Source: Abstract Linear Algebra, University of Illinois at Urbana-Champaign
Tags: spectral theorem, Schur's theorem, orthonormal eigenbasis, diagonalisation, normal operator, self-adjoint, upper triangular, eigenvalues, eigenvectors, linear algebra
Difficulty: Intermediate to Advanced Prerequisites: Adjoint operators, self-adjoint and normal operators, inner product spaces, characteristic polynomial, eigenvalues, Gram-Schmidt process, induction proofs.
Big picture: The Spectral Theorem is one of the headline results in linear algebra. It tells you exactly when a linear operator can be diagonalised using an orthonormal basis of eigenvectors, and the answer differs depending on whether you work over ℂ or ℝ. Schur's Theorem is a stepping stone: it guarantees you can at least get an upper-triangular representation (weaker than diagonal, but still very useful) whenever the characteristic polynomial splits. If you are comfortable with adjoint, self-adjoint, and normal operators, you are ready.
Over ℂ, an operator has an orthonormal eigenbasis if and only if it is normal (TT* = TT). Over ℝ, the condition is stronger: the operator must be self-adjoint (T = T). Schur's Theorem says that whenever the characteristic polynomial splits, you can find an orthonormal basis that makes the matrix upper triangular, which always holds over ℂ.
Spectral Theorem (Complex)
If 𝔽 = ℂ, there exists an orthonormal basis of V consisting of eigenvectors of T if and only if T is normal. In simple terms, over the complex numbers, "commutes with its adjoint" is the exact condition for a clean diagonal decomposition.
Spectral Theorem (Real)
If 𝔽 = ℝ, there exists an orthonormal basis of V consisting of eigenvectors of T if and only if T is self-adjoint. In simple terms, over the reals you need the stronger condition that T equals its adjoint (i.e. the matrix is symmetric).
Schur's Theorem
If the characteristic polynomial of T: V → V splits (factors completely into linear factors), then there exists an orthonormal basis β of V such that [T]_β is upper triangular. Think of it as: you might not get a diagonal matrix, but you can always get the next best thing (upper triangular) as long as all eigenvalues live in your field.
Characteristic polynomial splits
The characteristic polynomial det(A − tI) factors as a product of linear terms (t − λ₁)^n₁ ··· (t − λ_k)^n_k with all roots in 𝔽. Over ℂ this always happens (by the Fundamental Theorem of Algebra). Over ℝ it may not, since some roots could be complex.
Statement: There is an orthonormal basis of V consisting of eigenvectors of T if and only if T is normal.
Forward direction (⇒): Suppose an orthonormal eigenbasis β exists.
Then [T]_β = diag(λ₁, …, λₙ), a diagonal matrix of eigenvalues.
Since β is orthonormal, [T*]_β = ([T]_β)* = diag(λ̄₁, …, λ̄ₙ).
Multiplying: [TT*]_β = diag(λ₁λ̄₁, …, λₙλ̄ₙ) = diag(|λ₁|², …, |λₙ|²).
Equally, [T*T]_β = diag(λ̄₁λ₁, …, λ̄ₙλₙ) = the same diagonal matrix.
So TT* = T*T, meaning T is normal.
Remark: The same argument works for 𝔽 = ℝ (the forward direction does not depend on the field).
Backward direction (⇐): The proof that normality implies the existence of an orthonormal eigenbasis uses Schur's Theorem as an input (see below) plus additional arguments not fully covered in this lecture excerpt.
Statement: There is an orthonormal basis of V consisting of eigenvectors of T if and only if T is self-adjoint.
The real case requires the stronger condition (self-adjoint rather than just normal) because a normal real operator can have complex eigenvalues that prevent diagonalisation over ℝ.
Example illustrating the difference: the rotation matrix [[0, −1], [1, 0]] is normal (and even orthogonal), but its eigenvalues are ±i, so no real eigenvectors exist at all.
Statement: If the characteristic polynomial of T: V → V splits over 𝔽, then there exists an orthonormal basis β of V such that [T]_β is upper triangular.
Over ℂ, the characteristic polynomial always splits, so Schur's Theorem always applies.
Over ℝ, it applies only when all eigenvalues happen to be real.
Schur's Theorem is weaker than the Spectral Theorem (upper triangular rather than diagonal), but it requires fewer conditions on T.
Base case (n = 1): Any 1×1 matrix is already upper triangular. (Exercise.)
Inductive step: Assume the result holds for all inner product spaces of dimension n − 1.
Since the characteristic polynomial splits, T has at least one eigenvalue λ₁ with eigenvector w₁ ≠ 0. So Tw₁ = λ₁w₁.
Extend {w₁} to a basis {w₁, …, wₙ} of V, then apply Gram-Schmidt to produce an orthonormal basis β₁ = {u₁, …, uₙ}.
Because Gram-Schmidt processes vectors in order, u₁ = w₁/‖w₁‖, so u₁ is still an eigenvector: Tu₁ = λ₁u₁.
In the basis β₁, the matrix [T]_{β₁} has λ₁ in the top-left corner and zeros below it in the first column (because Tu₁ = λ₁u₁ contributes only to the u₁ component).
The remaining (n−1) × (n−1) block is A₁, a matrix representing the restricted and projected operator Proj_{U₁⊥} ∘ T|_{U₁⊥} on the subspace U₁⊥ = Span{u₂, …, uₙ}.
Applying the inductive hypothesis:
The characteristic polynomial of A₁ splits (it divides the characteristic polynomial of T, up to the factor (λ₁ − t)).
By the inductive hypothesis, there is an orthonormal basis α₂ of U₁⊥ making the matrix of the restricted operator upper triangular.
Setting β₂ = {u₁} ∪ α₂ gives an orthonormal basis of V in which [T]_{β₂} is upper triangular: λ₁ sits in the top-left, and the lower-right block is the upper-triangular matrix from the inductive step.
Spectral Theorem (ℂ): Orthonormal eigenbasis exists ⟺ TT* = T*T
Spectral Theorem (ℝ): Orthonormal eigenbasis exists ⟺ T = T*
Diagonal form from eigenbasis: [T]_β = diag(λ₁, …, λₙ), [T*]_β = diag(λ̄₁, …, λ̄ₙ)
Schur form: [T]_β is upper triangular with eigenvalues on the diagonal
Characteristic polynomial split: det(A − tI) = (λ₁ − t)^{n₁} ··· (λ_k − t)^{n_k}
The Spectral Theorem underpins principal component analysis (PCA) in statistics, where you diagonalise a symmetric covariance matrix to find the directions of greatest variance. In quantum mechanics, observable quantities correspond to self-adjoint operators, and the Spectral Theorem guarantees they have real eigenvalues and an orthonormal eigenbasis of measurement states. Schur decomposition is used numerically to compute eigenvalues (the QR algorithm produces a Schur form iteratively).
Students often confuse the complex and real versions of the Spectral Theorem. Over ℂ the condition is normality; over ℝ it is the strictly stronger condition of being self-adjoint. Mixing these up on an exam loses marks quickly.
Students sometimes think Schur's Theorem gives you a diagonal matrix. It does not: it gives upper triangular, which is a weaker conclusion.
A common error in the induction proof is forgetting why u₁ remains an eigenvector after Gram-Schmidt. The key point is that Gram-Schmidt processes w₁ first, so u₁ = w₁/‖w₁‖ inherits the eigenvector property.
Students sometimes assume Schur's Theorem applies over ℝ without checking that the characteristic polynomial splits. Over ℝ, it may not.
⚠️ "State the Spectral Theorem" is a standard exam question. Know both versions (ℂ and ℝ) and which condition goes with which field.
⚠️ The forward direction of the proof (orthonormal eigenbasis ⇒ normality) is clean and examinable. Be able to reproduce the diagonal-matrix argument.
⚠️ Schur's Theorem and its induction proof are commonly examined. The key steps to remember: find an eigenvector, normalise it, split V = Span{u₁} ⊕ U₁⊥, restrict and project, apply the inductive hypothesis.
⚠️ Expect a question asking you to explain why the Spectral Theorem fails for a specific operator (e.g. a rotation in ℝ² by a non-trivial angle: normal but not self-adjoint, and eigenvalues are complex).
True or false: Over ℂ, a self-adjoint operator has an orthonormal eigenbasis.
Fill in the blank: Over ℝ, the Spectral Theorem requires T to be ______.
True or false: Schur's Theorem guarantees a diagonal matrix representation.
True or false: Over ℂ, Schur's Theorem always applies (no extra conditions on T).
Fill in the blank: In the induction proof of Schur's Theorem, after finding eigenvector w₁, you apply ______ to build an orthonormal basis.
Answers: 1. True (self-adjoint ⇒ normal, and the complex Spectral Theorem applies). 2. Self-adjoint. 3. False (upper triangular). 4. True (because the characteristic polynomial always splits over ℂ). 5. Gram-Schmidt.
Q: State the Spectral Theorem for 𝔽 = ℂ.
A: There is an orthonormal basis of V consisting of eigenvectors of T if and only if T is normal (TT* = T*T).
Q: Why does the real Spectral Theorem require self-adjointness rather than just normality?
A: A normal operator over ℝ can have non-real eigenvalues, which means it has no real eigenvectors for those eigenvalues. Self-adjointness forces all eigenvalues to be real, ensuring the eigenvectors live in the real vector space.
Q: Explain the base case of the induction proof of Schur's Theorem.
A: When dim V = 1, any linear operator T has a 1×1 matrix representation, which is automatically upper triangular.
Q: In the inductive step of Schur's proof, why does the characteristic polynomial of A₁ split?
A: The characteristic polynomial of [T]{β₁} has the form (λ₁ − t) · det(A₁ − tI{n−1}). Since the full characteristic polynomial splits over 𝔽 and (λ₁ − t) is already a linear factor, the remaining factor det(A₁ − tI_{n−1}) must also split over 𝔽.
Q: Give an operator that is normal but does not satisfy the real Spectral Theorem. Explain why.
A: Rotation by π/2 in ℝ², matrix [[0, −1], [1, 0]]. It is orthogonal, so AA^T = I = A^TA (normal). But A ≠ A^T, so it is not self-adjoint. Its eigenvalues are ±i, which are not real, so no real orthonormal eigenbasis exists.
The Spectral Theorem builds on the definitions of adjoint, self-adjoint, and normal operators (covered in the companion notes). Schur's Theorem connects back to the Gram-Schmidt process and the theory of characteristic polynomials. Looking ahead, the Spectral Theorem leads into spectral decomposition and the singular value decomposition (SVD), which generalises these ideas to non-square and non-normal operators.
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