• An eigenvector is a non-zero vector whose direction is unchanged by a transformation — its image is just a scaled copy of itself.
• The scale factor is the eigenvalue, λ. The equation Ae = λe defines this relationship.
• Eigenvalues are found by solving the characteristic equation det(A − λI) = 0.
• For a 2×2 matrix this gives a quadratic in λ; for a 3×3 matrix, a cubic.
• Orthogonal matrices are built from mutually perpendicular unit vectors — their inverse equals their transpose.
• Eigenvectors of a symmetric matrix (belonging to different eigenvalues) are always orthogonal — this is the key theorem of the chapter.
• Diagonalising A means finding an orthogonal matrix P (columns = normalised eigenvectors) so that P^TAP = D, a diagonal matrix of eigenvalues.
• The whole method scales up from 2×2 to 3×3 matrices with no new ideas — just more algebra.

1. Eigenvectors — the "unmoved directions"

Picture a transformation T stretching, squashing and rotating the whole plane (or space). Most vectors get sent off in some completely new direction. But for a handful of special directions, T doesn't rotate the vector at all — it just makes it longer, shorter, or flips it. That special vector e is an eigenvector of T.

If e is an eigenvector, its image e′ must be parallel to e — so e′ is just a multiple of e:

Here λ (lambda) is the eigenvalue — the scale factor along that direction. Geometrically, e defines a whole line through the origin, and that entire line maps onto itself (it's called an invariant line). If λ = 1, every individual point on that line stays exactly where it is — a line of invariant points.

2. The characteristic equation

Starting from Ae = λe, rearrange to get everything on one side:

Since eigenvectors are defined to be non-zero, this equation must have a non-zero solution for e. A matrix equation Mx = 0 only has non-zero solutions for x when M is singular (det(M) = 0). So:

Solving this equation (a quadratic for a 2×2 matrix, a cubic for a 3×3 matrix) gives the eigenvalues. Each eigenvalue is then substituted back into (A − λI)e = 0 to find its corresponding eigenvector.

The method is always the same three steps:

1. Write down det(A − λI) = 0 and solve for λ (this gives the eigenvalues).
2. For each λ, substitute into Ax = λx and solve the resulting simultaneous equations for x and y — this gives the eigenvector's direction.
3. Choose the simplest possible vector in that direction (small whole numbers, no need to normalise unless asked).

Worked example

Find the eigenvalues and eigenvectors for A = (1  1; −2  4).

Normalised eigenvectors

A normalised eigenvector is simply an eigenvector scaled to have length 1. You get one by dividing each component by the vector's magnitude.

For example, from the worked example above, e₁ = (1,1) has magnitude √2, and e₂ = (1,2) has magnitude √5, so:

Orthogonal vectors

"Orthogonal" is just the posh word for perpendicular. Two vectors are orthogonal exactly when their scalar (dot) product is zero.

Orthogonal matrices

A matrix is orthogonal if its columns (read as vectors) are:

• mutually orthogonal (every pair perpendicular to every other pair), and
• each of length 1 (i.e. already normalised).

For example, ( 1/√5 , 2/√5 ) and ( −2/√5 , 1/√5 ) are both unit vectors, and their scalar product is −2/5 + 2/5 = 0, so they're orthogonal. That means

The key algebraic payoff is this: for any orthogonal matrix M,

In other words, for an orthogonal matrix you never need the tedious cofactor method to find the inverse — you just transpose it! This is exactly what makes diagonalisation so efficient in the next section.

Let A be a 2×2 matrix with eigenvalues λ₁, λ₂ and eigenvectors e₁ = (u₁, v₁), e₂ = (u₂, v₂). By definition, Ae₁ = λ₁e₁ and Ae₂ = λ₂e₂. Stack the eigenvectors as the columns of a matrix P, and stack the eigenvalues along the diagonal of a matrix D:

Multiplying A by each column of P and comparing with P multiplied by D shows that:

This is true for any matrix that has enough independent eigenvectors — but finding P⁻¹ in general means the full cofactor method. In this course, we only diagonalise symmetric matrices, because there P⁻¹ is free — it's just P^T, thanks to the orthogonal matrix property above.

Eigenvectors of symmetric matrices are orthogonal

This is the single most important result in the chapter, so it's worth seeing the proof once. First, a useful fact linking scalar products to matrix multiplication:

Theorem: if A is symmetric (A^T = A) and Ae₁ = λ₁e₁, Ae₂ = λ₂e₂ with λ₁ ≠ λ₂, then e₁ . e₂ = 0.

Diagonalising a symmetric matrix — the method

1. Find the eigenvalues λ₁, λ₂ (solve the characteristic equation) and their eigenvectors e₁, e₂.
2. Normalise the eigenvectors to get ê₁, ê₂ (unit length).
3. Write P with ê₁, ê₂ as columns. Since they're automatically orthogonal (theorem above) and now length 1, P is guaranteed to be an orthogonal matrix, so P⁻¹ = P^T — no cofactor work required.
4. Then P^TAP = D, the diagonal matrix of eigenvalues.

Worked example

Diagonalise the symmetric matrix A = (6  −2; −2  9).

Here's the good news: every result above extends to 3×3 matrices (and in general, n×n matrices) with no new ideas required. The proofs are identical or extend naturally into extra dimensions. The characteristic equation det(A − λI) = 0 is now a cubic in λ, so you'll typically get three eigenvalues (and hence three eigenvectors) instead of two.

Often in FP3, you're given one eigenvalue and asked to find its eigenvector — this saves you from the cubic. The method: substitute λ into Ax = λx, write out the three resulting equations, then eliminate one variable at a time (exactly like solving simultaneous equations), and since the system is genuinely singular, you'll always be free to set one variable equal to a parameter.

Worked example

Given that λ = 5 is an eigenvalue of M = (3 −1 2; −2 1 −1; 4 −1 −2), find the corresponding eigenvector.

Same four-step method as the 2×2 case, just with three eigenvalues and three eigenvectors instead of two. Because A is symmetric, all three eigenvectors will automatically come out mutually orthogonal — guaranteed by the same theorem proved earlier (it applies to any pair of eigenvectors from different eigenvalues, in any dimension).

Worked example

Diagonalise A = (2 −2 0; −2 1 2; 0 2 0).

Step 1 — Eigenvalues

Step 2 — Normalised eigenvectors

Step 3 & 4 — Build P and D

In the exam you're very unlikely to be asked to do this whole problem in one go — but every individual step (finding a root of a cubic by inspection, solving for one eigenvector, normalising, checking orthogonality) is fair game on its own.

• I can explain what an eigenvector and eigenvalue represent geometrically
• I can derive and use the characteristic equation det(A − λI) = 0
• I can find eigenvalues and eigenvectors of a 2×2 matrix
• I can normalise a vector and test two vectors for orthogonality
• I know the defining property of an orthogonal matrix, M^TM = I
• I understand the general diagonalisation relationship AP = PD
• I can prove (or at least reproduce) that eigenvectors of a symmetric matrix are orthogonal
• I can diagonalise a symmetric 2×2 matrix, finding P and D
• I can find an eigenvector of a 3×3 matrix given its eigenvalue
• I can find all eigenvalues of a 3×3 matrix by solving a cubic (spotting a root by inspection)
• I can fully diagonalise a symmetric 3×3 matrix, building P and D correctly