MTH2021 · Linear Algebra with Applications
Orthogonal Diagonalisation & the Spectral Theorem
Week 11 specialises diagonalisation to symmetric matrices. Orthogonal matrices satisfy A⁻¹ = Aᵀ and preserve lengths and dot products; the Spectral Theorem says a matrix is orthogonally diagonalisable (A = QDQᵀ) if and only if it is symmetric, with an orthonormal eigenbasis. The spectral decomposition A = Σ λᵢuᵢuᵢᵀ and quadratic forms follow. Symmetric-matrix arguments are prime handwritten-section material in the comprehensive 50% final.
What this chapter covers
- 01Orthogonal matrix: A⁻¹ = Aᵀ; rows (and columns) form an orthonormal basis
- 02Orthogonal matrices preserve length and dot product (‖Ax‖ = ‖x‖, Ax·Ay = x·y); det = ±1; products of orthogonal matrices are orthogonal
- 03Change-of-basis between orthonormal bases is an orthogonal matrix; rotations R_θ and reflections are examples
- 04Orthogonally diagonalisable: A = PDPᵀ with P orthogonal
- 05Spectral Theorem: A is orthogonally diagonalisable ⇔ A is symmetric ⇔ A has an orthonormal set of n eigenvectors
- 06Symmetric matrices: eigenvectors for distinct eigenvalues are automatically orthogonal
- 07Algorithm: basis per eigenspace → Gram–Schmidt within each eigenspace → orthonormal P
- 08Spectral decomposition A = Σᵢ λᵢuᵢuᵢᵀ; connection to quadratic forms
Orthogonal diagonalisation and spectral decomposition of a symmetric matrix
- +1Eigenvalues: det(A − λI) = (3 − λ)² − 1 = λ² − 6λ + 8 = (λ − 2)(λ − 4), so λ₁ = 2 and λ₂ = 4.
- +1Eigenvectors: for λ = 2, (A − 2I) = [[1, 1], [1, 1]] gives (1, −1); for λ = 4, (A − 4I) = [[−1, 1], [1, −1]] gives (1, 1). As predicted for a symmetric matrix, (1, −1) ⊥ (1, 1).
- +1Normalise to an orthonormal eigenbasis: u₁ = (1/√2)(1, −1) and u₂ = (1/√2)(1, 1). Form Q = (1/√2)[[1, 1], [−1, 1]] (columns u₁, u₂) and D = diag(2, 4).
- +1Then A = QDQᵀ (for orthogonal Q, Q⁻¹ = Qᵀ). Spot-check: AQ = QD since A(1,−1)ᵀ = (2,−2)ᵀ = 2(1,−1)ᵀ and A(1,1)ᵀ = (4,4)ᵀ = 4(1,1)ᵀ ✓.
- +1Spectral decomposition A = 2·u₁u₁ᵀ + 4·u₂u₂ᵀ. Here u₁u₁ᵀ = (1/2)[[1, −1], [−1, 1]] and u₂u₂ᵀ = (1/2)[[1, 1], [1, 1]], so A = 2·(1/2)[[1,−1],[−1,1]] + 4·(1/2)[[1,1],[1,1]] = [[1,−1],[−1,1]] + [[2,2],[2,2]] = [[3,1],[1,3]] ✓.
Key terms
- Orthogonal matrix
- A square matrix with A⁻¹ = Aᵀ; its rows and columns form orthonormal bases, it preserves lengths and dot products, and det = ±1.
- Orthogonally diagonalisable
- A = PDPᵀ for an orthogonal P and diagonal D; equivalently A has an orthonormal basis of eigenvectors.
- Spectral Theorem
- A real matrix is orthogonally diagonalisable if and only if it is symmetric; then its eigenvectors can be chosen orthonormal.
- Orthogonal eigenvectors
- For a symmetric matrix, eigenvectors belonging to distinct eigenvalues are automatically orthogonal, so only normalisation is needed.
- Spectral decomposition
- A = Σᵢ λᵢ uᵢuᵢᵀ, expressing a symmetric matrix as a weighted sum of rank-one projections onto its orthonormal eigenvectors.
- Quadratic form
- An expression xᵀAx with A symmetric; orthogonal diagonalisation rotates coordinates to remove cross terms, revealing the form's signature.
Orthogonal Diagonalisation & the Spectral Theorem FAQ
What makes a matrix orthogonally diagonalisable?
The Spectral Theorem gives a clean answer: a real matrix is orthogonally diagonalisable (A = QDQᵀ with Q orthogonal) if and only if it is symmetric. So you never have to hunt — check symmetry, and if Aᵀ = A you are guaranteed an orthonormal eigenbasis.
Why are the eigenvectors of a symmetric matrix orthogonal?
For distinct eigenvalues it is automatic: if Ax = λx and Ay = μy with λ ≠ μ and A symmetric, then λ(x·y) = (Ax)·y = x·(Ay) = μ(x·y), forcing x·y = 0. For a repeated eigenvalue you apply Gram–Schmidt within its eigenspace to get an orthonormal set.
What is the difference between diagonalisation and orthogonal diagonalisation?
Ordinary diagonalisation A = PDP⁻¹ only needs n independent eigenvectors and P need not be orthogonal. Orthogonal diagonalisation A = QDQᵀ additionally makes the eigenvectors orthonormal, so P⁻¹ = Qᵀ. It is available exactly for symmetric matrices, which is why the two chapters are kept separate.
What is the spectral decomposition good for?
Writing A = Σλᵢuᵢuᵢᵀ expresses a symmetric matrix as a sum of rank-one pieces, each an eigenvalue times a projection onto an eigenvector. It makes powers, functions of A and quadratic-form analysis transparent, and reconstructing A from the sum is a reliable self-check on your eigenvectors.
How is this examined?
Week 11 lands after Test 2's window, so orthogonal diagonalisation and the spectral theorem are assessed in the weekly quiz and the comprehensive 50% final. The final's handwritten section often asks for a symmetric-matrix proof (orthogonality of eigenvectors) alongside an orthogonal-diagonalisation computation, so rehearse both.
Exam move
Use symmetry as your shortcut: if Aᵀ = A, the eigenvectors of distinct eigenvalues are already orthogonal, so just normalise them into Q and use Qᵀ in A = QDQᵀ. Reserve Gram–Schmidt for repeated eigenvalues. Practise writing and reconstructing the spectral decomposition A = Σλᵢuᵢuᵢᵀ as a check, and rehearse the proof that symmetric matrices have orthogonal eigenvectors. This is examined in the 50% final. When Q doesn't reproduce A, ask Sia to verify the orthonormalisation and the ordering of eigenvalues.
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