Difficulty: Advanced | Prerequisites: Inner product spaces, self-adjoint operators, eigenvalues and eigenvectors, Cayley–Hamilton theorem, orthogonal complements.
This lecture proves the Spectral Theorem over ℝ: a linear operator on a real inner product space has an orthonormal basis of eigenvectors if and only if it is self-adjoint (T = T*). This is one of the most important results in linear algebra. It tells you exactly when a real matrix can be orthogonally diagonalised, and it underpins applications across physics, statistics, and engineering (principal component analysis, quantum mechanics, vibration analysis). You should already be comfortable with inner products, the adjoint of a linear map, eigenvalues, and the Cayley–Hamilton theorem.
A linear operator T on a finite-dimensional real inner product space can be diagonalised by an orthonormal basis of eigenvectors precisely when T equals its adjoint T*. The proof works by showing that self-adjointness forces at least one real eigenvalue to exist, then peeling off one eigenvector at a time via orthogonal complements, each of which inherits self-adjointness.
Self-adjoint operator (symmetric operator)
An operator T on an inner product space satisfying T = T*, i.e. ⟨Tv, w⟩ = ⟨v, Tw⟩ for all v, w. In simple terms, this means the matrix of T with respect to any orthonormal basis is symmetric (equal to its own transpose).
Orthonormal basis
A basis {u₁, …, uₙ} where every vector has unit length and distinct vectors are orthogonal: ⟨uᵢ, uⱼ⟩ = δᵢⱼ. Think of it as a set of perfectly perpendicular unit-length "axes" for the space.
Spectral decomposition
The expression T = λ̃₁T̃₁ + … + λ̃ₖT̃ₖ, where the λ̃ⱼ are the distinct eigenvalues and each T̃ⱼ is the orthogonal projection onto the corresponding eigenspace. In simple terms, this breaks the operator into a weighted sum of projections, one per eigenvalue.
Orthogonal complement
For a subspace U of V, the set U⊥ = {v ∈ V : ⟨v, u⟩ = 0 for all u ∈ U}. Think of it as everything in the space that is perpendicular to U.
T-invariant subspace
A subspace W such that T(W) ⊆ W. The operator maps every vector in W back into W.
Let T: (Vⁿ, ⟨ , ⟩) → (Vⁿ, ⟨ , ⟩) be a linear operator on a finite-dimensional real inner product space.
There exists an orthonormal basis of V consisting of eigenvectors of T if and only if T = T* (T is self-adjoint).
Goal 1: A self-adjoint operator has at least one real eigenvalue.
By Lemma A, the characteristic polynomial of T over ℝ factors as:
det([T] − tIₙ) = c(t − λ₁)···(t − λₘ)(t² + b₁t + c₁)···(t² + bₗt + cₗ)
where λ₁, …, λₘ are the real roots and each quadratic factor has no real roots (bⱼ² < 4cⱼ).
Goal 1 reduces to showing m ≥ 1 (at least one linear factor exists).
By the Cayley–Hamilton theorem, plugging T into its own characteristic polynomial gives the zero map.
Lemma B shows each quadratic factor T² + bⱼT + cⱼI is invertible when T = T* and bⱼ² < 4cⱼ. So the product of all quadratic factors is invertible, which forces the product of the linear factors to be the zero map, meaning m ≥ 1.
Lemma B: if T = T and b² < 4c, then N(T² + bT + cI) = {0ᵥ}.*
Need to show (T² + bT + cI)v ≠ 0ᵥ for every v ≠ 0ᵥ.
Compute the inner product ⟨(T² + bT + cI)v, v⟩ and show it is strictly positive:
⟨(T² + bT + cI)v, v⟩ = ⟨T²v, v⟩ + b⟨Tv, v⟩ + c‖v‖²
Because T = T*, the term ⟨T²v, v⟩ = ⟨Tv, Tv⟩ = ‖Tv‖².
Apply Cauchy–Schwarz: ⟨Tv, v⟩ ≥ −‖Tv‖‖v‖, so b⟨Tv, v⟩ ≥ −|b|·‖Tv‖·‖v‖.
Complete the square:
‖Tv‖² + c‖v‖² + b⟨Tv, v⟩ ≥ (‖Tv‖ − |b|‖v‖/2)² + ‖v‖²(c − b²/4) > 0
The last inequality holds because c − b²/4 > 0 and v ≠ 0ᵥ.
Goal 2: Peel off eigenspaces via orthogonal complements (induction).
T = T* implies T has a real eigenvalue λ₁. Choose a unit eigenvector u₁ with Tu₁ = λ₁u₁.
Set U₁ = Span{u₁}.
Key lemma: since T = T*, the orthogonal complement U₁⊥ is T-invariant, and the restriction T|_{U₁⊥} : U₁⊥ → U₁⊥ is itself self-adjoint.
Proof of the key lemma (two parts):
Part 1 – U₁⊥ is T-invariant:
For v ∈ U₁⊥, need ⟨Tv, u₁⟩ = 0.
⟨Tv, u₁⟩ = ⟨v, T*u₁⟩ = ⟨v, Tu₁⟩ = ⟨v, λ₁u₁⟩ = λ̄₁⟨v, u₁⟩ = 0. ✓
Part 2 – T|_{U₁⊥} is self-adjoint:
For u, v ∈ U₁⊥: ⟨T|{U₁⊥} u, v⟩ = ⟨Tu, v⟩ = ⟨u, T*v⟩ = ⟨u, Tv⟩ = ⟨u, T|{U₁⊥} v⟩. ✓
Completion by induction:
V = U₁ ⊕ U₁⊥. Apply the same argument to T|_{U₁⊥} to get U₂, then repeat.
After n steps: V = U₁ ⊕ U₂ ⊕ … ⊕ Uₙ, where Uⱼ = Span{uⱼ} and Tuⱼ = λⱼuⱼ.
β = {u₁, …, uₙ} is an orthonormal basis of eigenvectors.
Setting Tᵢ = Proj_{Uᵢ} (orthogonal projection onto Uᵢ):
T = λ₁T₁ + … + λₙTₙ
Grouping by distinct eigenvalues λ̃₁, …, λ̃ₖ:
T = λ̃₁T̃₁ + … + λ̃ₖT̃ₖ
where each T̃ⱼ is the orthogonal projection onto the full eigenspace of λ̃ⱼ.
Characteristic polynomial factorisation (over ℝ):
det([T] − tIₙ) = c(t − λ₁)···(t − λₘ)(t² + b₁t + c₁)···(t² + bₗt + cₗ)
Completing-the-square inequality (Lemma B):
⟨(T² + bT + cI)v, v⟩ ≥ (‖Tv‖ − |b|‖v‖/2)² + ‖v‖²(c − b²/4) > 0
Spectral decomposition:
T = λ̃₁T̃₁ + … + λ̃ₖT̃ₖ (sum of eigenvalue × orthogonal projection)
Consider L_A : ℝ³ → ℝ³ for A = [[2,1,1],[1,2,1],[1,1,2]].
A is symmetric (A = Aᵀ), so L_A is self-adjoint. The Spectral Theorem guarantees an orthonormal eigenbasis.
Characteristic polynomial: det(A − tI₃) = (−1)³(t − 1)²(t − 4).
Eigenvalues: λ = 1 (multiplicity 2), λ = 4 (multiplicity 1).
Eigenspace E₁:
Basis before normalisation: w₁ = (−1, 1, 0)ᵀ, w₂ = (−1, 0, 1)ᵀ.
After Gram–Schmidt and normalisation:
u₁ = (1/√2)(−1, 1, 0)ᵀ
u₂ = √(2/3)(1/2, −1/2, 1)ᵀ
Eigenspace E₄:
Basis: (1, 1, 1)ᵀ. Normalised: u₃ = (1/√3)(1, 1, 1)ᵀ.
Result:
β = {u₁, u₂, u₃} is an orthonormal eigenbasis.
[L_A]_β = diag(1, 1, 4).
A = P⁻¹ · diag(1, 1, 4) · P, where P = [u₁ u₂ u₃] is orthogonal (P⁻¹ = Pᵀ).
Orthogonal diagonalisation of symmetric matrices is the engine behind principal component analysis (PCA) in statistics and data science. It is also how physicists identify the principal axes of inertia for rigid bodies and how engineers analyse stress tensors in materials.
Students often think every operator can be orthogonally diagonalised. It cannot: the operator must be self-adjoint (symmetric matrix) for this to hold over ℝ.
Students confuse "diagonalisable" with "orthogonally diagonalisable." A matrix can be diagonalisable by some basis without that basis being orthonormal. The Spectral Theorem gives the stronger orthonormal version, but only for self-adjoint operators.
Students sometimes forget that the eigenvalues of a self-adjoint operator are guaranteed to be real. This is precisely what Goal 1 establishes.
Students overlook the role of the orthogonal complement lemma. The induction step only works because U₁⊥ inherits self-adjointness; without that, the argument collapses.
⚠️ You must be able to state the Spectral Theorem precisely: orthonormal eigenbasis exists if and only if T = T*.
⚠️ Expect to reproduce or outline the proof that T = T* implies T has a real eigenvalue (Lemma B + Cayley–Hamilton argument).
⚠️ The orthogonal complement invariance lemma is a classic exam proof. Know both parts: T-invariance and inherited self-adjointness.
⚠️ Worked examples asking you to find the orthonormal eigenbasis and spectral decomposition of a symmetric matrix are very likely.
True or False: Every diagonalisable operator on a real inner product space is self-adjoint.
Fill in the blank: The Spectral Theorem states there is an orthonormal basis of eigenvectors if and only if T = ______.
True or False: If T is self-adjoint on a real inner product space, all eigenvalues of T are real.
Fill in the blank: In the induction step, the key property that U₁⊥ inherits from T is ______.
True or False: The spectral decomposition writes T as a sum of eigenvalues times orthogonal projections onto eigenspaces.
Q: State the Spectral Theorem for a linear operator on a finite-dimensional real inner product space.
A: There exists an orthonormal basis of V consisting of eigenvectors of T if and only if T is self-adjoint (T = T*).
Q: Why does the proof of Goal 1 need Lemma B?
A: Lemma B shows that each irreducible quadratic factor T² + bT + cI (with b² < 4c) is invertible when T = T*. This means the product of all quadratic factors is invertible, so for the Cayley–Hamilton equation to hold (the whole product equals the zero map), there must be at least one linear factor, i.e. at least one real eigenvalue.
Q: Prove that if T = T and v ∈ U₁⊥ (where U₁ = Span{u₁} for an eigenvector u₁), then Tv ∈ U₁⊥.*
A: ⟨Tv, u₁⟩ = ⟨v, T*u₁⟩ = ⟨v, Tu₁⟩ = ⟨v, λ₁u₁⟩ = λ₁⟨v, u₁⟩ = 0, since v ∈ U₁⊥. Hence Tv ∈ U₁⊥.
Q: For A = [[2,1,1],[1,2,1],[1,1,2]], find the eigenvalues and verify A is orthogonally diagonalisable.
A: The characteristic polynomial is −(t − 1)²(t − 4), giving eigenvalues 1 (multiplicity 2) and 4 (multiplicity 1). Since A is symmetric (A = Aᵀ), the Spectral Theorem guarantees an orthonormal eigenbasis exists, confirming orthogonal diagonalisability.
Q: What is the spectral decomposition of a self-adjoint operator T with distinct eigenvalues λ̃₁, …, λ̃ₖ?
A: T = λ̃₁T̃₁ + … + λ̃ₖT̃ₖ, where T̃ⱼ is the orthogonal projection onto the eigenspace of λ̃ⱼ.
This connects directly to the complex Spectral Theorem (for normal operators on complex inner product spaces), which relaxes self-adjointness to normality (TT* = T*T). It also sets the stage for the next topic in this lecture, Jordan Canonical Form, which addresses what happens when an operator is not diagonalisable. In applied mathematics, the spectral theorem underpins singular value decomposition (SVD) and quadratic form classification.
spectral theorem, real spectral theorem, self-adjoint operator, symmetric matrix, orthogonal diagonalisation, orthonormal eigenbasis, spectral decomposition, orthogonal complement, T-invariant subspace, Cayley–Hamilton theorem, completing the square, Cauchy–Schwarz inequality, eigenvalue, eigenvector, inner product space, adjoint operator, orthogonal matrix, PCA, principal axes, MATH 416DE, abstract linear algebra