Abstract Linear Algebra, University of Illinois at Urbana-Champaign
Difficulty: Advanced | Prerequisites: Inner product spaces, eigenvalues and eigenvectors, diagonalisability, orthonormal bases, conjugate transpose (A*). Part 1 of these notes (Least Squares Approximation) covers the supporting material.
Earlier in the course you learned when a linear map T: V → V can be diagonalised, i.e. when V has a basis of eigenvectors of T. This lecture asks a sharper question: when does V have an orthonormal basis of eigenvectors of T? The answer depends on the underlying field. Over ℂ the condition is that T is normal (TT* = T*T). Over ℝ the condition is stricter: T must be self-adjoint (T = T*). These are the Spectral Theorems, and they sit behind principal component analysis, quantum mechanics, and any application where you need to decompose an operator into independent, orthogonal modes.
The adjoint T* of a linear map T on an inner product space is the unique map satisfying ⟨T(x), y⟩ = ⟨x, T*(y)⟩. Over ℂ, an orthonormal eigenbasis exists if and only if T is normal. Over ℝ, the condition tightens to T being self-adjoint (symmetric).
Adjoint (T*)
Given T: (V, ⟨·,·⟩) → (V, ⟨·,·⟩), the adjoint T*: V → V is the map satisfying ⟨T(x), y⟩ = ⟨x, T*(y)⟩ for all x, y ∈ V. Think of it as "moving T to the other side of the inner product." In matrix terms, if β is an orthonormal basis, then [T*]_β = ([T]_β)*, i.e. the matrix of the adjoint is the conjugate transpose of the matrix of T.
Self-adjoint (Hermitian)
A map T (or matrix A) is self-adjoint if T = T* (equivalently A = A*). Over ℝ this simply means A = Aᵀ, i.e. A is symmetric. In simple terms, the operator equals its own adjoint, so ⟨Tx, y⟩ = ⟨x, Ty⟩ for all x, y.
Normal operator
A map T is normal if TT* = T*T, i.e. T and its adjoint commute. Think of it as a weaker condition than self-adjoint: every self-adjoint operator is normal, but not every normal operator is self-adjoint.
Orthonormal eigenbasis
A basis of V consisting of eigenvectors of T that are mutually orthogonal and each of unit length. This is the "nicest possible" basis for T: it simultaneously diagonalises T and respects the geometry of the inner product.
Fact: If dim(V) = ∞, the adjoint T* need not exist.
Theorem: If dim(V) < ∞, then T* exists and is unique.
The matrix representation satisfies [T*]_β = ([T]_β)* when β is an orthonormal basis.
Given T, U: V → V and c a scalar:
(T + U)* = T* + U*
(cT)* = c̄ T* (the scalar gets conjugated)
(TU)* = U*T* (order reverses, like transpose)
(T*)* = T (double adjoint returns you to T)
(I_V)* = I_V (the identity is its own adjoint)
These mirror the familiar rules for the conjugate transpose of matrices. The reversal in property 3 and the conjugation in property 2 are the two to watch.
Definition: T is self-adjoint if T = T*.
Matrix version: A ∈ M_{n×n}(ℝ) is self-adjoint iff A = Aᵀ (symmetric). A ∈ M_{n×n}(ℂ) is self-adjoint iff A = A* (Hermitian).
Self-adjoint implies normal (since TT* = TT = T*T), but the converse is false.
Definition: T is normal if TT* = T*T (equivalently, AA* = A*A for its matrix).
Every self-adjoint operator is normal.
Example of normal but not self-adjoint: The rotation matrix
A = | cos θ −sin θ |
| sin θ cos θ |
satisfies AAᵀ = AᵀA = I, so A is normal (in fact, orthogonal). But A ≠ Aᵀ unless θ is a multiple of π, so it is not self-adjoint in general.
Statement: Let T: (V, ⟨·,·⟩) → (V, ⟨·,·⟩) with 𝔽 = ℂ. There is an orthonormal basis of V consisting of eigenvectors of T if and only if T is normal.
Over ℂ, every operator has eigenvalues (the characteristic polynomial always splits), so the condition reduces to whether eigenvectors from different eigenspaces are orthogonal and span V. Normality is exactly what guarantees this.
Statement: Let T: (V, ⟨·,·⟩) → (V, ⟨·,·⟩) with 𝔽 = ℝ. There is an orthonormal basis of V consisting of eigenvectors of T if and only if T is self-adjoint.
Over ℝ the condition is stricter because not every polynomial splits over ℝ. Self-adjointness guarantees that all eigenvalues are real, which is necessary for the characteristic polynomial to factor completely over ℝ.
Proof sketch (forward direction): If there is an orthonormal eigenbasis β, then [T]_β is a diagonal matrix with the eigenvalues on the diagonal. Its conjugate transpose (here just transpose, since 𝔽 = ℝ) is the same diagonal matrix. So [T*]_β = [T]_β, meaning T = T*.
Name | Formula / Condition |
|---|---|
Adjoint defining property | ⟨T(x), y⟩ = ⟨x, T*(y)⟩ for all x, y ∈ V |
Matrix of adjoint (orthonormal basis) | [T*]_β = ([T]_β)* |
Self-adjoint | T = T* (equivalently A = A*) |
Normal | TT* = T*T |
Spectral Theorem (ℂ) | Orthonormal eigenbasis exists ⟺ T is normal |
Spectral Theorem (ℝ) | Orthonormal eigenbasis exists ⟺ T is self-adjoint |
The Spectral Theorem over ℝ is why symmetric matrices appear everywhere in applications: covariance matrices in statistics, Hessians in optimisation, and stiffness matrices in engineering are all symmetric, guaranteeing real eigenvalues and orthogonal eigenvectors.
In quantum mechanics, observables are self-adjoint operators on a complex Hilbert space. The Spectral Theorem guarantees that measurement outcomes (eigenvalues) are real numbers and that measurement states (eigenvectors) are orthogonal.
"Normal and self-adjoint are the same thing." They are not. Self-adjoint is the special case where T = T*. Normal only requires that T and T* commute. Over ℂ the distinction matters: the Spectral Theorem needs normality, not self-adjointness. Over ℝ, you do need the stronger self-adjoint condition.
"Every diagonalisable operator has an orthonormal eigenbasis." Diagonalisability gives you a basis of eigenvectors, but that basis need not be orthogonal. The Spectral Theorem tells you exactly when the basis can be chosen to be orthonormal.
"The adjoint always exists." Only in finite dimensions. In infinite-dimensional spaces, T* may fail to exist.
"(cT)* = cT*." The scalar conjugates: (cT)* = c̄T*. Easy to forget when working over ℂ.
⚠️ Know both versions of the Spectral Theorem and how they differ: normal (ℂ) vs self-adjoint (ℝ).
⚠️ Be able to verify whether a given matrix is self-adjoint, normal, or neither.
⚠️ The five properties of the adjoint (sum, scalar, composition, double adjoint, identity) are standard exam material. The composition rule (TU)* = U*T* and the scalar rule (cT)* = c̄T* are the ones students most often get wrong.
⚠️ The rotation matrix example (normal but not self-adjoint) is a classic illustration. Be able to reproduce it.
True or False: If T is self-adjoint, then T is normal.
Fill in the blank: (TU)* = _____.
True or False: Over ℝ, a normal operator always has an orthonormal eigenbasis.
Fill in the blank: The adjoint T* is defined by the equation ⟨T(x), y⟩ = ⟨x, ___⟩.
True or False: (cT)* = cT* for any scalar c.
Q: State the Spectral Theorem for 𝔽 = ℂ.
A: Let T: (V, ⟨·,·⟩) → (V, ⟨·,·⟩) with 𝔽 = ℂ. There exists an orthonormal basis of V consisting of eigenvectors of T if and only if T is normal (TT* = T*T).
Q: Give an example of a matrix that is normal but not self-adjoint.
A: The 2×2 rotation matrix A = [[cos θ, −sin θ], [sin θ, cos θ]] for θ not a multiple of π. It satisfies AAᵀ = AᵀA = I but A ≠ Aᵀ.
Q: Why does the Spectral Theorem over ℝ require self-adjointness rather than just normality?
A: Over ℝ, the characteristic polynomial may not split (it can have complex roots). Self-adjointness guarantees all eigenvalues are real, ensuring the polynomial factors completely over ℝ and an orthonormal eigenbasis exists.
Q: What is [T*]_β when β is an orthonormal basis?
A: [T*]_β = ([T]_β)*, the conjugate transpose of the matrix of T with respect to β.
Q: Prove that (T*)* = T using the defining property of the adjoint.
A: For all x, y: ⟨T*(x), y⟩ = ⟨x, (T*)*(y)⟩ (applying the adjoint definition to T*). But also ⟨T*(x), y⟩ = ⟨y, T*(x)⟩̄ = (⟨T(y), x⟩)̄ = ⟨x, T(y)⟩. Since this holds for all x and y, (T*)*(y) = T(y) for all y, so (T*)* = T.
The adjoint is the abstract version of the conjugate transpose from Lecture 33, Part 1. Lemma 1 (⟨Ax, y⟩ = ⟨x, A*y⟩) is the matrix-level statement of the adjoint defining property.
The Spectral Theorem generalises the earlier diagonalisability results. Diagonalisability asks for a basis of eigenvectors; the Spectral Theorem asks for an orthonormal one, which is strictly stronger.
Over ℂ, normal operators include unitary operators (U*U = UU* = I) and self-adjoint operators as special cases. The classification of normal operators via the Spectral Theorem is the starting point for functional calculus in analysis.
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