mathbf{P}^{-1}\mathbf{A}\mathbf{P} =\begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}.$$ Here, \(\mathbf{P} = (\mathbf{x}_1 \mid \mathbf{x}_2)\) is the matrix whose first column is \(\mathbf{x}_1\) and whose second column is \(\mathbf{x}_2\).

Let \(\lambda_1, \lambda_2\) be the eigenvalues, \(\mathbf{x_1}, \mathbf{x_2}\) associated eigenvectors, \(\mathbf{P} = (\mathbf{x_1} | \mathbf{x_2})\) the matrix with these columns, and \(\mathbf{D}\) the diagonal matrix of the eigenvalues. Computing the product column by column: $$\mathbf{A}\mathbf{P} = (\mathbf{A}\mathbf{x_1} \,|\, \mathbf{A}\mathbf{x_2}) = (\lambda_1\mathbf{x_1} \,|\, \lambda_2\mathbf{x_2}) = \mathbf{P}\mathbf{D}.$$ Since the eigenvalues are distinct, the two eigenvectors are not proportional: if \(\mathbf{x_2} = c\,\mathbf{x_1}\) with \(c \neq 0\), applying \(\mathbf{A}\) would give \(\lambda_2 c\,\mathbf{x_1} = c\lambda_1\mathbf{x_1}\), hence \(c(\lambda_2 - \lambda_1)\mathbf{x_1} = \mathbf{0}\) — impossible. So \(\det(\mathbf{P}) \neq 0\) and \(\mathbf{P}\) is invertible. Multiplying \(\mathbf{A}\mathbf{P} = \mathbf{P}\mathbf{D}\) on the left by \(\mathbf{P}^{-1}\) gives \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P} = \mathbf{D}\).

Not every matrix is diagonalisable. The matrix \(\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\) has the repeated eigenvalue \(\lambda = 1\), but all its eigenvectors are multiples of \(\begin{pmatrix} 1 \\ 0 \end{pmatrix}\): two proportional columns cannot form an invertible \(\mathbf{P}\). (A matrix \(a\mathbf{I}\), by contrast, is already diagonal, even though its eigenvalue \(a\) is repeated.) A characteristic polynomial can also have no real root at all: for the rotation matrix \(\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\), it is \(\lambda^2 + 1\) — the eigenvalues are complex conjugates, and no direction of the plane is preserved.
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Example
The matrix$$ \mathbf{A} = \begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix} $$ has eigenvalues \(\lambda_1 = 5\) and \(\lambda_2 = 2\) with corresponding eigenvectors$$\mathbf{x}_1 = \begin{pmatrix} 2 \\ 1 \end{pmatrix}\quad\text{and}\quad\mathbf{x}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}.$$ Show that \(\mathbf{P} = (\mathbf{x}_1 \mid \mathbf{x}_2)\) diagonalises \(\mathbf{A}\).

To show that \(\mathbf{P}\) diagonalises \(\mathbf{A}\), we must show that \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\) is a diagonal matrix.
  1. Form the matrix \(\mathbf{P}\): $$ \mathbf{P} = \begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix}. $$
  2. Find the inverse \(\mathbf{P}^{-1}\): $$ \det(\mathbf{P}) = (2)(-1) - (1)(1) = -3, $$ so $$ \mathbf{P}^{-1} = \frac{1}{-3} \begin{pmatrix} -1 & -1 \\ -1 & 2 \end{pmatrix} = \frac{1}{3} \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix}. $$
  3. Calculate \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\): $$ \begin{aligned} \mathbf{P}^{-1}\mathbf{A}\mathbf{P} &= \frac{1}{3} \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix} \begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix} \begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix} \\ &= \frac{1}{3} \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix} \begin{pmatrix} 10 & 2 \\ 5 & -2 \end{pmatrix} \\ &= \frac{1}{3} \begin{pmatrix} 15 & 0 \\ 0 & 6 \end{pmatrix} \\ &= \begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix}. \end{aligned} $$
The result is the diagonal matrix$$ \mathbf{D} = \begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix}, $$ which contains the eigenvalues on the diagonal. Thus, \(\mathbf{P}\) diagonalises \(\mathbf{A}\).

Skills to practice
  • Verifying and Deciding Diagonalisation
  • Performing Full Matrix Diagonalisation
IV Powers by Diagonalisation
IV.1 Calculating Matrix Powers
Calculating powers of a matrix \(\mathbf{A}\) (such as \(\mathbf{A}^{10}\)) by repeated multiplication is tedious. However, if we diagonalise the matrix first, the process becomes much simpler.
If \(\mathbf{A}\) is diagonalisable, we can write$$ \mathbf{A} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}, $$ where \(\mathbf{D}\) is diagonal. To calculate \(\mathbf{A}^n\) for an integer \(n \geq 1\) (with the convention \(\mathbf{A}^0 = \mathbf{I}\)):$$\begin{aligned}\mathbf{A}^n &= \underbrace{\mathbf{A}\mathbf{A}\dots \mathbf{A}}_{n \text{ times}}\\ &= \underbrace{(\mathbf{P}\mathbf{D}\mathbf{P}^{-1}) (\mathbf{P}\mathbf{D}\mathbf{P}^{-1}) \dots (\mathbf{P}\mathbf{D}\mathbf{P}^{-1})}_{n \text{ times}} && \text{(substitute } \mathbf{A} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1})\\ &= \mathbf{P}\mathbf{D}(\mathbf{P}^{-1}\mathbf{P})\mathbf{D} (\mathbf{P}^{-1}\mathbf{P})\dots\mathbf{D}\mathbf{P}^{-1} && \text{(regroup terms)}\\ &= \mathbf{P}\mathbf{D}\mathbf{I}\mathbf{D}\mathbf{I}\dots\mathbf{D}\mathbf{P}^{-1} && \text{(since } \mathbf{P}^{-1}\mathbf{P} = \mathbf{I})\\ &= \mathbf{P}(\mathbf{D}\mathbf{D}\dots\mathbf{D})\mathbf{P}^{-1}\\ &= \mathbf{P}\mathbf{D}^n\mathbf{P}^{-1}.\end{aligned}$$ It is easy to raise a diagonal matrix to a power. If$$\mathbf{D} = \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix},$$ then$$\begin{aligned}\mathbf{D}^2 &=\begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}\begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}= \begin{pmatrix} \lambda_1^2 & 0 \\ 0 & \lambda_2^2 \end{pmatrix},\\ \mathbf{D}^3 &= \mathbf{D}^2\mathbf{D}= \begin{pmatrix} \lambda_1^2 & 0 \\ 0 & \lambda_2^2 \end{pmatrix} \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}= \begin{pmatrix} \lambda_1^3 & 0 \\ 0 & \lambda_2^3 \end{pmatrix},\\ &\vdots\\ \mathbf{D}^n &= \begin{pmatrix} \lambda_1^n & 0 \\ 0 & \lambda_2^n \end{pmatrix}.\end{aligned}$$
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Method — Calculating Matrix Powers
To calculate a high power \(\mathbf{A}^n\) of a diagonalisable matrix \(\mathbf{A}\):
  1. Find the eigenvalues and eigenvectors of \(\mathbf{A}\).
  2. Form \(\mathbf{P}\) from the eigenvectors and \(\mathbf{D}\) from the eigenvalues so that \(\mathbf{A} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}\).
  3. Compute \(\mathbf{D}^n\) by raising each diagonal entry to the power \(n\).
  4. Use $$ \mathbf{A}^n = \mathbf{P}\mathbf{D}^n\mathbf{P}^{-1}. $$
Example
The matrix$$ \mathbf{P} = \begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix} $$ diagonalises$$ \mathbf{A} = \begin{pmatrix} 4 & 2 \\ 1 & 3 \end{pmatrix} $$ with$$ \mathbf{P}^{-1} = \frac{1}{3} \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix}\quad\text{and}\quad\mathbf{D} = \begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix}. $$ Calculate the matrix \(\mathbf{A}^5\).

$$\begin{aligned}\mathbf{A}^5 &= \mathbf{P}\mathbf{D}^5\mathbf{P}^{-1} \\ &= \begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 5^5 & 0 \\ 0 & 2^5 \end{pmatrix} \left( \frac{1}{3}\begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix} \right) \\ &= \frac{1}{3} \begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 3125 & 0 \\ 0 & 32 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix} \\ &= \frac{1}{3} \begin{pmatrix} 6250 & 32 \\ 3125 & -32 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix} \\ &= \frac{1}{3} \begin{pmatrix} 6250(1) + 32(1) & 6250(1) + 32(-2) \\ 3125(1) - 32(1) & 3125(1) - 32(-2) \end{pmatrix} \\ &= \frac{1}{3} \begin{pmatrix} 6282 & 6186 \\ 3093 & 3189 \end{pmatrix} \\ &= \begin{pmatrix} 2094 & 2062 \\ 1031 & 1063 \end{pmatrix}.\end{aligned}$$

Example
Two towns, \(X\) and \(Y\), have \(10\,000\) inhabitants in all. Each year, \(20\pourcent\) of the inhabitants of \(X\) move to \(Y\) and \(30\pourcent\) of the inhabitants of \(Y\) move to \(X\); nobody else arrives or leaves. After \(n\) years, \(X\) has \(x_n\) inhabitants and \(Y\) has \(y_n\), with \(x_0 = 4000\) and \(y_0 = 6000\); the model gives expected populations, which need not be whole numbers. Thus \(\mathbf{U_{n+1}} = \mathbf{A}\mathbf{U_n}\) where $$\mathbf{U_n} = \begin{pmatrix} x_n \\ y_n \end{pmatrix} \qquad\text{and}\qquad \mathbf{A} = \begin{pmatrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{pmatrix}.$$
  1. Find the eigenvalues of \(\mathbf{A}\) and an eigenvector for each.
  2. Write \(\mathbf{A} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}\), then give \(\mathbf{A}^n\) for every integer \(n \geq 1\).
  3. Deduce \(x_n\) and \(y_n\), and the populations of the two towns in the long run.

  1. By the Method “Finding Eigenvalues and Eigenvectors”: $$\begin{aligned}[t] \det(\mathbf{A} - \lambda\mathbf{I}) &= (0.8 - \lambda)(0.7 - \lambda) - 0.3 \times 0.2 \\ &= \lambda^2 - 1.5\lambda + 0.56 - 0.06 \\ &= \lambda^2 - 1.5\lambda + 0.5 \\ &= (\lambda - 1)(\lambda - 0.5). \end{aligned}$$ The eigenvalues are \(\lambda_1 = 1\) and \(\lambda_2 = 0.5\). For \(\lambda_1 = 1\), with \(\mathbf{x} = \begin{pmatrix} x \\ y \end{pmatrix}\), \((\mathbf{A} - \mathbf{I})\mathbf{x} = \mathbf{0}\) gives \(-0.2x + 0.3y = 0\), so \(2x = 3y\): \(\mathbf{x}_1 = \begin{pmatrix} 3 \\ 2 \end{pmatrix}\). For \(\lambda_2 = 0.5\), \((\mathbf{A} - 0.5\mathbf{I})\mathbf{x} = \mathbf{0}\) gives \(0.3x + 0.3y = 0\), so \(y = -x\): \(\mathbf{x}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}\).
  2. The eigenvalues are distinct and real, so, by the Proposition “Diagonalising a \(2 \times 2\) Matrix”, \(\mathbf{P} = \begin{pmatrix} 3 & 1 \\ 2 & -1 \end{pmatrix}\) and \(\mathbf{D} = \begin{pmatrix} 1 & 0 \\ 0 & 0.5 \end{pmatrix}\) give \(\mathbf{A} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}\), with \(\det(\mathbf{P}) = -3 - 2 = -5\) and $$\mathbf{P}^{-1} = \frac{1}{-5}\begin{pmatrix} -1 & -1 \\ -2 & 3 \end{pmatrix} = \frac{1}{5}\begin{pmatrix} 1 & 1 \\ 2 & -3 \end{pmatrix}.$$ By the Method “Calculating Matrix Powers”, for every integer \(n \geq 1\), $$\begin{aligned}[t] \mathbf{A}^n = \mathbf{P}\mathbf{D}^n\mathbf{P}^{-1} &= \begin{pmatrix} 3 & 1 \\ 2 & -1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 0.5^n \end{pmatrix} \times \frac{1}{5}\begin{pmatrix} 1 & 1 \\ 2 & -3 \end{pmatrix} \\ &= \frac{1}{5}\begin{pmatrix} 3 & 0.5^n \\ 2 & -0.5^n \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 2 & -3 \end{pmatrix} \\ &= \frac{1}{5}\begin{pmatrix} 3 + 2 \times 0.5^n & 3 - 3 \times 0.5^n \\ 2 - 2 \times 0.5^n & 2 + 3 \times 0.5^n \end{pmatrix}. \end{aligned}$$ Check for \(n = 1\): \(\frac{1}{5}\begin{pmatrix} 4 & 1.5 \\ 1 & 3.5 \end{pmatrix} = \begin{pmatrix} 0.8 & 0.3 \\ 0.2 & 0.7 \end{pmatrix} = \mathbf{A}\).
  3. Each year multiplies the column \(\mathbf{U_n}\) by \(\mathbf{A}\) once more, so \(\mathbf{U_n} = \mathbf{A}^n\mathbf{U_0}\): $$\begin{aligned}[t] x_n &= \frac{1}{5}\Big((3 + 2 \times 0.5^n) \times 4000 + (3 - 3 \times 0.5^n) \times 6000\Big) \\ &= \frac{1}{5}\left(30\,000 - 10\,000 \times 0.5^n\right) \\ &= 6000 - 2000 \times 0.5^n, \end{aligned}$$ and, since the total stays \(10\,000\), \(y_n = 10\,000 - x_n = 4000 + 2000 \times 0.5^n\). Check for \(n = 1\): \(x_1 = 0.8 \times 4000 + 0.3 \times 6000 = 5000 = 6000 - 2000 \times 0.5\). Since \(0 < 0.5 < 1\), \(0.5^n\) approaches \(0\) as \(n\) increases: in the long run, \(X\) has about \(6000\) inhabitants and \(Y\) about \(4000\). These are the entries of the eigenvector \(\begin{pmatrix} 3 \\ 2 \end{pmatrix}\) for the eigenvalue \(1\), scaled to the total \(10\,000\).

The powers \(\mathbf{T}^n\) of a transition matrix and the steady state — an eigenvector for the eigenvalue \(1\), normalised so its entries sum to \(1\) — are the engine of Markov chains, studied in a dedicated chapter.
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Skills to practice
  • Calculating Matrix Powers using Diagonalisation
V Going further
Skills to practice
  • Solving Problems with Eigenvalues