Source: MATH 416DE, Abstract Linear Algebra, University of Illinois at Urbana-Champaign
Tags: spectral theorem, normal operator, Schur's theorem, orthonormal basis, eigenvectors, diagonalisation, upper triangular matrix, adjoint, inner product space, complex field
Difficulty: Advanced | Prerequisites: Inner product spaces, adjoint operators (T*), orthonormal bases, characteristic polynomial, eigenvalues and eigenvectors, Gram-Schmidt process.
This lecture proves the Spectral Theorem over ℂ, one of the centrepiece results in linear algebra. It tells you exactly when a linear operator on a complex inner product space can be diagonalised by an orthonormal basis of eigenvectors: precisely when the operator is normal (TT = TT). The proof leans on Schur's Theorem, which gives you an upper-triangular representation first, and then uses normality to force all off-diagonal entries to zero. If you are coming in cold, make sure you are comfortable with adjoint operators and what it means for a basis to be orthonormal.
A linear operator on a finite-dimensional complex inner product space has an orthonormal basis of eigenvectors if and only if it is normal (commutes with its adjoint). The proof uses Schur's Theorem to get an upper-triangular matrix, then shows normality collapses it to a diagonal.
Normal operator
A linear operator T on an inner product space satisfying TT = TT, where T* is the adjoint of T. In simple terms, T and its adjoint commute, so applying them in either order gives the same result.
Adjoint (T)*
The unique operator satisfying ⟨Tv, w⟩ = ⟨v, T*w⟩ for all v, w in V. Think of it as the "transpose-conjugate" generalised to abstract inner product spaces.
Orthonormal basis
A basis {u₁, …, uₙ} where every vector has unit norm and distinct vectors are orthogonal: ⟨uᵢ, uⱼ⟩ = δᵢⱼ. In simple terms, the basis vectors are mutually perpendicular and each has length 1.
Schur's Theorem
If the characteristic polynomial of T splits (which always happens over ℂ), then there exists an orthonormal basis β = {u₁, …, uₙ} such that [T]_β is upper triangular.
Characteristic polynomial splitting
A polynomial splits over a field F if it factors completely into linear factors in F[x]. Over ℂ, every polynomial splits, by the Fundamental Theorem of Algebra.
Statement: If the characteristic polynomial of T splits, then there is an orthonormal basis β = {u₁, …, uₙ} of V such that the matrix [T]_β is upper triangular.
Over ℂ, the characteristic polynomial always splits (Fundamental Theorem of Algebra), so Schur's Theorem applies to every operator on a complex inner product space.
The upper-triangular matrix has entries aᵢⱼ = ⟨Tuⱼ, uᵢ⟩ = ⟨uⱼ, T*uᵢ⟩, because β is orthonormal and Tuⱼ = Σᵢ ⟨Tuⱼ, uᵢ⟩ uᵢ.
Statement: There is an orthonormal basis of V consisting of eigenvectors of T if and only if T is normal (TT = TT).
(⇐) direction is the substantial part of the proof. The (⇒) direction is straightforward: if T is diagonalised by an orthonormal basis, checking TT = TT is a direct matrix computation.
Lemma: If T is normal, then ‖Tv‖ = ‖T*v‖ for all v ∈ V.
Proof outline:
‖Tv‖² = ⟨Tv, Tv⟩ = ⟨v, TTv⟩ = ⟨v, TTv⟩ = ⟨Tv, Tv⟩ = ‖T*v‖²
The crucial swap from TT to TT uses the normality hypothesis.
Start with the upper-triangular matrix [T]_β from Schur's Theorem.
For the first basis vector u₁: Tu₁ = a₁₁u₁, so ‖Tu₁‖² = |a₁₁|².
Compute [Tu₁]_β by taking the conjugate-transpose of [T]_β and applying it to e₁. The result is the first column of ([T]_β), which gives ‖T*u₁‖² = |a₁₁|² + |a₁₂|² + … + |a₁ₙ|².
By the lemma, ‖Tu₁‖² = ‖T*u₁‖², so |a₁₁|² = |a₁₁|² + |a₁₂|² + … + |a₁ₙ|².
This forces a₁₂ = a₁₃ = … = a₁ₙ = 0. The entire first row (off-diagonal) is zero.
The same argument applied to u₂ eliminates the second row's off-diagonal entries, and so on by induction down the rows.
The upper-triangular matrix becomes diagonal, with eigenvalues a₁₁, a₂₂, …, aₙₙ on the diagonal.
Each basis vector uⱼ satisfies Tuⱼ = aⱼⱼuⱼ, so β is an orthonormal basis of eigenvectors.
Matrix entry formula (orthonormal basis):
aᵢⱼ = ⟨Tuⱼ, uᵢ⟩ = ⟨uⱼ, T*uᵢ⟩
Norm-preservation lemma:
T normal ⟹ ‖Tv‖ = ‖T*v‖ for all v ∈ V
Proof chain:
‖Tv‖² = ⟨Tv, Tv⟩ = ⟨v, TTv⟩ = ⟨v, TTv⟩ = ⟨Tv, Tv⟩ = ‖T*v‖²
Normal operators appear throughout quantum mechanics, where observables are self-adjoint (a special case of normal) and the Spectral Theorem guarantees they have a complete set of orthonormal eigenstates. The result also underpins principal component analysis in data science: symmetric (hence normal) covariance matrices can always be orthogonally diagonalised.
"Every operator on a complex vector space is diagonalisable." Not true. Every operator has an upper-triangular form (Schur), but diagonalisability requires normality. A non-normal operator (e.g. a non-trivial nilpotent) is not diagonalisable.
"Schur's Theorem gives you eigenvectors." It does not. It gives an orthonormal basis making [T]_β upper triangular. Eigenvectors only emerge after you additionally use normality to zero out the off-diagonal.
"Normality is just a technical convenience." It is the exact, necessary and sufficient condition for orthonormal diagonalisation over ℂ. You cannot weaken or remove it.
"The (⇒) direction of the Spectral Theorem is the hard part." The reverse: if you already have an orthonormal eigenbasis, verifying TT = TT is a short computation with diagonal matrices. The (⇐) direction, building the eigenbasis from normality, is where the work happens.
⚠️ Be prepared to state the Spectral Theorem precisely, including the "if and only if" and the condition that T is normal (not just self-adjoint, which is only required in the real case).
⚠️ The proof strategy (Schur first, then norm argument to kill off-diagonals) is a common exam outline question. Know the lemma ‖Tv‖ = ‖T*v‖ and be able to prove it in two lines.
⚠️ Know why Schur's Theorem applies over ℂ without extra hypotheses: the characteristic polynomial always splits there.
True or false: Every normal operator on a finite-dimensional complex inner product space is diagonalisable by an orthonormal basis.
Fill in the blank: Schur's Theorem requires that the ______ of T splits over the field.
True or false: If T is normal, then ‖Tv‖ = ‖T*v‖ for all v.
True or false: The (⇒) direction of the complex Spectral Theorem is the harder direction to prove.
Fill in the blank: If [T]_β is upper triangular with respect to an orthonormal basis and T is normal, the off-diagonal entries must all be ______.
Answers: 1. True. 2. Characteristic polynomial. 3. True. 4. False (the (⇐) direction is harder). 5. Zero.
Q: State the Spectral Theorem for operators on a finite-dimensional complex inner product space.
A: There exists an orthonormal basis of V consisting of eigenvectors of T if and only if T is normal, i.e. TT = TT.
Q: Why does Schur's Theorem apply to every operator on a complex inner product space?
A: Over ℂ, the Fundamental Theorem of Algebra guarantees that every polynomial splits into linear factors, so the characteristic polynomial of T always splits.
Q: Prove that if T is normal, then ‖Tv‖ = ‖T*v‖ for all v ∈ V.
A: ‖Tv‖² = ⟨Tv, Tv⟩ = ⟨v, TTv⟩ = ⟨v, TTv⟩ (by normality) = ⟨Tv, Tv⟩ = ‖T*v‖². Taking square roots gives the result.
Q: In the proof of the Spectral Theorem (⇐ direction), how do you show the first row of [T]_β (above the diagonal) is zero?
A: Tu₁ = a₁₁u₁ gives ‖Tu₁‖² = |a₁₁|². The conjugate-transpose calculation gives ‖T*u₁‖² = |a₁₁|² + |a₁₂|² + … + |a₁ₙ|². The lemma forces these equal, so |a₁₂|² + … + |a₁ₙ|² = 0, hence a₁₂ = … = a₁ₙ = 0.
Q: Give an example of an operator on a complex inner product space that is upper-triangularisable but not diagonalisable.
A: Any non-trivial nilpotent operator, e.g. the shift operator on ℂ² defined by T(e₁) = 0, T(e₂) = e₁. It is upper-triangularisable (strictly upper triangular) but not diagonalisable, and it is not normal.
This connects to the real Spectral Theorem (same lecture), which requires self-adjointness (T = T*) rather than normality, and needs extra work because the characteristic polynomial may not split over ℝ.
The Spectral Theorem is the foundation for the singular value decomposition (SVD), which applies to any operator by working with T*T (always self-adjoint, hence spectrally decomposable).
In quantum mechanics, the Spectral Theorem guarantees that measurement outcomes (eigenvalues of self-adjoint operators) are real and the corresponding states are orthogonal.
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