Linear Algebra · step 11 of 13

Diagonalisation & powers of a matrix

A is diagonalisable when A = PDP⁻¹ with D diagonal — which happens exactly when there are enough independent eigenvectors to fill the columns of P. The payoff is Aⁿ = PDⁿP⁻¹, and Dⁿ just raises the diagonal entries to the nth power, turning an infeasible computation into a trivial one.

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Method: how to approach it

The order below is what examiners expect to see, and each step carries its own marks.

Find all eigenvalues and eigenvectorsThe eigenvectors become the columns of P and the corresponding eigenvalues the diagonal of D, IN MATCHING ORDER.
Check diagonalisabilityYou need n independent eigenvectors. n distinct eigenvalues guarantees this; a repeated eigenvalue needs its eigenspace dimension checked.
Assemble P and D and invert PVerify with AP = PD, which avoids computing P⁻¹ just to check.
Compute powersAⁿ = PDⁿP⁻¹, with Dⁿ obtained by raising each diagonal entry to the power n.

Worked example

Diagonalise A = [[2, 1], [0, 3]] and give a formula for Aⁿ.

A is triangular, so the eigenvalues are the diagonal entries: 2 and 3.
For λ = 2: (A − 2I) = [[0,1],[0,1]], giving y = 0 and v = (1,0).
For λ = 3: (A − 3I) = [[−1,1],[0,0]], giving x = y and v = (1,1).
So P = [[1,1],[0,1]], D = diag(2,3), and P⁻¹ = [[1,−1],[0,1]].

Answer. Aⁿ = P·diag(2ⁿ, 3ⁿ)·P⁻¹ = [[2ⁿ, 3ⁿ − 2ⁿ], [0, 3ⁿ]]. Check n = 1 gives back A ✓.

Where marks get dropped

These are the specific errors that cost credit on diagonalisation & powers of a matrix questions — QED's rubric penalises each of them separately.

Mismatching the order of eigenvalues in D and eigenvectors in P. Column k of P must correspond to entry k of D.
Assuming every matrix is diagonalisable. [[1,1],[0,1]] has only one independent eigenvector and is not.
Computing Aⁿ as PⁿDⁿ(P⁻¹)ⁿ. The intermediate P⁻¹P pairs cancel, which is exactly why Aⁿ = PDⁿP⁻¹ and nothing more.

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Diagonalisation & powers of a matrix — frequently asked questions

When is a matrix guaranteed diagonalisable?

When it has n distinct eigenvalues, or when it is symmetric — the spectral theorem gives real symmetric matrices an orthonormal eigenbasis.

What if it is not diagonalisable?

Use the Jordan normal form, which is block-diagonal with 1s on a superdiagonal. Powers are still computable, just messier.

Why does this matter for recurrences?

A linear recurrence in matrix form has solution vₙ = Aⁿv₀, and diagonalising A gives the closed form — which is precisely how Binet’s Fibonacci formula arises.

The rest of Linear Algebra

Matrices, systems, determinants, eigenvalues. Each subtopic below has its own method, worked example and mark-losing traps.

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