Source: MATH 416DE, Abstract Linear Algebra, University of Illinois at Urbana-Champaign
Tags: spectral theorem real, self-adjoint operator, real eigenvalue, Cayley-Hamilton theorem, polynomial factorisation over reals, orthonormal eigenbasis, symmetric matrix, inner product space
Difficulty: Advanced | Prerequisites: Complex Spectral Theorem (Part 1 of these notes), Cayley-Hamilton Theorem, polynomial factorisation, null space and invertibility, induction arguments.
The real Spectral Theorem mirrors the complex version but with a harder hypothesis: over ℝ, normality alone is not enough because the characteristic polynomial might not split. The theorem says an operator on a real inner product space has an orthonormal eigenbasis if and only if it is self-adjoint (T = T*). The main obstacle is proving that a self-adjoint operator has at least one real eigenvalue, since real polynomials can have complex roots. This lecture develops two lemmas to clear that obstacle, then finishes the proof by induction on the dimension.
Over a real inner product space, T has an orthonormal basis of eigenvectors if and only if T = T* (self-adjoint). The hard part is showing self-adjoint operators always have real eigenvalues, despite the characteristic polynomial not necessarily splitting over ℝ. Lemmas about polynomial factorisation and invertibility of certain quadratic expressions handle this.
Self-adjoint operator (symmetric operator)
A linear operator T satisfying T = T*. In matrix terms, [T]_β is equal to its own conjugate transpose with respect to any orthonormal basis. Over ℝ, this simply means the matrix is symmetric. Think of it as the real-field version of "normal," but stricter.
Characteristic polynomial splitting
A polynomial in ℝ[x] splits over ℝ if all its roots are real. Over ℝ this is not guaranteed, which is the core difficulty of the real Spectral Theorem. Over ℂ, every polynomial splits (Fundamental Theorem of Algebra).
Irreducible quadratic factor
A factor of the form x² + bx + c with b² < 4c, meaning it has no real roots (its discriminant is negative). Every real polynomial factors uniquely into linear factors and such irreducible quadratics.
Cayley-Hamilton Theorem
Every linear operator satisfies its own characteristic polynomial. If det([T] - tI) = f(t), then f(T) = 0 (the zero operator). This is used in the proof to transfer polynomial factorisation into a statement about compositions of operators.
T-invariant subspace
A subspace W of V such that T(W) ⊆ W. In the proof, the orthogonal complement of an eigenvector's span is shown to be T-invariant when T is self-adjoint, enabling the induction step.
The complex Spectral Theorem uses Schur's Theorem, which requires the characteristic polynomial to split.
Over ℝ, the characteristic polynomial need not split. A real matrix can have complex eigenvalues (e.g. a 2×2 rotation matrix).
Worse: it is not even obvious that a self-adjoint operator has any eigenvalues at all, since we are restricted to ℝ. The lecture proves this from scratch.
Statement: Every f(x) = a₀ + a₁x + … + aₙxⁿ ∈ Pₙ(ℝ) has a unique factorisation of the form f(x) = c(x - λ₁)···(x - λₘ)(x² + b₁x + c₁)···(x² + bₘx + cₘ) where bⱼ² < 4cⱼ for each j. (m may be zero, meaning no real roots.)
Proof idea:
Treat f as a polynomial over ℂ, where it splits completely by the Fundamental Theorem of Algebra.
For a real polynomial, complex roots come in conjugate pairs: if μ is a root, so is μ̄.
Pair each non-real root μ with μ̄ to get the real quadratic factor (x - μ)(x - μ̄) = x² - 2Re(μ)x + |μ|².
Here b = -2Re(μ) and c = |μ|², and b² < 4c follows from μ having a non-zero imaginary part.
The remaining (real) roots give the linear factors (x - λⱼ).
Statement: If b² < 4c and T is self-adjoint, then T² + bT + cI is invertible.
Proof idea:
Need to show N(T² + bT + cI) = {0ᵥ}, i.e. (T² + bT + cI)v ≠ 0ᵥ for all v ≠ 0ᵥ.
Compute ⟨(T² + bT + cI)v, v⟩:
= ⟨T²v, v⟩ + b⟨Tv, v⟩ + c‖v‖²
= ⟨Tv, Tv⟩ + b⟨Tv, v⟩ + c‖v‖² (using T = T)
= ‖T*v‖² + b⟨Tv, v⟩ + c‖v‖²
Apply the completing-the-square trick:
= (‖T*v‖ - |b|‖Tv‖‖v‖/2)² + ‖v‖²(c - b²/4)
Since b² < 4c, the term (c - b²/4) > 0, so the whole expression is strictly positive for v ≠ 0.
This means ⟨(T² + bT + cI)v, v⟩ > 0, so (T² + bT + cI)v ≠ 0ᵥ, hence the operator is invertible.
Statement: If T = T*, then T has at least one real eigenvalue.
Proof:
By Lemma A, factor the characteristic polynomial: det([T] - tI) = c(t - λ₁)···(t - λₘ)(t² + b₁t + c₁)···(t² + bₘt + cₘ).
By Cayley-Hamilton, substituting T for t gives: (T - λ₁I) ∘ ··· ∘ (T - λₘI) ∘ (T² + b₁T + c₁I) ∘ ··· ∘ (T² + bₘT + cₘI) = T₀ (the zero operator).
By Lemma B, each quadratic factor (T² + bⱼT + cⱼI) is invertible, so the zero on the right cannot come from those factors.
Therefore there must be at least one linear factor, i.e. m ≥ 1, and N(T - λⱼI) ≠ {0ᵥ} for some j.
That λⱼ is a real eigenvalue of T.
Statement: There is an orthonormal basis of V consisting of eigenvectors of T if and only if T = T*.
(⇐) direction (T = T* implies orthonormal eigenbasis):
T has a real eigenvalue λ₁ (by the theorem above). Choose a unit eigenvector u₁.
Set U₁ = span{u₁}. The key claim is that U₁⊥ (the orthogonal complement) is T-invariant.
For any w ∈ U₁⊥: ⟨Tw, u₁⟩ = ⟨w, T*u₁⟩ = ⟨w, Tu₁⟩ = ⟨w, λ₁u₁⟩ = λ₁⟨w, u₁⟩ = 0. So Tw ∈ U₁⊥.
The restriction T|_{U₁⊥} is also self-adjoint (inherits from T).
By induction on dim(V), U₁⊥ has an orthonormal basis β₁⊥ of eigenvectors of T|_{U₁⊥}.
Then {u₁} ∪ β₁⊥ is an orthonormal basis of V consisting of eigenvectors of T.
(⇒) direction is straightforward: if T has an orthonormal eigenbasis, its matrix is real diagonal, which is symmetric, so T = T*.
Real polynomial factorisation:
f(x) = c(x - λ₁)···(x - λₘ)(x² + b₁x + c₁)···(x² + bₘx + cₘ), with bⱼ² < 4cⱼ
Conjugate root pairing:
(z - μ)(z - μ̄) = z² - 2Re(μ)z + |μ|²
Completing the square (Lemma B):
⟨(T² + bT + cI)v, v⟩ = (‖T*v‖ - |b|‖v‖/2)² + ‖v‖²(c - b²/4) > 0 when b² < 4c and v ≠ 0
Cayley-Hamilton application:
f(T) = (T - λ₁I) ∘ ··· ∘ (T - λₘI) ∘ (T² + b₁T + c₁I) ∘ ··· = 0
Self-adjoint operators (symmetric matrices in the real case) arise whenever a physical system is described by a real quadratic form: vibrating systems (stiffness and mass matrices), covariance matrices in statistics, and moment-of-inertia tensors in mechanics. The real Spectral Theorem guarantees these always have real eigenvalues and orthogonal eigenvectors, which correspond to natural frequencies, principal components, and principal axes respectively.
"Normal and self-adjoint are the same thing over ℝ." They are not. A rotation by 90° is normal (orthogonal, so TT = TT = I) but not self-adjoint, and it has no real eigenvalues. Self-adjointness is strictly stronger over ℝ for the purpose of this theorem.
"Since every polynomial splits over ℂ, the real case should follow immediately." It does not, because we need real eigenvalues and an orthonormal basis over ℝ. Complex eigenvalues of a real operator do not help directly in the real inner product space setting.
"Cayley-Hamilton is only used for computing matrix powers." Here it plays a structural role: it transfers polynomial factorisation into a composition of operators, which combined with the invertibility lemma forces the existence of a real eigenvalue.
"The induction step in the proof is the hard part." The induction is relatively clean once you have the real eigenvalue. The genuinely hard part is establishing that eigenvalue exists.
⚠️ Be able to state the real Spectral Theorem precisely. The condition is T = T* (self-adjoint), not TT = TT (normal). Know why normality alone is insufficient over ℝ.
⚠️ The two-lemma structure (Lemma A for factorisation, Lemma B for invertibility) is a common exam proof outline. Be ready to reproduce the completing-the-square argument in Lemma B.
⚠️ Know why U₁⊥ is T-invariant when T is self-adjoint. This one-line calculation (⟨Tw, u₁⟩ = ⟨w, Tu₁⟩ = λ₁⟨w, u₁⟩ = 0) is frequently tested on its own.
⚠️ The Cayley-Hamilton step is the bridge between polynomial algebra and operator theory. Be comfortable applying it.
True or false: Over ℝ, every normal operator has an orthonormal basis of eigenvectors.
Fill in the blank: The real Spectral Theorem requires T to be ______, not merely normal.
True or false: If b² < 4c and T is self-adjoint, then T² + bT + cI is invertible.
Fill in the blank: Complex roots of a real polynomial come in ______ pairs.
True or false: The orthogonal complement of an eigenspace of a self-adjoint operator is always T-invariant.
Answers: 1. False (it must be self-adjoint). 2. Self-adjoint (T = T*). 3. True. 4. Conjugate. 5. True.
Q: State the Spectral Theorem for operators 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 can't we just apply Schur's Theorem directly in the real case?
A: Schur's Theorem requires the characteristic polynomial to split over the base field. Over ℝ, the characteristic polynomial may have non-real complex roots, so it need not split.
Q: Explain why a self-adjoint operator on a real inner product space must have at least one real eigenvalue.
A: Factor the characteristic polynomial into linear and irreducible quadratic factors (Lemma A). By Cayley-Hamilton, the composition of the corresponding operators is zero. Lemma B shows each quadratic operator T² + bⱼT + cⱼI is invertible, so the product of linear factors alone must produce the zero operator. This requires at least one linear factor (T - λⱼI) with a non-trivial null space, giving a real eigenvalue λⱼ.
Q: In the proof of Lemma B, why is b² < 4c essential?
A: The completing-the-square step produces the term ‖v‖²(c - b²/4). This is strictly positive only when c - b²/4 > 0, i.e. b² < 4c. Without this, the inner product ⟨(T² + bT + cI)v, v⟩ could be zero for a non-zero v, and invertibility would fail.
Q: Show that U₁⊥ is T-invariant when T = T and u₁ is an eigenvector with eigenvalue λ₁.*
A: For w ∈ U₁⊥, compute ⟨Tw, u₁⟩ = ⟨w, T*u₁⟩ = ⟨w, Tu₁⟩ = ⟨w, λ₁u₁⟩ = λ₁⟨w, u₁⟩ = 0, since w ⊥ u₁. So Tw ∈ U₁⊥.
Q: Give an example of a normal operator over ℝ that is not self-adjoint and has no real eigenvalues.
A: The rotation matrix R = [[0, -1], [1, 0]] on ℝ². It satisfies RR = RR = I (so it is normal/orthogonal), but R ≠ R*, and its eigenvalues are ±i, which are not real.
This connects to the complex Spectral Theorem (Part 1 of these notes): over ℂ, normal suffices; over ℝ, self-adjoint is needed. Understanding why reveals how the base field shapes spectral theory.
The factorisation of real polynomials into linear and irreducible quadratic factors reappears in partial fraction decomposition (calculus/complex analysis) and in the classification of real Jordan normal forms.
The induction argument using T-invariant orthogonal complements is a technique that generalises to the proof of the singular value decomposition and to the decomposition of representations in algebra.
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