Existence and Uniqueness of the Adjoint

Derivation

Existence and Uniqueness of the Adjoint

D-077 Home PU-104 Threads symmetry Depends on Cauchy–Schwarz and the Triangle Inequality, Riesz Representation in Finite Dimensions

Statement

Let \(V\) be a finite-dimensional inner-product space over \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{C}\), with inner product \(\langle\,\cdot\,,\,\cdot\,\rangle\) linear in its first slot and conjugate-linear in its second. For every linear operator \(A:V\to V\) there exists one and only one linear operator \(A^{\dagger}:V\to V\), the adjoint, satisfying \(\langle Au,v\rangle=\langle u,A^{\dagger}v\rangle\) for all \(u,v\in V\).

Why it matters

The adjoint is the algebraic engine of every symmetry statement in quantum mechanics and linear analysis. Self-adjoint operators (\(A=A^{\dagger}\)) carry real spectra and orthonormal eigenbases, which is exactly why observables are represented by them; unitary operators (\(A^{\dagger}=A^{-1}\)) implement symmetry transformations that preserve probabilities. None of this even has a meaning until we know the adjoint exists and is unique, so that \(A^{\dagger}\) is a genuine function of \(A\) rather than a matter of choice.

The result also cleanly separates the geometry (the inner product) from the coordinates (a basis). We prove existence abstractly through Riesz representation, so that \(A^{\dagger}\) is basis-independent; the familiar matrix formula \(A^{\dagger}=\overline{A}^{\mathsf T}\) then drops out only after we pick an orthonormal basis, revealing it as a computational convenience rather than a definition.

Assumptions

The space \(V\) is finite-dimensional.Riesz representation can fail on incomplete inner-product spaces; on a general (non-complete) infinite-dimensional space some functionals \(v\mapsto\langle Au,v\rangle\) need not be represented by a vector, and the adjoint may not exist as an everywhere-defined bounded operator. \(\langle\,\cdot\,,\,\cdot\,\rangle\) is a genuine inner product (positive definite).Uniqueness rests on non-degeneracy: if some nonzero \(w\) had \(\langle x,w\rangle=0\) for all \(x\), the defining relation would fix \(A^{\dagger}\) only up to that \(w\), destroying uniqueness. The form is sesquilinear with the stated conjugate-linear second slot.If one instead makes the first slot conjugate-linear (the pure-mathematics convention), every explicit conjugation swaps sides; the abstract statement is unchanged but the matrix bookkeeping flips, a frequent source of sign and bar errors. \(A\) is linear.For a merely additive or antilinear map the functional \(u\mapsto\langle Au,v\rangle\) is not linear in \(u\), so Riesz does not apply and no linear adjoint is produced.

Derivation

\[ v\in V \text{ fixed}, \qquad \varphi_v:V\to\mathbb{F}, \quad \varphi_v(u):=\langle Au,v\rangle. \]

Freeze \(v\) and read the defining relation as a map of \(u\). We must find a vector to represent this map. A

\[ \varphi_v(\alpha u_1+\beta u_2)=\langle A(\alpha u_1+\beta u_2),v\rangle=\alpha\langle Au_1,v\rangle+\beta\langle Au_2,v\rangle=\alpha\,\varphi_v(u_1)+\beta\,\varphi_v(u_2). \]

Linearity of \(A\) and linearity of the inner product in its first slot make \(\varphi_v\) a linear functional. A

\[ \exists!\, w_v\in V \ \text{ such that } \ \varphi_v(u)=\langle u,w_v\rangle \quad \forall u\in V. \]

Riesz representation on the finite-dimensional \(V\): every linear functional is \(\langle\,\cdot\,,w\rangle\) for a unique \(w\). This is the one nontrivial input and the sole use of finite dimension. C

\[ \text{Define } \ A^{\dagger}v:=w_v, \qquad \text{so } \ \langle Au,v\rangle=\langle u,A^{\dagger}v\rangle \quad \forall u,v. \]

Step 3 assigns exactly one \(w_v\) to each \(v\); calling it \(A^{\dagger}v\) defines a single-valued map \(A^{\dagger}:V\to V\) that satisfies the required identity by construction. A

\[ \langle u,A^{\dagger}(\alpha v_1+\beta v_2)\rangle=\langle Au,\alpha v_1+\beta v_2\rangle=\overline{\alpha}\langle Au,v_1\rangle+\overline{\beta}\langle Au,v_2\rangle. \]

Conjugate-linearity of the inner product in the second slot; the bars appear because we expand in the second argument. B

\[ =\overline{\alpha}\langle u,A^{\dagger}v_1\rangle+\overline{\beta}\langle u,A^{\dagger}v_2\rangle=\langle u,\alpha A^{\dagger}v_1+\beta A^{\dagger}v_2\rangle \quad \forall u. \]

Apply the defining identity to each term, then repack using conjugate-linearity of the second slot in reverse — the bars cancel and \(\alpha,\beta\) come back out unconjugated. B

\[ \langle u,\;A^{\dagger}(\alpha v_1+\beta v_2)-\alpha A^{\dagger}v_1-\beta A^{\dagger}v_2\rangle=0 \ \ \forall u \ \Longrightarrow\ A^{\dagger}(\alpha v_1+\beta v_2)=\alpha A^{\dagger}v_1+\beta A^{\dagger}v_2. \]

A vector orthogonal to every \(u\in V\) (take \(u\) equal to it) has zero norm, hence is \(0\) by positive-definiteness. So \(A^{\dagger}\) is linear. B

\[ \text{Suppose } B,C \text{ both satisfy } \langle Au,v\rangle=\langle u,Bv\rangle=\langle u,Cv\rangle \ \ \forall u,v. \]

Set up uniqueness: assume two adjoints and compare. A

\[ \langle u,(B-C)v\rangle=0 \ \ \forall u \ \Longrightarrow\ (B-C)v=0 \ \ \forall v \ \Longrightarrow\ B=C. \]

Non-degeneracy again: only the zero vector is orthogonal to all of \(V\). Hence the adjoint is unique. B

Result

\[ \boxed{\ \exists!\ A^{\dagger}:V\to V \text{ linear},\qquad \langle Au,v\rangle=\langle u,A^{\dagger}v\rangle\ \ \forall u,v\in V.\ } \]

Reading. Moving \(A\) across the inner product costs exactly one dagger, and the operator you land on is completely determined by \(A\) and the inner product — no choices, no basis. In an orthonormal basis the dagger is realized as conjugate-transpose, \((A^{\dagger})_{ij}=\overline{A_{ji}}\), but the operator itself predates any basis.

Units check. The construction is purely algebraic and dimensionless in structure: \(A^{\dagger}\) carries the same physical dimensions as \(A\), since \(\langle Au,v\rangle\) and \(\langle u,A^{\dagger}v\rangle\) are the same scalar and \(\langle u,\cdot\rangle\) contributes identically on both sides. If \(A\) is a momentum operator with units of \(\mathrm{kg\,m\,s^{-1}}\), so is \(A^{\dagger}\).

Limiting cases

\(A=I\): the identity is self-adjoint, \(I^{\dagger}=I\), since \(\langle u,v\rangle=\langle u,v\rangle\).
\(A=\lambda I\) with \(\lambda\in\mathbb{C}\): scalar multiples dagger to their conjugate, \((\lambda I)^{\dagger}=\overline{\lambda}\,I\); real scalars are self-adjoint, giving the real-spectrum intuition for observables.
Real inner-product space (\(\mathbb{F}=\mathbb{R}\)), orthonormal basis: conjugation is trivial, \(A^{\dagger}=A^{\mathsf T}\), and the adjoint reduces to the ordinary transpose.
\(\dim V=1\): \(A\) is multiplication by a scalar \(a\), and \(A^{\dagger}\) is multiplication by \(\overline{a}\) — the whole theorem collapses to complex conjugation.

Breaks when

Infinite dimensions without completeness. On a non-complete inner-product (pre-Hilbert) space, Riesz can fail: the functional \(u\mapsto\langle Au,v\rangle\) may not be represented by any vector, so \(A^{\dagger}v\) is undefined. Even on a full Hilbert space an unbounded operator has an adjoint only on a restricted domain \(D(A^{\dagger})\subsetneq V\), and existence becomes a delicate domain question rather than an automatic one.
Degenerate or indefinite forms. If \(\langle\,\cdot\,,\,\cdot\,\rangle\) is only a bilinear form with a nontrivial radical (some \(w\neq0\) orthogonal to everything), uniqueness fails — \(A^{\dagger}\) is fixed only modulo the radical. On indefinite (Minkowski/Krein) spaces one gets a well-defined but different object, the \(\eta\)-adjoint \(A^{\ddagger}=\eta^{-1}A^{\dagger}\eta\), which is what actually appears in relativistic field theory.
Nonlinear or antilinear \(A\). For an antilinear \(A\) (as in time reversal) the map \(u\mapsto\langle Au,v\rangle\) is conjugate-linear, not linear, so Riesz does not apply and one must instead define the antilinear adjoint via \(\langle Au,v\rangle=\overline{\langle u,A^{\dagger}v\rangle}\).

Failure modes

Slot confusion. Writing \(\langle A^{\dagger}u,v\rangle=\langle u,Av\rangle\) and treating it as a new fact — it is the same statement read backwards, valid only because \(A^{\dagger\dagger}=A\), yet students use it to "prove" \(A=A^{\dagger}\).
Dropping the conjugate on scalars. Claiming \((\lambda A)^{\dagger}=\lambda A^{\dagger}\). The scalar conjugates: \((\lambda A)^{\dagger}=\overline{\lambda}\,A^{\dagger}\), because \(\lambda\) sits in the first slot on the left and must cross to the second.
Transpose without conjugate. Using \(A^{\dagger}=A^{\mathsf T}\) over \(\mathbb{C}\). The correct matrix is \(\overline{A}^{\mathsf T}\); forgetting the bar makes "self-adjoint" wrongly mean "symmetric."
Non-orthonormal basis. Applying \((A^{\dagger})_{ij}=\overline{A_{ji}}\) in a basis that is not orthonormal. Then the correct relation is \(A^{\dagger}=G^{-1}\overline{A}^{\mathsf T}G\) with \(G\) the Gram matrix; the naive conjugate-transpose is simply the wrong operator.
Assuming existence infinitely. Carrying the finite-dimensional theorem verbatim to differential operators without checking domains or boundary terms — the integration-by-parts "adjoint" hides boundary conditions that decide whether \(A^{\dagger}\) even equals \(A\).

Discussion

The proof is a template worth internalizing: to define an operator abstractly, characterize how it must pair against every test vector, then invoke Riesz to convert that pairing into an actual vector. The adjoint is the first and cleanest instance, but the same move produces the gradient of a functional, the transpose of a linear map between different spaces, and the Hilbert-space adjoint in quantum mechanics. The essential content is that a finite-dimensional inner-product space is self-dual in a canonical way, and the adjoint is nothing but this duality applied to carry the dual map back into \(V\).

Physically, the adjoint encodes which "measurement pairing" is preserved. Self-adjoint operators are exactly those for which the pairing is symmetric, \(\langle Au,v\rangle=\langle u,Av\rangle\), which by the spectral theorem forces real eigenvalues and an orthonormal eigenbasis — the mathematical reason observables have real, definite measured values with orthogonal outcome states. Unitary operators \(U^{\dagger}U=I\) are those that preserve the inner product itself, \(\langle Uu,Uv\rangle=\langle u,v\rangle\), which is why time evolution \(e^{-iHt/\hbar}\) with self-adjoint \(H\) conserves probability.

The map \(A\mapsto A^{\dagger}\) is itself a rich structure: it is conjugate-linear, \((\lambda A)^{\dagger}=\overline{\lambda}A^{\dagger}\), order-reversing on products, \((AB)^{\dagger}=B^{\dagger}A^{\dagger}\), and involutive, \(A^{\dagger\dagger}=A\). These make the operator algebra a \(*\)-algebra, and adding the norm identity \(\|A^{\dagger}A\|=\|A\|^{2}\) makes it a \(C^{*}\)-algebra — the abstract setting in which all of quantum theory can be phrased without ever choosing a basis.

The deepest reading is categorical: \(\dagger\) is a contravariant functor making finite-dimensional inner-product spaces into a dagger category, and "unitary," "self-adjoint," "isometry," and "projection" all become statements purely about how \(\dagger\) interacts with composition, independent of vectors entirely. This is the viewpoint under which the same word "adjoint" recurs for bounded operators, for the four fundamental subspaces of a matrix, and (loosely) for adjoint functors — the finite-dimensional theorem here is the concrete shadow of a very general principle.

Common misconceptions. The adjoint is not defined as the conjugate-transpose; that is a theorem that holds only in orthonormal coordinates. And "adjoint" here (Hermitian adjoint, \(A^{\dagger}\)) is unrelated to the classical "adjugate/adjoint matrix" \(\operatorname{adj}(A)\) from Cramer's rule — a genuinely different object that unfortunately shares the name.

Worked examples

\[ V=\mathbb{C}^{2},\quad A=\begin{pmatrix} 2 & 1+i \\ 0 & 3i \end{pmatrix},\quad \text{standard inner product } \langle x,y\rangle=\textstyle\sum_k x_k\overline{y_k}. \]

The standard basis is orthonormal, so the adjoint is the conjugate-transpose. A

\[ A^{\dagger}=\overline{A}^{\mathsf T}=\begin{pmatrix} 2 & 0 \\ 1-i & -3i \end{pmatrix}. \]

Transpose, then conjugate each entry: \(1+i\to 1-i\) moves off-diagonal, \(3i\to-3i\). A

\[ \text{Take } u=\begin{pmatrix}1\\0\end{pmatrix},\ v=\begin{pmatrix}0\\1\end{pmatrix}:\quad Au=\begin{pmatrix}2\\0\end{pmatrix},\qquad \langle Au,v\rangle=2\cdot\overline{0}+0\cdot\overline{1}=0. \]

Compute the left side of the defining identity for a probing pair. A

\[ A^{\dagger}v=\begin{pmatrix}0\\-3i\end{pmatrix},\qquad \langle u,A^{\dagger}v\rangle=1\cdot\overline{0}+0\cdot\overline{-3i}=0=\langle Au,v\rangle.\ \checkmark \]

Right side matches; the identity holds for this pair (and, by linearity, all pairs). B

\[ A^{\dagger}=\begin{pmatrix} 2 & 0 \\ 1-i & -3i \end{pmatrix}. \]

Reading. A dimensionless matrix operator; \(A\neq A^{\dagger}\), so \(A\) is not self-adjoint, and indeed its diagonal entry \(3i\) is not real.

Units check. Pure complex numbers throughout; both sides of the test are scalars of the same (dimensionless) type.

\[ V=\mathbb{R}^{2} \text{ with weighted inner product } \langle x,y\rangle=x_1y_1+2x_2y_2,\quad A=\begin{pmatrix}0&1\\0&0\end{pmatrix}. \]

Non-orthonormal geometry: Gram matrix \(G=\operatorname{diag}(1,2)\neq I\), so the plain transpose is the wrong answer. B

\[ A^{\dagger}=G^{-1}A^{\mathsf T}G,\qquad G=\begin{pmatrix}1&0\\0&2\end{pmatrix},\ G^{-1}=\begin{pmatrix}1&0\\0&\tfrac12\end{pmatrix}. \]

In a non-orthonormal basis the defining identity \(x^{\mathsf T}G\,(A^{\dagger}y)=(Ax)^{\mathsf T}Gy\) rearranges to \(A^{\dagger}=G^{-1}A^{\mathsf T}G\) (real entries, so no conjugation). C

\[ A^{\mathsf T}G=\begin{pmatrix}0&0\\1&0\end{pmatrix}\begin{pmatrix}1&0\\0&2\end{pmatrix}=\begin{pmatrix}0&0\\1&0\end{pmatrix},\quad A^{\dagger}=\begin{pmatrix}1&0\\0&\tfrac12\end{pmatrix}\begin{pmatrix}0&0\\1&0\end{pmatrix}=\begin{pmatrix}0&0\\ \tfrac12&0\end{pmatrix}. \]

Multiply the three matrices in order; the weight injects the factor \(\tfrac12\) that a naive transpose would miss. B

\[ \text{Check } u=e_1,\ v=e_2:\quad \langle Au,v\rangle=\langle 0,\,e_2\rangle=0,\qquad A^{\dagger}v=\tfrac12 e_2,\ \ \langle u,A^{\dagger}v\rangle=\langle e_1,\tfrac12 e_2\rangle=0.\ \checkmark \]

Confirm with one pair. The naive transpose \(\left(\begin{smallmatrix}0&0\\1&0\end{smallmatrix}\right)\) would give \(A^{\mathsf T}v=e_1\) and \(\langle e_1,e_1\rangle=1\neq0\), the wrong value. B

\[ A^{\dagger}=\begin{pmatrix}0&0\\ \tfrac12&0\end{pmatrix}\neq A^{\mathsf T}=\begin{pmatrix}0&0\\1&0\end{pmatrix}. \]

Reading. With a non-trivial metric the adjoint is metric-conjugation of the transpose; forgetting \(G\) is the single most common concrete error.

Units check. \(G\) and \(G^{-1}\) carry inverse dimensions that cancel, so \(A^{\dagger}\) has the same dimension as \(A\) — dimensionless here.

Problems

(A) On \(\mathbb{C}^2\) with the standard inner product, let \(A=\begin{pmatrix}2&1-i\\0&3\end{pmatrix}\). Write \(A^{\dagger}\) and state whether \(A\) is self-adjoint.
Solution
\(A^{\dagger}=\overline{A}^{\mathsf T}=\begin{pmatrix}2&0\\1+i&3\end{pmatrix}\). Self-adjointness needs \(A_{21}=\overline{A_{12}}=1+i\), but \(A_{21}=0\). So \(A\neq A^{\dagger}\): \(A\) is not self-adjoint.
(A) Prove \((A^{\dagger})^{\dagger}=A\) directly from the defining identity.
Solution
For all \(u,v\): \(\langle A^{\dagger}u,v\rangle=\overline{\langle v,A^{\dagger}u\rangle}=\overline{\langle Av,u\rangle}=\langle u,Av\rangle\), using conjugate symmetry twice and the definition of \(A^{\dagger}\) in the middle. Thus \(A\) satisfies the adjoint identity for \(A^{\dagger}\); by uniqueness \((A^{\dagger})^{\dagger}=A\).
(B) Prove \((AB)^{\dagger}=B^{\dagger}A^{\dagger}\).
Solution
For all \(u,v\): \(\langle ABu,v\rangle=\langle Bu,A^{\dagger}v\rangle=\langle u,B^{\dagger}A^{\dagger}v\rangle\), peeling \(A\) off first, then \(B\). So \(B^{\dagger}A^{\dagger}\) satisfies the defining identity for \(AB\); by uniqueness it is \((AB)^{\dagger}\). The order reversal is the hallmark of the involution.
(B) Let \(U=\dfrac{1}{\sqrt2}\begin{pmatrix}1&-1\\1&1\end{pmatrix}\) on \(\mathbb{R}^2\) (standard inner product). Verify \(U^{\dagger}U=I\) and conclude \(U^{\dagger}=U^{-1}\).
Solution
Real orthonormal basis, so \(U^{\dagger}=U^{\mathsf T}=\tfrac1{\sqrt2}\begin{pmatrix}1&1\\-1&1\end{pmatrix}\). Then \(U^{\dagger}U=\tfrac12\begin{pmatrix}1&1\\-1&1\end{pmatrix}\begin{pmatrix}1&-1\\1&1\end{pmatrix}=\tfrac12\begin{pmatrix}2&0\\0&2\end{pmatrix}=I\). Since \(V\) is finite-dimensional, a left inverse is a two-sided inverse, so \(U^{-1}=U^{\dagger}\): \(U\) is orthogonal (unitary) and preserves all inner products.
(C) On \(\mathbb{R}^2\) with inner product \(\langle x,y\rangle=x^{\mathsf T}Gy\), \(G=\begin{pmatrix}2&1\\1&1\end{pmatrix}\), find the adjoint of \(A=\begin{pmatrix}1&1\\0&1\end{pmatrix}\).
Solution
Use \(A^{\dagger}=G^{-1}A^{\mathsf T}G\). Here \(\det G=1\), so \(G^{-1}=\begin{pmatrix}1&-1\\-1&2\end{pmatrix}\), and \(A^{\mathsf T}=\begin{pmatrix}1&0\\1&1\end{pmatrix}\). Then \(A^{\mathsf T}G=\begin{pmatrix}1&0\\1&1\end{pmatrix}\begin{pmatrix}2&1\\1&1\end{pmatrix}=\begin{pmatrix}2&1\\3&2\end{pmatrix}\), and \(A^{\dagger}=G^{-1}\begin{pmatrix}2&1\\3&2\end{pmatrix}=\begin{pmatrix}1&-1\\-1&2\end{pmatrix}\begin{pmatrix}2&1\\3&2\end{pmatrix}=\begin{pmatrix}-1&-1\\4&3\end{pmatrix}\). Sanity check: \(\operatorname{tr}A^{\dagger}=2=\operatorname{tr}A\) and \(\det A^{\dagger}=(-1)(3)-(-1)(4)=1=\det A\), as a \(G\)-conjugate of \(A^{\mathsf T}\) must preserve.