The Adjoint of a Linear Map

Suppose you have a linear map TT between two inner product spaces. You want to understand it through the lens of the inner product itself. If T:V→WT: V \to W scales, rotates, and shears vectors, is there a companion map going the other way? This companion map, W→VW \to V, would capture exactly how TT interacts with angles and lengths. This companion, called the adjoint of TT, turns out to exist always. It is also unique, and it encodes almost every structural fact we care about. This includes symmetry, orthogonality, positivity, and eventually the singular value decomposition itself. The adjoint matters because it is the correct linear-algebraic generalization of the conjugate transpose of a matrix. It also matters because so many important classes of operators (self-adjoint, normal, unitary/orthogonal) are defined purely in terms of how TT relates to T∗T^*. Before we can even define T∗T^* rigorously, we need a tool. That tool converts every linear functional on an inner product space into an inner-product-with-a-fixed-vector. This is the Riesz Representation Theorem, and it is the engine that makes the whole construction work. The big picture: an inner product space lets you measure lengths and angles. A linear functional is a way of extracting a number from a vector. Riesz representation says that on a finite-dimensional (or complete) inner product space, every functional is secretly "take the inner product with some fixed vector." Once we know this, we can define T∗wT^*w by declaring that ⟨Tv,w⟩=⟨v,T∗w⟩\langle Tv, w\rangle = \langle v, T^*w\rangle for all vv. This works because the left side is a linear functional of vv for each fixed ww. So Riesz representation produces the vector T∗wT^*w automatically. This article proves existence and uniqueness of T∗T^* via Riesz representation. It establishes the basic algebraic identities ((S+T)∗=S∗+T∗ (S+T)^* = S^*+T^*, (ST)∗=T∗S∗(ST)^*=T^*S^*, (T∗)∗=T(T^*)^*=T, etc.). It also proves that in orthonormal bases the matrix of T∗T^* is the conjugate transpose of the matrix of TT. It sets up the language needed for Self-Adjoint Operators, Normal Operators, and eventually the Singular Value Decomposition.

Key takeaways
  • Every linear functional φ\varphi on a finite-dimensional inner product space has a unique vector uu with φ(v)=⟨v,u⟩\varphi(v) = \langle v, u\rangle for all vv — this is Riesz representation.
  • For T:V→WT: V \to W linear, the adjoint T∗:W→VT^*: W \to V is the unique linear map satisfying ⟨Tv,w⟩W=⟨v,T∗w⟩V\langle Tv, w\rangle_W = \langle v, T^*w\rangle_V for all v∈V,w∈Wv \in V, w \in W.
  • Existence and uniqueness of T∗T^* both come directly from Riesz representation, applied one ww at a time.
  • The adjoint is conjugate-linear-friendly: (S+T)∗=S∗+T∗(S+T)^*=S^*+T^*, (λT)∗=λˉT∗(\lambda T)^* = \bar\lambda T^*, (ST)∗=T∗S∗(ST)^*=T^*S^*, (T∗)∗=T(T^*)^*=T.
  • In orthonormal bases, the matrix of T∗T^* is the conjugate transpose M(T)‾⊤\overline{M(T)}^{\top} of the matrix of TT.
  • Key idea: the adjoint is the abstract, basis-free generalization of "flip the matrix and conjugate," produced entirely by the Riesz representation theorem.

Where this lives today

The adjoint is not a curiosity; it is the hinge on which the entire structure theory of operators on inner product spaces turns. Self-adjoint operators (T=T∗T = T^*) generalize real symmetric and complex Hermitian matrices, and by the Spectral Theorem (built later using Self-Adjoint Operators and Normal Operators) they are exactly diagonalizable with real eigenvalues in an orthonormal basis. Normal operators (TT∗=T∗TTT^* = T^*T) are exactly the operators diagonalizable by a unitary/orthonormal change of basis, unifying Hermitian, skew-Hermitian, and unitary operators under one umbrella. The Singular Value Decomposition, arguably the most computationally important factorization in applied linear algebra, is built by analyzing the self-adjoint, positive operator T∗TT^*T. Beyond pure linear algebra: in quantum mechanics, observables are modeled as self-adjoint operators on a Hilbert space precisely because the adjoint machinery guarantees real "eigenvalues" (measurement outcomes). In statistics and data science, the normal equations for least squares are literally T∗Tx=T∗bT^*T x = T^* b, an adjoint identity in disguise, underlying linear regression and the Moore–Penrose pseudoinverse. In numerical linear algebra and engineering, adjoint methods power efficient gradient computations (adjoint-state methods in PDE-constrained optimization, and backpropagation itself can be viewed as repeated adjoint computation through a composition of linear maps). In signal processing, the discrete Fourier transform matrix is unitary, meaning its inverse equals its adjoint, which is why Fourier synthesis and analysis are adjoint operations of each other.

Ask the tutor: How does the adjoint T∗TT^*T being self-adjoint and positive lead directly into constructing the Singular Value Decomposition?

DefinitionAdjoint of a linear map

Let VV and WW be finite-dimensional inner product spaces over F\mathbb{F} (where F=R\mathbb{F} = \mathbb{R} or F=C\mathbb{F}=\mathbb{C}), and let T:V→WT: V \to W be linear. The adjoint of TT is the map T∗:W→VT^*: W \to V satisfying

⟨Tv,w⟩W=⟨v,T∗w⟩Vfor all v∈V, w∈W.\langle Tv, w \rangle_W = \langle v, T^*w \rangle_V \qquad \text{for all } v \in V,\ w \in W.

(Below we prove that such a map T∗T^* exists, is unique, and is itself linear.)

Intuition

Think of TT as a machine converting inputs in VV to outputs in WW. The defining equation of the adjoint says: measuring TvTv against ww in WW is the same number as measuring vv against T∗wT^*w in VV. The adjoint is the operator that lets you "move TT across the inner product" from one side to the other, at the cost of moving it from VV-side to WW-side and back. The key structural fact underneath this is the Riesz representation theorem: linear functionals on an inner product space are nothing but "inner product with a fixed vector." A linear functional φ:V→F\varphi: V \to \mathbb{F} eats a vector and returns a scalar; Riesz representation says every such φ\varphi is φ(v)=⟨v,uφ⟩\varphi(v) = \langle v, u_\varphi\rangle for a unique uφ∈Vu_\varphi \in V. Once you fix w∈Ww \in W, the map v↦⟨Tv,w⟩Wv \mapsto \langle Tv, w\rangle_W is exactly such a functional on VV, so it must equal ⟨v,u⟩V\langle v, u\rangle_V for a unique uu depending on ww. We define T∗w:=uT^*w := u. This is the entire proof strategy: Riesz representation is invoked once for every ww, and linearity of T∗T^* in ww is then checked separately. Geometrically, in the real, finite-dimensional case with orthonormal bases, TT is a matrix AA, and T∗T^* is A⊤A^\top; over C\mathbb{C} it is A‾⊤\overline{A}^\top. The picture below shows this "transpose and conjugate the matrix" idea alongside the abstract "move across the inner product" idea, which are the same fact viewed from coordinates versus viewed intrinsically.

Figure
The adjoint moves T across the inner product, and in coordinates this becomes conjugate-transpose
Abstract picture: ⟨Tv, w⟩_W = ⟨v, T*w⟩_V V v T*w W Tv w T In coordinates (orthonormal basis): if M(T) = A, then M(T*) = A̅ᵀ

Why finite-dimensionality is used

Everything below assumes VV and WW are finite-dimensional. This hypothesis is used twice: once to guarantee that orthonormal bases exist (needed for Riesz representation's proof), and once to guarantee that a linear functional is automatically continuous/well-behaved, which is automatic in finite dimensions but requires extra care (completeness) in infinite dimensions.

TheoremRiesz Representation Theorem

Let VV be a finite-dimensional inner product space over F\mathbb{F}, and let φ:V→F\varphi: V \to \mathbb{F} be a linear functional. Then there exists a unique vector u∈Vu \in V such that

φ(v)=⟨v,u⟩for all v∈V.\varphi(v) = \langle v, u \rangle \qquad \text{for all } v \in V.
Proof

Existence. Since VV is finite-dimensional, it has an orthonormal basis e1,…,ene_1, \dots, e_n (this uses Existence of Orthonormal Bases, e.g. via Gram–Schmidt). Define

u:=∑i=1nφ(ei)‾ ei∈V.u := \sum_{i=1}^n \overline{\varphi(e_i)}\, e_i \in V.

We check that uu represents φ\varphi. Let v∈Vv \in V be arbitrary. Because e1,…,ene_1,\dots,e_n is an orthonormal basis, we can write v=∑i=1naieiv = \sum_{i=1}^n a_i e_i where ai=⟨v,ei⟩a_i = \langle v, e_i\rangle (this is the standard orthonormal-basis coordinate formula, itself a direct computation using ⟨ei,ej⟩=δij\langle e_i, e_j\rangle = \delta_{ij}). Then by linearity of φ\varphi,

φ(v)=φ(∑iaiei)=∑iaiφ(ei).\varphi(v) = \varphi\Big(\sum_i a_i e_i\Big) = \sum_i a_i \varphi(e_i).

On the other hand, using linearity of the inner product in the first slot and conjugate-linearity in the second slot,

⟨v,u⟩=⟨∑iaiei, ∑jφ(ej)‾ej⟩=∑i,jaiφ(ej)‾‾ ⟨ei,ej⟩=∑i,jaiφ(ej) δij=∑iaiφ(ei).\langle v, u \rangle = \Big\langle \sum_i a_i e_i,\ \sum_j \overline{\varphi(e_j)} e_j \Big\rangle = \sum_{i,j} a_i \overline{\overline{\varphi(e_j)}}\, \langle e_i, e_j\rangle = \sum_{i,j} a_i \varphi(e_j)\, \delta_{ij} = \sum_i a_i \varphi(e_i).

Both expressions equal ∑iaiφ(ei)\sum_i a_i \varphi(e_i), so φ(v)=⟨v,u⟩\varphi(v) = \langle v, u \rangle for all v∈Vv \in V. This proves existence.

Uniqueness. Suppose u1,u2∈Vu_1, u_2 \in V both satisfy φ(v)=⟨v,u1⟩=⟨v,u2⟩\varphi(v) = \langle v, u_1\rangle = \langle v, u_2 \rangle for all v∈Vv \in V. Then for all v∈Vv \in V,

⟨v,u1−u2⟩=⟨v,u1⟩−⟨v,u2⟩=0.\langle v, u_1 - u_2 \rangle = \langle v, u_1\rangle - \langle v, u_2\rangle = 0.

Taking v=u1−u2v = u_1 - u_2 gives ⟨u1−u2,u1−u2⟩=0\langle u_1 - u_2, u_1 - u_2\rangle = 0, i.e. ∥u1−u2∥2=0\|u_1 - u_2\|^2 = 0. By positive-definiteness of the inner product, u1−u2=0u_1 - u_2 = 0, so u1=u2u_1 = u_2. ■\blacksquare

∎
TheoremExistence and uniqueness of the adjoint

Let V,WV, W be finite-dimensional inner product spaces over F\mathbb{F}, and let T:V→WT: V \to W be linear. Then there exists a unique linear map T∗:W→VT^*: W \to V such that

⟨Tv,w⟩W=⟨v,T∗w⟩Vfor all v∈V, w∈W.\langle Tv, w \rangle_W = \langle v, T^*w \rangle_V \qquad \text{for all } v \in V,\ w \in W.
Proof

Existence. Fix w∈Ww \in W. Define φw:V→F\varphi_w: V \to \mathbb{F} by

φw(v):=⟨Tv,w⟩W.\varphi_w(v) := \langle Tv, w \rangle_W.

This is linear in vv: for v1,v2∈Vv_1, v_2 \in V and λ∈F\lambda \in \mathbb{F},

φw(v1+λv2)=⟨T(v1+λv2),w⟩W=⟨Tv1+λTv2,w⟩W=⟨Tv1,w⟩W+λ⟨Tv2,w⟩W=φw(v1)+λφw(v2),\varphi_w(v_1 + \lambda v_2) = \langle T(v_1 + \lambda v_2), w\rangle_W = \langle Tv_1 + \lambda Tv_2, w\rangle_W = \langle Tv_1, w\rangle_W + \lambda \langle Tv_2, w \rangle_W = \varphi_w(v_1) + \lambda \varphi_w(v_2),

using linearity of TT and linearity of the inner product in its first argument. So φw\varphi_w is a linear functional on VV. By Riesz Representation Theorem applied to VV, there is a unique vector, call it T∗w∈VT^*w \in V, such that

φw(v)=⟨v,T∗w⟩Vfor all v∈V,\varphi_w(v) = \langle v, T^*w\rangle_V \qquad \text{for all } v \in V,

which is exactly the required identity ⟨Tv,w⟩W=⟨v,T∗w⟩V\langle Tv, w\rangle_W = \langle v, T^*w\rangle_V. This defines a function T∗:W→VT^*: W \to V (well-defined precisely because Riesz representation produces a unique vector for each ww).

T∗T^* is linear. Let w1,w2∈Ww_1, w_2 \in W and μ∈F\mu \in \mathbb{F}. For every v∈Vv \in V,

⟨v,T∗(w1+μw2)⟩V=⟨Tv,w1+μw2⟩W=⟨Tv,w1⟩W+μ‾⟨Tv,w2⟩W\langle v, T^*(w_1 + \mu w_2)\rangle_V = \langle Tv, w_1 + \mu w_2\rangle_W = \langle Tv, w_1\rangle_W + \overline{\mu}\langle Tv, w_2\rangle_W

using conjugate-linearity of the inner product in the second slot. Continuing,

=⟨v,T∗w1⟩V+μ‾⟨v,T∗w2⟩V=⟨v,T∗w1⟩V+⟨v,μT∗w2⟩V=⟨v,T∗w1+μT∗w2⟩V,= \langle v, T^*w_1\rangle_V + \overline{\mu}\langle v, T^*w_2\rangle_V = \langle v, T^*w_1\rangle_V + \langle v, \mu T^*w_2\rangle_V = \langle v, T^*w_1 + \mu T^*w_2\rangle_V,

again using conjugate-linearity in the second slot in reverse. Since this holds for all v∈Vv \in V, the uniqueness clause of Riesz representation (applied to the functional v↦⟨v,T∗(w1+μw2)⟩Vv \mapsto \langle v, T^*(w_1+\mu w_2)\rangle_V, which is represented by two candidate vectors) forces

T∗(w1+μw2)=T∗w1+μT∗w2.T^*(w_1 + \mu w_2) = T^*w_1 + \mu T^*w_2.

Hence T∗T^* is linear.

Uniqueness of T∗T^* as a map. Suppose S:W→VS: W \to V is any linear map with ⟨Tv,w⟩W=⟨v,Sw⟩V\langle Tv, w\rangle_W = \langle v, Sw\rangle_V for all v,wv, w. Fix w∈Ww \in W. Then for all v∈Vv \in V,

⟨v,Sw⟩V=⟨Tv,w⟩W=⟨v,T∗w⟩V,\langle v, Sw \rangle_V = \langle Tv, w \rangle_W = \langle v, T^*w \rangle_V,

so ⟨v,Sw−T∗w⟩V=0\langle v, Sw - T^*w\rangle_V = 0 for all v∈Vv \in V. Taking v=Sw−T∗wv = Sw - T^*w gives ∥Sw−T∗w∥2=0\|Sw - T^*w\|^2 = 0, so Sw=T∗wSw = T^*w. Since ww was arbitrary, S=T∗S = T^*. ■\blacksquare

∎
LemmaBasic algebraic properties of the adjoint

Let V,W,UV, W, U be finite-dimensional inner product spaces over F\mathbb{F}. Let S,T:V→WS, T: V \to W and R:W→UR: W \to U be linear, and let λ∈F\lambda \in \mathbb{F}. Then: 1. (S+T)∗=S∗+T∗(S + T)^* = S^* + T^*. 2. (λT)∗=λ‾ T∗(\lambda T)^* = \overline{\lambda}\, T^*. 3. (T∗)∗=T(T^*)^* = T. 4. (RT)∗=T∗R∗(RT)^* = T^* R^* (note the order reversal), where RT:V→URT: V \to U. 5. (idV)∗=idV(\mathrm{id}_V)^* = \mathrm{id}_V.

Proof

Each identity is proved by exhibiting the defining inner-product identity and invoking uniqueness of the adjoint.

(1) For all v∈V,w∈Wv \in V, w \in W,

⟨(S+T)v,w⟩W=⟨Sv,w⟩W+⟨Tv,w⟩W=⟨v,S∗w⟩V+⟨v,T∗w⟩V=⟨v,(S∗+T∗)w⟩V.\langle (S+T)v, w\rangle_W = \langle Sv, w\rangle_W + \langle Tv, w\rangle_W = \langle v, S^*w\rangle_V + \langle v, T^*w\rangle_V = \langle v, (S^*+T^*)w\rangle_V.

By the uniqueness half of the adjoint theorem (applied to the map S+TS+T), the unique operator satisfying this identity is (S+T)∗(S+T)^*; since S∗+T∗S^*+T^* also satisfies it, (S+T)∗=S∗+T∗(S+T)^* = S^*+T^*.

(2) For all v∈V,w∈Wv \in V, w \in W,

⟨(λT)v,w⟩W=λ⟨Tv,w⟩W=λ⟨v,T∗w⟩V=⟨v,λ‾T∗w⟩V,\langle (\lambda T)v, w\rangle_W = \lambda \langle Tv, w\rangle_W = \lambda \langle v, T^*w\rangle_V = \langle v, \overline{\lambda} T^* w\rangle_V,

using conjugate-linearity in the second slot to move λ\lambda across as λ‾\overline{\lambda}. By uniqueness, (λT)∗=λ‾T∗(\lambda T)^* = \overline\lambda T^*.

(3) By definition, T∗:W→VT^*: W \to V satisfies ⟨Tv,w⟩W=⟨v,T∗w⟩V\langle Tv, w\rangle_W = \langle v, T^*w \rangle_V for all v,wv, w. Taking complex conjugates of both sides (both sides are scalars),

⟨Tv,w⟩W‾=⟨v,T∗w⟩V‾  ⟺  ⟨w,Tv⟩W=⟨T∗w,v⟩V,\overline{\langle Tv, w \rangle_W} = \overline{\langle v, T^*w\rangle_V} \iff \langle w, Tv\rangle_W = \langle T^*w, v\rangle_V,

using conjugate symmetry of the inner product. Relabeling (w∈W,v∈Vw \in W, v \in V swapped in role), this says exactly that for all w∈W,v∈Vw \in W, v \in V,

⟨T∗w,v⟩V=⟨w,Tv⟩W,\langle T^*w, v \rangle_V = \langle w, Tv\rangle_W,

which is precisely the defining property of the adjoint of T∗T^* applied to TT: it says TT satisfies the identity that defines (T∗)∗(T^*)^*. By uniqueness of the adjoint of T∗T^*, we conclude (T∗)∗=T(T^*)^* = T.

(4) RT:V→URT: V \to U is linear (composite of linear maps). For all v∈V,u∈Uv \in V, u \in U,

⟨(RT)v,u⟩U=⟨R(Tv),u⟩U=⟨Tv,R∗u⟩W=⟨v,T∗(R∗u)⟩V=⟨v,(T∗R∗)u⟩V,\langle (RT)v, u \rangle_U = \langle R(Tv), u\rangle_U = \langle Tv, R^*u \rangle_W = \langle v, T^*(R^*u)\rangle_V = \langle v, (T^*R^*)u \rangle_V,

using the defining property of R∗R^* first, then of T∗T^*. By uniqueness, (RT)∗=T∗R∗(RT)^* = T^*R^*.

(5) For all v1,v2∈Vv_1, v_2 \in V, trivially ⟨idVv1,v2⟩V=⟨v1,v2⟩V=⟨v1,idVv2⟩V\langle \mathrm{id}_V v_1, v_2\rangle_V = \langle v_1, v_2\rangle_V = \langle v_1, \mathrm{id}_V v_2\rangle_V, so idV\mathrm{id}_V satisfies the defining property of its own adjoint; by uniqueness (idV)∗=idV(\mathrm{id}_V)^* = \mathrm{id}_V. ■\blacksquare

∎
TheoremMatrix of the adjoint is the conjugate transpose

Let V,WV, W be finite-dimensional inner product spaces over F\mathbb{F} with orthonormal bases e1,…,ene_1, \dots, e_n of VV and f1,…,fmf_1, \dots, f_m of WW respectively. Let T:V→WT: V \to W be linear, and let A=(aij)A = (a_{ij}) be the matrix of TT with respect to these bases, meaning

Tej=∑i=1maijfi,1≤j≤n.T e_j = \sum_{i=1}^m a_{ij} f_i, \qquad 1 \le j \le n.

Then the matrix B=(bkl)B = (b_{kl}) of T∗:W→VT^*: W \to V with respect to the bases f1,…,fmf_1,\dots,f_m (domain) and e1,…,ene_1,\dots,e_n (codomain) satisfies

bkl=alk‾,i.e.B=A‾⊤.b_{kl} = \overline{a_{lk}}, \qquad \text{i.e.} \qquad B = \overline{A}^{\top}.
Proof

Because the bases are orthonormal, matrix entries can be extracted by taking inner products against basis vectors: for any linear map S:X→YS: X \to Y between spaces with orthonormal bases g1,…g_1,\dots (domain) and h1,…h_1,\dots (codomain), if Sgj=∑icijhiSg_j = \sum_i c_{ij} h_i, then taking ⟨⋅,hk⟩\langle \cdot, h_k\rangle of both sides and using orthonormality gives

ckj=⟨Sgj,hk⟩.c_{kj} = \langle Sg_j, h_k \rangle.

Apply this to TT: aij=⟨Tej,fi⟩Wa_{ij} = \langle Te_j, f_i \rangle_W. Apply this to T∗T^*: bkl=⟨T∗fl,ek⟩Vb_{kl} = \langle T^*f_l, e_k\rangle_V. Now compute bklb_{kl} using the defining property of the adjoint and conjugate symmetry:

bkl=⟨T∗fl,ek⟩V=⟨ek,T∗fl⟩V‾=⟨Tek,fl⟩W‾=alk‾.b_{kl} = \langle T^*f_l, e_k \rangle_V = \overline{\langle e_k, T^*f_l\rangle_V} = \overline{\langle Te_k, f_l \rangle_W} = \overline{a_{lk}}.

Thus bkl=alk‾b_{kl} = \overline{a_{lk}} for all k,lk, l, which is exactly the statement that B=A‾⊤B = \overline{A}^{\top}, the conjugate transpose of AA. ■\blacksquare

∎

Worked examples

Example 1 (an explicit non-square, non-symmetric complex map). Let V=C3V = \mathbb{C}^3 and W=C2W = \mathbb{C}^2, each with the standard inner product ⟨x,y⟩=∑xiyi‾\langle x, y\rangle = \sum x_i \overline{y_i}, and standard orthonormal bases. Let T:C3→C2T: \mathbb{C}^3 \to \mathbb{C}^2 be given by the matrix

A=(12i03−i1+i),A = \begin{pmatrix} 1 & 2i & 0 \\ 3 & -i & 1+i \end{pmatrix},

so T(z1,z2,z3)=(z1+2iz2, 3z1−iz2+(1+i)z3)T(z_1,z_2,z_3) = (z_1 + 2iz_2,\ 3z_1 - iz_2 + (1+i)z_3). By the theorem just proved, the matrix of T∗:C2→C3T^*: \mathbb{C}^2 \to \mathbb{C}^3 with respect to the standard bases is B=A‾⊤B = \overline{A}^\top:

A‾=(1−2i03i1−i),B=A‾⊤=(13−2ii01−i).\overline{A} = \begin{pmatrix} 1 & -2i & 0 \\ 3 & i & 1-i \end{pmatrix}, \qquad B = \overline{A}^\top = \begin{pmatrix} 1 & 3 \\ -2i & i \\ 0 & 1-i \end{pmatrix}.

So T∗(w1,w2)=(w1+3w2, −2iw1+iw2, (1−i)w2)T^*(w_1, w_2) = (w_1 + 3w_2,\ -2iw_1 + iw_2,\ (1-i)w_2).

Sanity check via the defining identity. Take v=(1,0,1)∈C3v = (1, 0, 1) \in \mathbb{C}^3 and w=(i,1)∈C2w = (i, 1) \in \mathbb{C}^2. Compute Tv=(1+2i⋅0, 3⋅1−i⋅0+(1+i)⋅1)=(1, 4+i)Tv = (1 + 2i\cdot 0,\ 3\cdot 1 - i\cdot 0 + (1+i)\cdot 1) = (1,\ 4+i). Then

⟨Tv,w⟩W=1⋅i‾+(4+i)⋅1‾=−i+4+i=4.\langle Tv, w\rangle_W = 1\cdot \overline{i} + (4+i)\cdot \overline{1} = -i + 4 + i = 4.

Compute T∗w=(i+3⋅1, −2i⋅i+i⋅1, (1−i)⋅1)=(3+i, 2+i, 1−i)T^*w = (i + 3\cdot 1,\ -2i\cdot i + i\cdot 1,\ (1-i)\cdot 1) = (3+i,\ 2+i,\ 1-i). Then

⟨v,T∗w⟩V=1⋅(3+i)‾+0⋅(2+i)‾+1⋅(1−i)‾=(3−i)+0+(1+i)=4.\langle v, T^*w\rangle_V = 1\cdot\overline{(3+i)} + 0 \cdot \overline{(2+i)} + 1\cdot\overline{(1-i)} = (3-i) + 0 + (1+i) = 4.

Both sides equal 44, confirming ⟨Tv,w⟩W=⟨v,T∗w⟩V\langle Tv, w \rangle_W = \langle v, T^*w\rangle_V for this pair, consistent with the theorem.

Example 2 (adjoint of a differentiation-like operator via change of orthonormal basis, and composition). Let V=R2V = \mathbb{R}^2 with the standard inner product, and let T:R2→R2T: \mathbb{R}^2 \to \mathbb{R}^2 be given, in the non-orthonormal-looking but actually we will use an orthonormal basis e1=(1,0)e_1 = (1,0), e2=(0,1)e_2 = (0,1), by

A=(2503).A = \begin{pmatrix} 2 & 5 \\ 0 & 3 \end{pmatrix}.

Since F=R\mathbb{F} = \mathbb{R} here, conjugation is trivial, so T∗T^* has matrix A⊤=(2053)A^\top = \begin{pmatrix} 2 & 0 \\ 5 & 3\end{pmatrix}, i.e. T∗(x,y)=(2x+5y, 3y)T^*(x,y) = (2x + 5y,\ 3y). Now let's verify the composition law (T∗)∗=T(T^*)^* = T (property 3 above) and the composition-reversal law (RT)∗=T∗R∗(RT)^* = T^*R^* (property 4) with a second operator, to see the algebra in action on an asymmetric example. Let R:R2→R2R: \mathbb{R}^2 \to \mathbb{R}^2 have matrix C=(01−14)C = \begin{pmatrix} 0 & 1 \\ -1 & 4\end{pmatrix} (also real, orthonormal standard basis). Then R∗R^* has matrix C⊤=(0−114)C^\top = \begin{pmatrix} 0 & -1 \\ 1 & 4\end{pmatrix}. Compute RTRT directly: matrix of RTRT is CACA:

CA=(01−14)(2503)=(0⋅2+1⋅00⋅5+1⋅3−1⋅2+4⋅0−1⋅5+4⋅3)=(03−27).CA = \begin{pmatrix} 0 & 1 \\ -1 & 4 \end{pmatrix}\begin{pmatrix} 2 & 5 \\ 0 & 3\end{pmatrix} = \begin{pmatrix} 0\cdot2+1\cdot0 & 0\cdot5+1\cdot3 \\ -1\cdot2+4\cdot0 & -1\cdot5+4\cdot3 \end{pmatrix} = \begin{pmatrix} 0 & 3 \\ -2 & 7\end{pmatrix}.

So (RT)∗(RT)^* should have matrix (CA)⊤=(0−237)(CA)^\top = \begin{pmatrix} 0 & -2 \\ 3 & 7\end{pmatrix}. Now compute T∗R∗T^*R^* directly, matrix A⊤C⊤A^\top C^\top:

A⊤C⊤=(2053)(0−114)=(2⋅0+0⋅12⋅(−1)+0⋅45⋅0+3⋅15⋅(−1)+3⋅4)=(0−237).A^\top C^\top = \begin{pmatrix} 2 & 0 \\ 5 & 3 \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 4\end{pmatrix} = \begin{pmatrix} 2\cdot0+0\cdot1 & 2\cdot(-1)+0\cdot4 \\ 5\cdot0+3\cdot1 & 5\cdot(-1)+3\cdot4 \end{pmatrix} = \begin{pmatrix} 0 & -2 \\ 3 & 7 \end{pmatrix}.

Indeed A⊤C⊤=(CA)⊤A^\top C^\top = (CA)^\top, confirming (RT)∗=T∗R∗(RT)^* = T^*R^* numerically (this is of course just the familiar fact (CA)⊤=A⊤C⊤(CA)^\top = A^\top C^\top from matrix algebra, matching the abstract identity proved above). This also illustrates why the order must reverse: CA≠ACCA \ne AC in general (check: AC=(2503)(01−14)=(−522−312)≠CAAC = \begin{pmatrix}2&5\\0&3\end{pmatrix}\begin{pmatrix}0&1\\-1&4\end{pmatrix} = \begin{pmatrix}-5&22\\-3&12\end{pmatrix} \ne CA), so T∗R∗≠R∗T∗T^*R^* \ne R^*T^* in general either, and only the reversed order (RT)∗=T∗R∗(RT)^* = T^*R^* is correct.

CounterexampleWhy finite-dimensionality (or completeness) cannot be dropped

The Riesz representation theorem, and hence existence of the adjoint, can fail if VV is infinite-dimensional and merely an inner product space (not complete, i.e. not a Hilbert space). Here is a concrete failure. Let VV be the space of sequences x=(x1,x2,x3,… )x = (x_1, x_2, x_3, \dots) of real numbers with only finitely many nonzero terms, equipped with the inner product ⟨x,y⟩=∑n=1∞xnyn\langle x, y \rangle = \sum_{n=1}^\infty x_n y_n (a finite sum, since only finitely many terms are nonzero). This is a genuine inner product space — positive-definite, symmetric, bilinear — but it is infinite-dimensional and not complete. Define the linear functional

φ(x)=∑n=1∞xnn.\varphi(x) = \sum_{n=1}^\infty \frac{x_n}{n}.

This is well-defined (finite sum for each x∈Vx \in V) and linear.

Claim: there is no u∈Vu \in V with φ(x)=⟨x,u⟩\varphi(x) = \langle x, u \rangle for all x∈Vx \in V. Suppose such a u=(u1,u2,… )∈Vu = (u_1, u_2, \dots) \in V existed. Since u∈Vu \in V, only finitely many unu_n are nonzero; say un=0u_n = 0 for all n>Nn > N. Let eN+1e_{N+1} be the standard basis vector with a 11 in position N+1N+1 and 00 elsewhere; eN+1∈Ve_{N+1} \in V. Then

φ(eN+1)=1N+1≠0,\varphi(e_{N+1}) = \frac{1}{N+1} \ne 0,

but

⟨eN+1,u⟩=uN+1=0,\langle e_{N+1}, u \rangle = u_{N+1} = 0,

since N+1>NN+1 > N. This contradicts φ(eN+1)=⟨eN+1,u⟩\varphi(e_{N+1}) = \langle e_{N+1}, u\rangle. So no such uu exists, and φ\varphi has no Riesz representative. Consequently, if one tried to define an adjoint using this φ\varphi (e.g. as φw\varphi_w for some operator TT landing appropriately), the construction used in the existence proof breaks down entirely: there is no vector to call T∗wT^*w. This is exactly why the theorem restricts to finite-dimensional spaces (or, in the general theory, complete inner product spaces / Hilbert spaces) — dropping finite-dimensionality (without substituting completeness) can genuinely destroy existence of the adjoint.

Scope, edge cases & common traps

SituationWhat happens
V,WV, W both finite-dimensionalT∗T^* exists and is unique (main theorem)
VV infinite-dimensional, not completeRiesz representation can fail; adjoint may not exist (see counterexample)
VV infinite-dimensional but complete (Hilbert space)Riesz representation and adjoint existence still hold, with a more delicate proof (Cauchy sequences replace finite bases); beyond this article's scope
F=R\mathbb{F} = \mathbb{R}Adjoint matrix is the ordinary transpose A⊤A^\top (conjugation is trivial)
F=C\mathbb{F} = \mathbb{C}Adjoint matrix is A‾⊤\overline{A}^\top; forgetting the conjugate is a very common error
Non-orthonormal basis usedThe formula "matrix of T∗T^* = conjugate transpose of matrix of TT" is false in general; it requires orthonormal bases
V=WV = WT∗T^* is also an operator on VV; this is the setting for Self-Adjoint Operators and Normal Operators

The mathematician behind it

Origins · 1 minFrigyes Riesz and how this was discovered
FR
Frigyes Riesz
1880–1956 · Hungarian analyst, foundational figure in functional analysis

Riesz worked in Szeged and Budapest, and his representation theorems (for linear functionals on Hilbert spaces, and separately for measures) are cornerstones of functional analysis. The theorem bearing his name used here is the finite-dimensional shadow of a deeper Hilbert-space result; both express the same idea, that continuous linear functionals are inner products against a fixed vector, and both routinely appear disguised inside proofs about adjoints, dual spaces, and weak convergence.

History
- 1907 — Riesz and, independently, Fréchet, establish representation theorems for continuous linear functionals on L2L^2-type spaces.
- 1910s–1920s — von Neumann and others formalize Hilbert space theory, placing adjoint operators at the center of spectral theory.
- 1930s — the abstract Hilbert-space adjoint becomes standard machinery in the mathematical foundations of quantum mechanics.
- 1958 — Halmos's Finite-Dimensional Vector Spaces popularizes the finite-dimensional treatment of adjoints via Riesz representation used in modern linear algebra courses.
- Present — the adjoint remains the organizing concept behind self-adjoint, normal, and unitary operators across pure and applied linear algebra.
Does every linear map between inner product spaces have an adjoint?

Yes, provided both spaces are finite-dimensional (or more generally, complete Hilbert spaces). The existence proof relies on the Riesz Representation Theorem, which guarantees that every linear functional on the domain can be written as an inner product against a fixed vector. Applying this one output vector at a time produces T∗T^* automatically. If the space is infinite-dimensional and not complete, Riesz representation can fail, as the counterexample in this article shows, and then the construction used to build T∗T^* has no vector to output.

Is the adjoint the same as the inverse?

No, they are entirely different in general. The adjoint T∗T^* always exists for any linear map between finite-dimensional inner product spaces, while the inverse T−1T^{-1} only exists when TT is bijective. They coincide precisely for unitary (complex) or orthogonal (real) operators, where T∗T=TT∗=IT^*T = TT^* = I, meaning T∗=T−1T^* = T^{-1}. This special coincidence is exactly why unitary/orthogonal operators are so important: their adjoint doubles as their inverse.

Why does the matrix of the adjoint involve a conjugate, not just a transpose?

The conjugate appears because the complex inner product is conjugate-linear in one slot (by convention, the second slot here), not bilinear. When you move a scalar across the inner product from the second slot, it comes out complex-conjugated. Tracing through the proof that bkl=⟨T∗fl,ek⟩b_{kl} = \langle T^*f_l, e_k\rangle, conjugate symmetry of the inner product forces a conjugation, giving bkl=alk‾b_{kl} = \overline{a_{lk}}. Over R\mathbb{R}, conjugation is the identity, so the formula reduces to the familiar plain transpose.

Does the formula "matrix of adjoint = conjugate transpose" work for any bases?

No — it requires orthonormal bases in both the domain and codomain. The proof extracts matrix entries using the formula aij=⟨Tej,fi⟩a_{ij} = \langle Te_j, f_i\rangle, which is only valid when fif_i form an orthonormal basis (otherwise you would need to solve a linear system against the Gram matrix to extract coordinates). If you use a non-orthonormal basis, the matrix of T∗T^* is instead the conjugate transpose of AA conjugated by Gram matrices of the two bases, which generally is not simply A‾⊤\overline{A}^\top.

How does the adjoint relate to self-adjoint and normal operators?

Self-adjoint operators are defined by the equation T=T∗T = T^*, and normal operators by TT∗=T∗TTT^* = T^*T; both definitions are meaningless without first knowing T∗T^* exists and is unique, which is exactly what this article establishes. These operator classes generalize familiar matrix classes (symmetric, Hermitian, skew-Hermitian, unitary, orthogonal) to the operator level, and their spectral properties (real eigenvalues, orthogonal eigenspaces, diagonalizability by an orthonormal basis) are proved using the algebraic identities for T∗T^* developed here.

Can the adjoint change if I change the inner product on VV or WW?

Yes, absolutely. The adjoint is defined entirely in terms of the inner products on VV and WW, so changing either inner product generally changes T∗T^*, even though TT itself is unchanged. For instance, the matrix formula "conjugate transpose" is valid only in an orthonormal basis for the given inner product; switching to a different inner product changes which bases are orthonormal, and hence changes the adjoint's matrix representation, even in the same basis.

Is the map T↦T∗T \mapsto T^* itself linear?

Almost, but not quite in the naive sense: it is additive, (S+T)∗=S∗+T∗(S+T)^* = S^*+T^*, but it is conjugate-linear in scalars, since (λT)∗=λ‾ T∗(\lambda T)^* = \overline{\lambda}\,T^* rather than λT∗\lambda T^*. Over R\mathbb{R} this distinction disappears because conjugation is trivial, so the adjoint map is genuinely linear in the real case. Over C\mathbb{C}, this conjugate-linearity is an essential and often-tested feature, directly traceable to the conjugate-linearity of the inner product itself.

Before this

Where this leads