Diagonalisation

Diagonalisation is the process of expressing a matrix in the form $A = PDP^{-1}$A=PDP−1, where $D$D is a diagonal matrix of eigenvalues and $P$P is a matrix of eigenvectors. This is extremely useful for computing matrix powers and understanding the long-term behaviour of repeated transformations.

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The Diagonalisation Formula

If an $n \times n$n×n matrix $A$A has $n$n linearly independent eigenvectors, then:

$A = PDP^{-1}$A=PDP−1

where:

• $D = \begin{pmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n \end{pmatrix}$D=​λ1​0⋮0​0λ2​⋮0​⋯⋯⋱⋯​00⋮λn​​​ is the diagonal matrix of eigenvalues
• $P = \begin{pmatrix} \mathbf{v}_1 & \mathbf{v}_2 & \cdots & \mathbf{v}_n \end{pmatrix}$P=(v1​​v2​​⋯​vn​​) is the matrix whose columns are the corresponding eigenvectors

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When Is a Matrix Diagonalisable?

A matrix is diagonalisable if and only if it has n linearly independent eigenvectors (where $n$n is the matrix size).

This is guaranteed when:

• The matrix has $n$n distinct eigenvalues
• The matrix is symmetric (always diagonalisable with real eigenvalues)

It may fail when:

• There are repeated eigenvalues and insufficient independent eigenvectors

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2×2 Worked Example

Diagonalise $A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$A=(42​13​).

From the previous lesson: eigenvalues $\lambda_1 = 5$λ1​=5, $\lambda_2 = 2$λ2​=2; eigenvectors $\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$v1​=(11​), $\mathbf{v}_2 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$v2​=(1−2​).

$P = \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix}, \quad D = \begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix}$P=(11​1−2​),D=(50​02​)

$\det(P) = -2 - 1 = -3$det(P)=−2−1=−3. $P^{-1} = \frac{1}{-3}\begin{pmatrix} -2 & -1 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} 2/3 & 1/3 \\ 1/3 & -1/3 \end{pmatrix}$P−1=−31​(−2−1​−11​)=(2/31/3​1/3−1/3​)

Verification: $PDP^{-1} = \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix}\begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} 2/3 & 1/3 \\ 1/3 & -1/3 \end{pmatrix}$PDP−1=(11​1−2​)(50​02​)(2/31/3​1/3−1/3​)

$= \begin{pmatrix} 5 & 2 \\ 5 & -4 \end{pmatrix}\begin{pmatrix} 2/3 & 1/3 \\ 1/3 & -1/3 \end{pmatrix} = \begin{pmatrix} 12/3 & 3/3 \\ 6/3 & 9/3 \end{pmatrix} = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix} = A$=(55​2−4​)(2/31/3​1/3−1/3​)=(12/36/3​3/39/3​)=(42​13​)=A ✓

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Matrix Powers Using Diagonalisation

If $A = PDP^{-1}$A=PDP−1, then:

$A^n = PD^nP^{-1}$An=PDnP−1

And $D^n$Dn is simply:

$D^n = \begin{pmatrix} \lambda_1^n & 0 \\ 0 & \lambda_2^n \end{pmatrix}$Dn=(λ1n​0​0λ2n​​)

This makes computing large powers straightforward.

Worked Example: Find $A^{10}$A10 for $A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$A=(42​13​).