Source: MATH 416 Abstract Linear Algebra, UIUC Practice Exam
Tags: adjoint, conjugate transpose, normal operator, self-adjoint, Hermitian, unitary, orthogonal matrix, spectral theorem, spectral decomposition, eigenvalue, eigenvector, positive definite, isometry
Difficulty: Intermediate to Advanced Prerequisites: Inner product spaces, orthogonality, Gram-Schmidt (see Part 1 notes). Familiarity with eigenvalues and eigenvectors from a first linear algebra course.
Once you have an inner product, you can ask: given a linear operator T, is there a "mirror" operator that moves the inner product from one side to the other? That mirror is the adjoint T*. Everything in this section flows from the adjoint. Self-adjoint operators (T = T*) have real eigenvalues and orthogonal eigenvectors. Normal operators (TT* = T*T) can be diagonalised by an orthonormal basis. Unitary operators preserve lengths. The Spectral Theorem ties it all together: a self-adjoint operator on a real space (or a normal operator on a complex space) can always be decomposed into a sum of orthogonal projections scaled by eigenvalues. This decomposition is the foundation for PCA, quantum mechanics, and most of applied mathematics.
The adjoint T* is the unique operator satisfying ⟨T(x), y⟩ = ⟨x, T*(y)⟩. Operators that equal their adjoint (self-adjoint) have real eigenvalues and orthonormal eigenvector bases. Operators that commute with their adjoint (normal) share that diagonalisability. Unitary operators preserve norms and have eigenvalues of absolute value 1. The Spectral Theorem lets you write any self-adjoint operator as a weighted sum of orthogonal projections.
Adjoint operator (T)*
The unique linear operator satisfying ⟨T(x), y⟩ = ⟨x, T*(y)⟩ for all x, y in V. In matrix terms, T* corresponds to the conjugate transpose A* = (conjugate of A)ᵀ. Think of it as "moving T from one side of the inner product to the other."
Self-adjoint operator (Hermitian)
An operator where T = T*, meaning ⟨T(x), y⟩ = ⟨x, T(y)⟩ for all x, y. For a real matrix, this is the same as being symmetric (A = Aᵀ). In simple terms, the operator and its mirror image are the same thing.
Normal operator
An operator satisfying TT* = T*T, i.e. T commutes with its adjoint. Self-adjoint, unitary, and skew-adjoint operators are all special cases of normal operators. Think of it as the broadest class of operators that can be orthogonally diagonalised (over ℂ).
Unitary operator (orthogonal, over ℝ)
An operator satisfying AA = I (equivalently, AA = I). For real matrices, this means AᵀA = I. In simple terms, unitary operators preserve lengths and angles: ‖T(x)‖ = ‖x‖ for all x.
Isometry
A linear operator T satisfying ‖T(x)‖ = ‖x‖ for all x ∈ V. On finite-dimensional inner product spaces, isometries are exactly the unitary (or orthogonal) operators.
Positive definite operator
A self-adjoint operator T satisfying ⟨T(x), x⟩ > 0 for all nonzero x. In simple terms, T "stretches" every nonzero vector into a direction that makes a positive inner product with itself. All its eigenvalues are strictly positive real numbers.
Spectral Theorem
For a self-adjoint operator T on a finite-dimensional real inner product space (or a normal operator on a complex space), there exists an orthonormal basis of eigenvectors. Equivalently, T can be written as T = λ₁P₁ + λ₂P₂ + ... + λₖPₖ where each Pᵢ is the orthogonal projection onto the eigenspace for λᵢ.
Spectral decomposition
The expression A = λ₁P₁ + λ₂P₂ + ... + λₖPₖ, where each Pᵢ = uᵢuᵢᵀ is the rank-one orthogonal projection onto the normalised eigenvector uᵢ. Think of it as breaking a matrix into its fundamental "frequency components."
The adjoint T* is characterised by the single identity:
⟨T(x), y⟩ = ⟨x, T*(y)⟩ for all x, y ∈ V.
In matrix terms, if T is represented by A in the standard basis, then T* is represented by A* = (conjugate of A)ᵀ.
For a real matrix: A* = Aᵀ (just the transpose).
For a complex matrix: conjugate every entry, then transpose.
Worked example: T(z₁, z₂) = (2iz₁ + z₂, z₁ − iz₂) has matrix A = [[2i, 1], [1, −i]]. Conjugate: [[−2i, 1], [1, i]]. Transpose: A* = [[−2i, 1], [1, i]]. So T*(z₁, z₂) = (−2iz₁ + z₂, z₁ + iz₂).
T is self-adjoint when T = T*, i.e. A = A*.
Over ℝ: A is symmetric (A = Aᵀ).
Over ℂ: A is Hermitian (A = conjugate transpose of A).
Key properties of self-adjoint operators:
All eigenvalues are real.
Eigenvectors corresponding to distinct eigenvalues are orthogonal.
There exists an orthonormal basis of eigenvectors (Spectral Theorem).
T is normal when TT* = T*T.
Self-adjoint operators are normal (since T = T* implies TT* = T² = T*T).
Unitary operators are normal (since TT* = I = T*T).
The Spectral Theorem extends to normal operators over ℂ: they can be unitarily diagonalised.
Worked example: A = [[1, 1], [−1, 1]] is a real matrix. Compute Aᵀ = [[1, −1], [1, 1]]. Then AAᵀ = [[2, 0], [0, 2]] and AᵀA = [[2, 0], [0, 2]]. Since AAᵀ = AᵀA, the matrix is normal.
T is unitary when A*A = I. Equivalently:
The columns of A form an orthonormal basis.
T preserves inner products: ⟨T(x), T(y)⟩ = ⟨x, y⟩.
T preserves norms: ‖T(x)‖ = ‖x‖ (this is the isometry condition).
Eigenvalue property: if λ is an eigenvalue of a unitary operator, then |λ| = 1. The proof is short: ‖T(x)‖ = ‖x‖ and T(x) = λx gives |λ| · ‖x‖ = ‖x‖, so |λ| = 1.
T is positive definite when:
T is self-adjoint (T = T*), and
⟨T(x), x⟩ > 0 for all x ≠ 0.
Key property: all eigenvalues of a positive definite operator are strictly positive real numbers.
Positive semi-definite relaxes the condition to ⟨T(x), x⟩ ≥ 0.
For a self-adjoint operator T on a finite-dimensional real inner product space:
There exists an orthonormal basis {u₁, u₂, ..., uₙ} of eigenvectors.
T can be written as A = λ₁P₁ + λ₂P₂ + ... + λₙPₙ, where Pᵢ = uᵢuᵢᵀ is the orthogonal projection onto the eigenspace for λᵢ.
Worked example: A = [[3, 1], [1, 3]].
Characteristic polynomial: (3 − λ)² − 1 = λ² − 6λ + 8 = 0, giving λ₁ = 4, λ₂ = 2.
For λ = 4: eigenvector (1, 1), normalised to u₁ = (1/√2)(1, 1).
For λ = 2: eigenvector (1, −1), normalised to u₂ = (1/√2)(1, −1).
P₁ = u₁u₁ᵀ = [[0.5, 0.5], [0.5, 0.5]].
P₂ = u₂u₂ᵀ = [[0.5, −0.5], [−0.5, 0.5]].
A = 4P₁ + 2P₂.
Adjoint defining identity: ⟨T(x), y⟩ = ⟨x, T*(y)⟩
Matrix adjoint: A* = (conjugate of A)ᵀ
Normality condition: TT* = T*T
Unitary condition: AA = I (equivalently AA = I)
Positive definiteness: T = T* and ⟨T(x), x⟩ > 0 for all x ≠ 0
Spectral decomposition: A = Σᵢ λᵢ Pᵢ where Pᵢ = uᵢuᵢᵀ
The Spectral Theorem is the mathematical engine behind Principal Component Analysis (PCA) in statistics and data science: the principal components are eigenvectors of the covariance matrix (which is symmetric and positive semi-definite). Unitary operators describe symmetries in quantum mechanics, where physical observables are self-adjoint operators with real eigenvalues. Positive definite matrices appear everywhere in optimisation: a function has a local minimum at a critical point when the Hessian matrix is positive definite.
Students often confuse "self-adjoint" (T = T*) with "normal" (TT* = T*T). Every self-adjoint operator is normal, but not every normal operator is self-adjoint. Unitary operators are normal too.
A common mistake is thinking that normality means TT* = I. That condition defines a unitary operator, which is a stricter requirement.
Students sometimes forget that positive definiteness requires self-adjointness as a prerequisite. An operator can satisfy ⟨T(x), x⟩ > 0 for all x ≠ 0 without being self-adjoint, but the standard definition requires both conditions.
When computing the adjoint over ℂ, students often transpose without conjugating, or conjugate without transposing. You must do both.
⚠️ The adjoint definition ⟨T(x), y⟩ = ⟨x, T*(y)⟩ is a standard multiple-choice question. Do not confuse it with ⟨T(x), y⟩ = ⟨T*(x), y⟩ (which would mean T = T*, a different claim).
⚠️ Normal means TT* = T*T. This is tested directly. Know how to verify normality by computing both products and checking equality.
⚠️ Unitary eigenvalues have absolute value 1. This appears as a multiple-choice question. The proof (using norm preservation) is short enough to reproduce in an exam.
⚠️ The Spectral Theorem for self-adjoint operators guarantees an orthonormal basis of eigenvectors. This is a separate, stronger statement than mere diagonalisability.
⚠️ Full spectral decomposition (finding eigenvalues, normalised eigenvectors, projection matrices, and the sum A = Σ λᵢPᵢ) is a 10-mark comprehensive problem. Practise the complete workflow.
⚠️ Computing T* from the operator form (not just the matrix) is a 5-mark question. Write the matrix, conjugate, transpose, then convert back.
Fill in the blank: The adjoint T* satisfies ⟨T(x), y⟩ = ⟨x, ______⟩ for all x, y.
True or false: A normal operator must be self-adjoint.
True or false: If A is unitary, all its eigenvalues are real.
Fill in the blank: A positive definite operator is self-adjoint and satisfies ⟨T(x), x⟩ ______ 0 for all x ≠ 0.
True or false: The Spectral Theorem guarantees that every self-adjoint operator on a finite-dimensional real inner product space has an orthonormal basis of eigenvectors.
Answers: 1. T*(y). 2. False (unitary operators are normal but not self-adjoint in general). 3. False (they have absolute value 1 but can be complex, e.g. rotation matrices have eigenvalues e^{iθ}). 4. > (strictly greater than). 5. True.
Q: For a linear operator T on V, the adjoint T is uniquely defined by which relationship?*
A: ⟨T(x), y⟩ = ⟨x, T*(y)⟩ for all x, y ∈ V.
Q: What is the primary condition for an operator T to be considered normal?
A: TT* = T*T, i.e. T commutes with its adjoint.
Q: A matrix A is unitary (A*A = I). What does this imply about the eigenvalues of A?
A: They must have absolute value 1. Since unitary operators preserve norms, |λ| · ‖x‖ = ‖T(x)‖ = ‖x‖ forces |λ| = 1.
Q: The Spectral Theorem for a self-adjoint operator on a finite-dimensional real inner product space guarantees what?
A: The existence of an orthonormal basis of eigenvectors.
Q: Determine the adjoint T of T : ℂ² → ℂ² defined by T(z₁, z₂) = (2iz₁ + z₂, z₁ − iz₂).*
A: The matrix is A = [[2i, 1], [1, −i]]. Conjugate transpose: A* = [[−2i, 1], [1, i]]. So T*(z₁, z₂) = (−2iz₁ + z₂, z₁ + iz₂).
Q: Show that A = [[1, 1], [−1, 1]] is normal.
A: Compute Aᵀ = [[1, −1], [1, 1]]. Then AAᵀ = [[2, 0], [0, 2]] and AᵀA = [[2, 0], [0, 2]]. Since AAᵀ = AᵀA, A is normal.
Q: State the definition of a positive definite linear operator T on a finite-dimensional inner product space V and name one property of its eigenvalues.
A: T is self-adjoint (T = T*) and ⟨T(x), x⟩ > 0 for all nonzero x. All eigenvalues are strictly positive real numbers.
Q: Find the eigenvalues of A = [[3, 1], [1, 3]], an orthonormal basis of eigenvectors, and the spectral decomposition.
A: Characteristic polynomial: λ² − 6λ + 8 = 0, so λ₁ = 4, λ₂ = 2. Eigenvectors: u₁ = (1/√2)(1, 1) for λ = 4, u₂ = (1/√2)(1, −1) for λ = 2. Spectral decomposition: A = 4·[[0.5, 0.5], [0.5, 0.5]] + 2·[[0.5, −0.5], [−0.5, 0.5]].
The adjoint is defined using the inner product from Part 1. The Spectral Theorem relies on the existence of orthonormal bases, which connects back to Gram-Schmidt. The spectral decomposition feeds directly into the Singular Value Decomposition (see Part 3), since SVD uses eigenvalues of A*A. Positive definiteness connects to quadratic forms (Part 3): a quadratic form is positive definite precisely when its associated symmetric matrix is positive definite.
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