Diagonalise a matrix using its eigenvalues and eigenvectors and use the diagonal form to compute powers of the matrix

What this dot point is asking

SEAB wants you to diagonalise a matrix by writing it as $\mathbf{A} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}$, where $\mathbf{D}$ is diagonal, and to use this form to compute high powers $\mathbf{A}^n$ efficiently. You must also know when a matrix can be diagonalised and how the construction uses eigenvalues and eigenvectors.

The answer

The diagonalisation factorisation

If a square matrix $\mathbf{A}$ has enough linearly independent eigenvectors, it can be written as

where the columns of $\mathbf{P}$ are the eigenvectors and $\mathbf{D}$ is the diagonal matrix of the corresponding eigenvalues, listed in the same order as the eigenvector columns. Equivalently $\mathbf{P}^{-1}\mathbf{A}\mathbf{P} = \mathbf{D}$.

When a matrix is diagonalisable

An $n\times n$ matrix is diagonalisable if and only if it has $n$ linearly independent eigenvectors. Two sufficient guarantees:

• $n$ distinct eigenvalues always give $n$ independent eigenvectors (so distinct eigenvalues imply diagonalisable);
• a symmetric real matrix is always diagonalisable.

A defective matrix, with a repeated eigenvalue whose eigenspace is too small, cannot be diagonalised.

Why diagonal form computes powers

Because $\mathbf{P}^{-1}\mathbf{P} = \mathbf{I}$ cancels in the middle of repeated products,

and a diagonal matrix is trivial to raise to a power: $\mathbf{D}^{\,n}$ just raises each diagonal entry to the $n$th power. This turns an otherwise messy repeated multiplication into one easy step plus two matrix products.

The procedure

1. Find the eigenvalues and eigenvectors of $\mathbf{A}$.
2. Form $\mathbf{P}$ from the eigenvectors (as columns) and $\mathbf{D}$ from the matching eigenvalues.
3. Compute $\mathbf{P}^{-1}$.
4. For a power, evaluate $\mathbf{D}^{\,n}$ and multiply $\mathbf{P}\mathbf{D}^{\,n}\mathbf{P}^{-1}$.

Examples in context

Example 1. Closed form of a coupled recurrence. A pair of sequences updated by a matrix each step, $\mathbf{x}_{n} = \mathbf{A}\mathbf{x}_{n-1}$, has the explicit solution $\mathbf{x}_n = \mathbf{A}^n\mathbf{x}_0 = \mathbf{P}\mathbf{D}^n\mathbf{P}^{-1}\mathbf{x}_0$. Diagonalisation delivers the closed form directly, linking back to the characteristic-equation method for recurrences.

Example 2. Population growth models. In a stage-structured population model the transition matrix is diagonalised so that the dominant eigenvalue gives the long-term growth rate and its eigenvector the stable stage distribution, the standard tool of mathematical ecology.

Try this

Q1. State the diagonalisation factorisation and what $\mathbf{P}$ and $\mathbf{D}$ contain. [2 marks]

• Cue. $\mathbf{A} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}$, where $\mathbf{P}$ has the eigenvectors as columns and $\mathbf{D}$ the eigenvalues on its diagonal.

Q2. Write $\mathbf{A}^n$ in terms of $\mathbf{P}$ and $\mathbf{D}$. [1 mark]

• Cue. $\mathbf{A}^n = \mathbf{P}\mathbf{D}^{\,n}\mathbf{P}^{-1}$.

Q3. Why does having two distinct eigenvalues guarantee a $2\times2$ matrix is diagonalisable? [2 marks]

• Cue. Distinct eigenvalues have linearly independent eigenvectors, so $\mathbf{P}$ is invertible and $\mathbf{A} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}$ exists.