Source: Friedberg, Insel, Spence -- Linear Algebra 4th ed., Ch. 6.3-6.4
Tags: adjoint operator, T star, normal operator, self-adjoint, Hermitian, Schur's theorem, least squares, minimal solution, positive definite, diagonalizable, eigenvalues real, MATH 301, UIUC, abstract linear algebra
Difficulty: Advanced Prerequisites: Inner products, norms, orthonormal bases, Gram-Schmidt, orthogonal projections (Sections 6.1-6.2). Eigenvalues, eigenvectors, diagonalisation (Chapter 5).
Sections 6.1 and 6.2 gave you the geometric toolkit: inner products, norms, orthonormal bases, projections. Now the course turns to operators and asks: what extra structure does a linear operator inherit when the underlying space has an inner product? The central object is the adjoint T*, which is to operators what conjugation is to complex numbers. From the adjoint come two crucial classes of operator: normal operators (TT* = T*T) and self-adjoint operators (T = T*). The payoff is diagonalisability: a normal operator on a complex inner product space, or a self-adjoint operator on a real one, always has an orthonormal basis of eigenvectors. This is the bridge to the Spectral Theorem in Section 6.6.
Every linear operator T on a finite-dimensional inner product space has a unique adjoint T* satisfying ⟨T(x), y⟩ = ⟨x, T*(y)⟩. If T commutes with T* it is called normal, and if T equals T* it is called self-adjoint. Self-adjoint operators on real spaces and normal operators on complex spaces have orthonormal bases of eigenvectors, which is the main result that powers the rest of the chapter.
Adjoint of T, written T*
The unique linear operator on V satisfying ⟨T(x), y⟩ = ⟨x, T*(y)⟩ for all x, y in V. Exists and is unique whenever V is finite-dimensional. Think of it as the operator-level version of taking the conjugate transpose of a matrix.
Relationship to matrices
If β is an orthonormal basis, then [T*]β = [T]β*. That is, the matrix of the adjoint is the conjugate transpose of the matrix of T. This only works with an orthonormal basis.
Normal operator
A linear operator T such that TT* = T*T. Equivalently, [T]β is a normal matrix (AA* = A*A) for any orthonormal basis β. In simple terms, T and its adjoint commute.
Self-adjoint (Hermitian) operator
A linear operator T such that T = T*. For real inner product spaces, this corresponds to a symmetric matrix; for complex spaces, to a Hermitian matrix. Think of it as the operator that equals its own "conjugate."
Schur's theorem
If the characteristic polynomial of T splits over F, then there exists an orthonormal basis β such that [T]β is upper triangular. This always applies over C (where every polynomial splits), and over R when the characteristic polynomial happens to split.
Positive definite / positive semidefinite operator
T is positive definite if T is self-adjoint and ⟨T(x), x⟩ > 0 for all x ≠ 0. Positive semidefinite if ⟨T(x), x⟩ ≥ 0. In simple terms, a self-adjoint operator with all strictly positive eigenvalues (or nonnegative for semidefinite).
Theorem 6.8 (Riesz representation, finite-dimensional version): every linear functional g : V → F on a finite-dimensional inner product space can be written as g(x) = ⟨x, y⟩ for a unique y ∈ V.
Theorem 6.9 builds on this: for any linear operator T on V, there exists a unique linear operator T* such that ⟨T(x), y⟩ = ⟨x, T*(y)⟩ for all x, y.
To compute T* explicitly:
Pick an orthonormal basis β
Compute A = [T]β
Then [T*]β = A*, and you can read off T* from that matrix
For the corollary: if A is an n × n matrix, then (L_A)* = L_{A*}.
These mirror properties of complex conjugation:
(T + U)* = T* + U*
(cT)* = c̄ T*
(TU)* = U* T* (note the reversal of order)
T** = T
I* = I
The matrix versions follow directly: (AB)* = B*A*, (A*)* = A, etc.
Given an m × n matrix A and y ∈ F^m, the least squares problem is to find x₀ ∈ F^n minimising ‖Ax - y‖.
Theorem 6.12: such an x₀ exists and satisfies the normal equation A*Ax₀ = A*y. If rank(A) = n, the unique solution is:
x₀ = (A*A)⁻¹ A* y
This is used to find the best-fit line (or polynomial) to a set of data points.
Supporting facts:
Lemma 1: ⟨Ax, y⟩_m = ⟨x, A*y⟩_n
Lemma 2: rank(A*A) = rank(A), so A*A is invertible when A has full column rank
When Ax = b is consistent but has infinitely many solutions, the minimal solution s (the one with smallest norm) is unique, lies in R(L_{A*}), and can be found by solving AA*u = b, then computing s = A*u.
If the characteristic polynomial of T splits, then there exists an orthonormal basis β for V such that [T]β is upper triangular. Over C, this always applies.
The proof uses induction on dimension, relying on the fact that if T has an eigenvector, then T* also has one (the lemma before the theorem). The eigenspace of T* determines a one-dimensional invariant subspace whose orthogonal complement is T-invariant, and the induction proceeds on that complement.
For a normal operator T:
‖T(x)‖ = ‖T*(x)‖ for all x
T - cI is normal for every scalar c
If T(x) = λx, then T*(x) = λ̄x (eigenvectors of T are eigenvectors of T*)
Eigenvectors corresponding to distinct eigenvalues are orthogonal
Theorem 6.16 (complex case): T is normal on a finite-dimensional complex inner product space if and only if V has an orthonormal basis of eigenvectors of T.
Theorem 6.17 (real case): T is self-adjoint on a finite-dimensional real inner product space if and only if V has an orthonormal basis of eigenvectors of T.
Key lemma for the real case: if T is self-adjoint, then every eigenvalue is real and the characteristic polynomial splits over R.
Note the asymmetry: over C, normality suffices; over R, you need the stronger condition of self-adjointness. A real normal operator that is not self-adjoint (e.g. a rotation) need not have any eigenvectors at all.
Defining property of the adjoint: ⟨T(x), y⟩ = ⟨x, T*(y)⟩ for all x, y ∈ V
Matrix of the adjoint (orthonormal basis β): [T*]β = ([T]β)*
Normal equation (least squares): A*A x₀ = A*y
Least squares solution (full column rank): x₀ = (A*A)⁻¹ A* y
Minimal solution to Ax = b: Solve AA*u = b, then s = A*u
Least squares approximation is one of the most widely used tools in applied mathematics and statistics. Fitting a regression line to data, calibrating instruments, and estimating parameters in models all reduce to solving the normal equation A*Ax = A*y. Self-adjoint operators (symmetric matrices over R) appear throughout physics: the Hamiltonian in quantum mechanics is self-adjoint, and the fact that its eigenvalues are real corresponds to the requirement that observable quantities be real numbers.
Students often assume [T*]β = [T]β* for any basis β. This is only true when β is an orthonormal basis. With a non-orthonormal basis, the relationship is more complicated.
A normal operator on a real inner product space need not be diagonalisable. The classic counterexample is a rotation of R² by an angle other than 0 or π, which is normal but has no real eigenvectors.
Students sometimes confuse "self-adjoint" with "symmetric." Over R they coincide (self-adjoint = symmetric matrix), but over C, self-adjoint means Hermitian (A = A*), not symmetric (A = Aᵗ). A complex symmetric matrix can fail to be normal.
The adjoint depends on the inner product. Change the inner product and you change T*.
⚠️ Be able to compute T* from T by finding the matrix in an orthonormal basis and taking the conjugate transpose.
⚠️ The defining property ⟨T(x), y⟩ = ⟨x, T*(y)⟩ appears in many proofs on exams. Know how to use it.
⚠️ Know the statement and significance of Theorems 6.16 and 6.17 (the diagonalisation results). These are the theorems the Spectral Theorem builds on.
⚠️ Least squares problems are a common application question. Be comfortable setting up A, y, and solving the normal equation.
⚠️ Know why self-adjoint operators have real eigenvalues (Lemma before Theorem 6.17). This is a favourite short proof question.
True or False: Every linear operator on a finite-dimensional inner product space has a unique adjoint. True.
True or False: If T is normal, eigenvectors for distinct eigenvalues of T are orthogonal. True.
True or False: Every normal operator on a real inner product space is diagonalisable. False. Normality is sufficient over C but not over R. You need self-adjointness over R.
Fill in the blank: For a self-adjoint operator, every eigenvalue is ____. Real.
Fill in the blank: The least squares solution to an overdetermined system Ax = y is x₀ = ____ (assuming full column rank). (A*A)⁻¹ A* y
Q: Let T on C² be defined by T(a₁, a₂) = (2ia₁ + 3a₂, a₁ - a₂). Find T*.
A: With the standard basis, [T] = [[2i, 3], [1, -1]]. Then [T*] = [[-2i, 1], [3, -1]]. So T*(a₁, a₂) = (-2ia₁ + a₂, 3a₁ - a₂).
Q: Prove that the eigenvalues of a self-adjoint operator are real.
A: Let T(x) = λx with x ≠ 0. Since T is self-adjoint, it is normal, so T*(x) = λ̄x. But T = T* gives T*(x) = T(x) = λx. Hence λx = λ̄x, so λ = λ̄, meaning λ is real.
Q: Find the least squares line y = ct + d for the data (1, 2), (2, 3), (3, 5), (4, 7).
A: A = [[1,1],[2,1],[3,1],[4,1]], y = [2,3,5,7]ᵗ. A*A = [[30,10],[10,4]]. A*y = [49,17]ᵗ. Solving: (A*A)⁻¹ = (1/20)[[4,-10],[-10,30]]. x₀ = (1/20)[[4,-10],[-10,30]] · [49,17]ᵗ = [1.7, 0]ᵗ. The least squares line is y = 1.7t.
Q: Why does Theorem 6.16 require the space to be complex?
A: Over C, every polynomial splits, so the characteristic polynomial of T factors into linear terms. Schur's theorem then gives an orthonormal basis making [T]β upper triangular, and normality forces this to be diagonal. Over R, the characteristic polynomial might not split (e.g. a rotation of R²), so the argument breaks down.
Q: Give an example showing that a normal operator on a real inner product space can fail to have eigenvectors.
A: Rotation of R² by π/2: the matrix [[0, -1], [1, 0]] satisfies AA* = I = A*A, so it is normal. Its characteristic polynomial is t² + 1, which has no real roots, so L_A has no eigenvectors over R.
The adjoint is the operator-level analogue of the conjugate transpose, continuing the theme of Chapter 6 that structures on F^n (dot product, transpose) lift to abstract inner product spaces. The diagonalisation results here (Theorems 6.16 and 6.17) are the key inputs to the Spectral Theorem (Section 6.6). Positive definite operators connect to quadratic forms (Section 6.8) and appear throughout optimisation, statistics, and physics. Least squares connects back to the projection theorem (Section 6.2) and forward to the pseudoinverse (Section 6.7).
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