Source: Abstract Linear Algebra, University of Illinois at Urbana-Champaign
Tags: adjoint operator, self-adjoint, normal operator, Hermitian, symmetric matrix, inner product space, conjugate transpose, orthonormal basis, linear algebra
Difficulty: Intermediate Prerequisites: Inner product spaces, orthonormal bases, Gram-Schmidt, matrix representations of linear maps, conjugate transpose.
Big picture: This lecture sets up the machinery needed to answer a central question in linear algebra: when can a linear operator be diagonalised by an orthonormal basis of eigenvectors? Before reaching the Spectral Theorem itself, you need three definitions that describe how an operator relates to its "mirror image" under the inner product. These are the adjoint, self-adjoint, and normal operator. If you are comfortable with inner products and matrix representations, you are ready for this material. If not, review those topics first.
The adjoint T* of a linear operator T reverses T's role inside an inner product. A self-adjoint operator equals its own adjoint (the matrix version: A = Aᵀ over ℝ, or A = A* over ℂ). A normal operator commutes with its adjoint. Self-adjoint implies normal, but not the other way round.
Adjoint (T)*
Given T: V → V on a finite-dimensional inner product space, the adjoint is the unique operator T*: V → V satisfying ⟨Tx, y⟩ = ⟨x, T*y⟩ for all x, y ∈ V. Think of it as the operator that "moves T from one side of the inner product to the other."
Self-adjoint (Hermitian)
T is self-adjoint if T = T*. For a real matrix this means A = Aᵀ (symmetric). For a complex matrix this means A = A* (Hermitian, i.e. conjugate transpose equals itself). In simple terms, the operator is its own mirror image under the inner product.
Normal operator
T is normal if TT* = TT, i.e. T commutes with its adjoint. In matrix language, AA = A*A. Think of it as a weaker version of self-adjoint: the operator and its adjoint do not have to be equal, they just have to commute.
Conjugate transpose (A)*
For a matrix A ∈ M_n(𝔽), the conjugate transpose is obtained by transposing and then complex-conjugating every entry. Over ℝ this is just the ordinary transpose.
T: (Vⁿ, ⟨ , ⟩) → (Vⁿ, ⟨ , ⟩) is a linear operator on a finite-dimensional inner product space.
The adjoint T* is defined by the relation:
⟨Tx, y⟩ = ⟨x, T*y⟩ for all x, y ∈ V
This uniquely determines T* (existence and uniqueness follow from the Riesz representation theorem, covered earlier in the course).
Key fact: if β is an orthonormal basis of V, then
[T*]_β = ([T]_β)*
That is, the matrix of the adjoint is the conjugate transpose of the matrix of T.
Why orthonormality matters: when β is orthonormal, the inner product ⟨x, y⟩ equals [y]_β* · [x]_β (the standard dot product of coordinate vectors). This is Lemma 1 in the lecture.
The proof runs:
⟨Tx, y⟩ = [y]_β* [Tx]_β = [y]_β* [T]_β [x]_β
⟨x, Ty⟩ = [Ty]_β* [x]_β = ([T*]_β [y]_β)* [x]_β = [y]_β* ([T*]_β)* [x]_β
Setting these equal for all x, y gives [T]_β = ([T*]_β), hence [T]_β = ([T]_β)*.
If the basis is not orthonormal, this clean relationship breaks down.
T is self-adjoint when T = T*.
In matrix terms (orthonormal basis):
Over ℝ: A is self-adjoint iff A = Aᵀ (symmetric matrix).
Over ℂ: A is self-adjoint iff A = A* (Hermitian matrix).
T is normal when TT* = T*T.
Every self-adjoint operator is normal (since T = T* trivially gives TT* = T*T).
Normal does not imply self-adjoint.
The rotation matrix
A = [ cos θ, −sin θ ; sin θ, cos θ ]
is normal for all θ, but is self-adjoint only when θ = kπ (i.e. when sin θ = 0).
Check: A* = Aᵀ = [ cos θ, sin θ ; −sin θ, cos θ ], which equals A only if sin θ = 0.
Normality check: AA = I = AA, so A commutes with its transpose regardless of θ. This is because rotation matrices are orthogonal.
Adjoint definition: ⟨Tx, y⟩ = ⟨x, T*y⟩
Matrix of adjoint (orthonormal basis): [T*]_β = ([T]_β)*
Inner product via coordinates (orthonormal β): ⟨x, y⟩ = [y]_β* [x]_β
Self-adjoint condition: T = T* ⟺ A = A* (in orthonormal basis)
Normal condition: TT* = TT ⟺ AA = A*A
Self-adjoint (symmetric/Hermitian) matrices appear everywhere physical systems are modelled: the inertia tensor in mechanics, the Hamiltonian in quantum mechanics, covariance matrices in statistics. Normality is the broadest class of operators that can still be orthogonally diagonalised over ℂ, which matters in signal processing and control theory.
Students often think "normal" and "self-adjoint" are the same thing. They are not: every self-adjoint operator is normal, but a rotation matrix shows the converse fails.
Students sometimes apply the formula [T*]_β = ([T]_β)* when β is not orthonormal. The formula requires orthonormality.
Over ℝ, "self-adjoint" just means symmetric. Students sometimes forget to conjugate when working over ℂ and confuse transpose with conjugate transpose.
The adjoint depends on the inner product, not just the operator. Change the inner product and T* changes too.
⚠️ The relationship [T*]_β = ([T]_β)* is a standard exam calculation. Be ready to use it in both directions.
⚠️ Knowing that self-adjoint ⇒ normal (but not the reverse) is a common true/false question.
⚠️ The rotation matrix example is a classic illustration of normal-but-not-self-adjoint. Be able to verify both properties by direct computation.
True or false: Every normal operator is self-adjoint.
Fill in the blank: The adjoint T* is defined by the property ⟨Tx, y⟩ = ______ for all x, y ∈ V.
True or false: [T*]_β = ([T]_β)* holds for any basis β.
Fill in the blank: A real matrix A is self-adjoint if and only if A = ______.
True or false: The 2×2 rotation matrix by angle π/4 is normal.
Answers: 1. False. 2. ⟨x, T*y⟩. 3. False (requires orthonormal β). 4. Aᵀ. 5. True.
Q: Let T: ℝ³ → ℝ³ have matrix A = [[2, 1, 0], [1, 3, 1], [0, 1, 2]] with respect to the standard basis. Is T self-adjoint?
A: The standard basis is orthonormal, so check whether A = Aᵀ. Since A is symmetric, yes, T is self-adjoint.
Q: Give an example of an operator that is normal but not self-adjoint.
A: The rotation by π/2 in ℝ², with matrix [[0, −1], [1, 0]]. Its transpose is [[0, 1], [−1, 0]], which is not equal to A, so it is not self-adjoint. But AᵀA = I = AAᵀ, so it is normal.
Q: Suppose β is an orthonormal basis and [T]_β = [[1, i], [0, 2]]. What is [T]_β?*
A: Take the conjugate transpose: [T*]_β = [[1, 0], [−i, 2]].
Q: Why does the proof that [T]_β = ([T]_β) require β to be orthonormal?**
A: The proof uses the fact that ⟨x, y⟩ = [y]_β* [x]_β, which holds only when β is orthonormal. For a general basis, the inner product involves the Gram matrix of β, and the clean conjugate-transpose relationship no longer applies.
This material connects directly to the Spectral Theorem (next set of notes), which answers the question: when does an orthonormal eigenbasis exist? The answer depends on whether T is normal (over ℂ) or self-adjoint (over ℝ). It also ties back to Gram-Schmidt, which is used in the proof of Schur's Theorem to build orthonormal bases.
Related Terms / Search Tags: adjoint operator, conjugate transpose, Hermitian matrix, symmetric matrix, self-adjoint operator, normal operator, inner product space, orthonormal basis, T*, A*, matrix representation, MATH 416, abstract linear algebra, UIUC