1.8.2 Applications of Matrices

Diagonalisation

What is matrix diagonalisation?

A non-zero, square matrix is considered to be diagonal if all elements not along its leading diagonal are zero
A matrix $P$ can be said to diagonalise matrix $M$ , if $D$ is a diagonal matrix where $D = P^{- 1} M P$
If matrix $M$ has eigenvalues $λ_{1}$ , $λ_{2}$ and eigenvectors $x_{1}$ , $x_{2}$ and is diagonisable by $P$ , then
- $P = (x_{1} x_{2})$ , where the first column is the eigenvector $x_{1}$ and the second column is the eigenvector $x_{2}$
- $D = (\begin{matrix} λ_{1} & 0 \\ 0 & λ_{2} \end{matrix})$
You will only need to be able to diagonalise $2 \times 2$ matrices
You will only need to consider matrices with real, distinct eigenvalues
- If there is only one eigenvalue, the matrix is either already diagonalised or cannot be diagonalised
- Diagonalisation of matrices with complex or imaginary eigenvalues is outside the scope of the course

Exam Tip

Remember to use the formula booklet for the determinant and inverse of a matrix

Worked example

The matrix $M = (\begin{matrix} 5 & 4 \\ 3 & 1 \end{matrix})$ has the eigenvalues $λ_{1} = 7$ and $λ_{2} = - 1$ with eigenvectors $x_{1} = (\begin{matrix} 2 \\ 1 \end{matrix})$ and $x_{2} = (\begin{matrix} 2 \\ - 3 \end{matrix})$ respectively.

Show that $P_{1} = (x_{1} x_{2})$ and $P_{2} = (x_{2} x_{1})$ both diagonalise $M$ .

1-8-2-ib-ai-hl-applications-of-matrices-we-1-solution

Matrix Powers

One of the main applications of diagonalising a matrix is to make it easy to find powers of the matrix, which is useful when modelling transient situations such as the movement of populations between two towns.

How can the diagonalised matrix be used to find higher powers of the original matrix?

The equation to find the diagonalised matrix can be re-arranged for $M$ :

$D = P^{- 1} M P \Rightarrow M = P D P^{- 1}$

Finding higher powers of a matrix when it is diagonalised is straight forward:

${(\begin{matrix} a & 0 \\ 0 & b \end{matrix})}^{n} = (\begin{matrix} a^{n} & 0 \\ 0 & b^{n} \end{matrix})$

Therefore, we can easily find higher powers of the matrix using the power formula for a matrix found in the formula booklet:

$M^{n} = P D^{n} P^{- 1}$

Exam Tip

If you are asked to show this by hand, don’t forget to use your GDC to check your answer afterwards!

Worked example

The matrix $M = (\begin{matrix} 3 & - 2 \\ - 4 & 1 \end{matrix})$ has the eigenvalues $λ_{1} = - 1$ and $λ_{2} = 5$ with eigenvectors $x_{1} = (\begin{matrix} 1 \\ 2 \end{matrix})$ and $x_{2} = (\begin{matrix} 1 \\ - 1 \end{matrix})$ respectively.

Show that

M^{n}

can be expressed as

$M^{n} = - \frac{1}{3} (\begin{matrix} (- {(- 1)}^{n} - 2 {(5)}^{n}) & (- {(- 1)}^{n} + {(5)}^{n}) \\ (- 2 {(- 1)}^{n} + 2 {(5)}^{n}) & (- 2 {(- 1)}^{n} - {(5)}^{n}) \end{matrix})$

1-8-2-ib-ai-hl-applications-of-matrices-we-2a-solution