Dual Maps and the Transpose

Every linear map between vector spaces has a shadow map. This shadow map runs in the opposite direction between their dual spaces. It is called the dual map or transpose map. This map encodes exactly the same information as the original map. But it acts on linear functionals rather than on vectors. The question this article answers is deceptively simple. If T:V→WT:V\to W is linear, is there a natural linear map W′→V′W'\to V' built from TT? If so, what does it look like in coordinates? The surprising and beautiful answer is this. This abstractly-defined map, when written in matrices with respect to dual bases, is literally the transpose of the matrix of TT. It is the same transpose you learned to compute in a first course. It is now revealed as a coordinate expression of a coordinate-free construction. This matters because it explains why the transpose behaves the way it does. It explains why (AB)t=BtAt(AB)^t = B^tA^t. It explains why row rank equals column rank. It explains why the transpose of an invertible matrix is invertible, with inverse the transpose of the inverse. All of these facts look like bookkeeping identities when proved by direct computation with matrix entries. But they become nearly automatic once you see them as consequences of the functorial nature of duality. The dual map reverses arrows and reverses composition order. This is exactly what one would guess from the way functions pull back. The big picture is that duality gives every construction in linear algebra a "mirror" construction. The dual map is the mirror of a linear map itself. Annihilators are subspaces of the dual space consisting of functionals vanishing on a given subspace. They turn out to be exactly the right language for describing the kernel and image of the dual map. This description is in terms of the kernel and image of the original map. This sets up the machinery used later to prove that row rank and column rank of a matrix coincide. This is one of the most useful and initially mysterious facts in the subject. This article gives the precise definition of the dual map. It proves the transpose theorem in full. It works out two detailed examples with asymmetric non-square matrices. It develops the annihilator identities relating ker⁡T′\ker T', im⁡T′\operatorname{im}T' to im⁡T\operatorname{im}T, ker⁡T\ker T. It assumes familiarity with The Dual Space and Dual Bases and The Matrix of a Linear Map.

Key takeaways
  • For linear T:V→WT:V\to W, the dual map T′:W′→V′T':W'\to V' is defined by T′(φ)=φ∘TT'(\varphi) = \varphi\circ T, precomposition.
  • T′T' is always linear, regardless of what TT looks like; duality reverses the direction of arrows.
  • With respect to dual bases, the matrix of T′T' is the transpose of the matrix of TT.
  • Composition reverses: (ST)′=T′S′(ST)' = T'S', mirroring (AB)t=BtAt(AB)^t = B^tA^t.
  • Annihilators U0={φ∈V′:φ∣U=0}U^0 = \{\varphi \in V' : \varphi|_U = 0\} satisfy ker⁡T′=(im⁡T)0\ker T' = (\operatorname{im}T)^0 and im⁡T′=(ker⁡T)0\operatorname{im}T' = (\ker T)^0 (finite dimensions).
  • Key idea: the transpose of a matrix is not an ad hoc operation — it is the coordinate representation of the natural "pullback" map on dual spaces.

Where this lives today

The dual map and transpose underlie the notion of the adjoint operator in functional analysis and quantum mechanics: for a bounded linear operator TT between Hilbert spaces, the Hermitian adjoint T∗T^* is built from the same precomposition idea, combined with the Riesz representation identifying a Hilbert space with its own dual. In numerical linear algebra and statistics, the transpose of a design matrix appears throughout least-squares regression, where the normal equations AtAx^=AtbA^tA\hat{x} = A^t b arise directly from differentiating a quadratic form, and the annihilator perspective explains why the residual vector lies in the orthogonal complement (a metric cousin of the annihilator) of the column space. In control theory and signal processing, the transpose of a state matrix governs the "dual system," used in observability/controllability duality — itself a direct descendant of the annihilator identities ker⁡T′=(im⁡T)0\ker T' = (\operatorname{im}T)^0 and im⁡T′=(ker⁡T)0\operatorname{im}T' = (\ker T)^0. In algebraic geometry and representation theory, the dual (or "transpose") of a linear map between vector spaces is the basic building block of contravariant functors, and the transpose theorem is the shadow, in the category of vector spaces, of a much more general categorical duality. Finally, the annihilator machinery developed here is precisely the tool used to give a clean, basis-free proof that the row rank and column rank of a matrix are equal, connecting this article directly to Rank of a Matrix.

Ask the tutor: How does the annihilator identity im⁡T′=(ker⁡T)0\operatorname{im}T' = (\ker T)^0 give a basis-free proof that row rank equals column rank?

DefinitionDual map / transpose map

Let VV and WW be vector spaces over a field F\mathbb{F}, and let T:V→WT : V \to W be a linear map. Let V′V' and W′W' denote the dual spaces L(V,F)\mathcal{L}(V,\mathbb{F}) and L(W,F)\mathcal{L}(W,\mathbb{F}) respectively. The dual map (or transpose) of TT is the map

T′:W′→V′,T′(φ)=φ∘Tfor all φ∈W′.T' : W' \to V', \qquad T'(\varphi) = \varphi \circ T \quad \text{for all } \varphi \in W'.

That is, T′(φ)T'(\varphi) is the linear functional on VV defined by (T′φ)(v)=φ(T(v))(T'\varphi)(v) = \varphi(T(v)) for all v∈Vv \in V. When V,WV, W are finite-dimensional with ordered bases β=(v1,…,vn)\beta = (v_1,\dots,v_n) of VV and γ=(w1,…,wm)\gamma = (w_1,\dots,w_m) of WW, with dual bases β′=(v1′,…,vn′)\beta' = (v_1',\dots,v_n') of V′V' and γ′=(w1′,…,wm′)\gamma' = (w_1',\dots,w_m') of W′W', the matrix of the dual map is [T′]γ′β′[T']_{\gamma'}^{\beta'}, computed with respect to these dual bases in the same convention used for [T]γβ[T]_\gamma^\beta: the jj-th column of [T′]γ′β′[T']_{\gamma'}^{\beta'} records the coordinates of T′(wj′)T'(w_j') in the basis β′\beta'.

Intuition

Think of a linear map T:V→WT : V \to W as a machine that pushes vectors forward from VV into WW. A linear functional φ\varphi on WW is a machine that measures vectors in WW, producing a scalar. If you want to measure a vector v∈Vv \in V using φ\varphi, the natural thing to do is first push vv forward into WW via TT, and then measure the result with φ\varphi. This composite operation, v↦φ(T(v))v \mapsto \varphi(T(v)), is itself a linear functional on VV — call it T′(φ)T'(\varphi). So while TT pushes vectors forward from VV to WW, the process of building new functionals out of old ones runs backward: functionals on WW give rise to functionals on VV. This reversal of direction is the essential content of duality, and it is why T′:W′→V′T' : W' \to V' points the opposite way from T:V→WT : V \to W. The transpose-of-a-matrix phenomenon is the coordinate shadow of this reversal. If TT is represented by an m×nm \times n matrix AA (using coordinate vectors, an nn-dimensional input turning into an mm-dimensional output), the dual map takes a functional on the mm-dimensional target — represented by a row vector, or equivalently a column vector of length mm once we fix the dual basis coordinates — and returns a functional on the nn-dimensional source. The bookkeeping of matching up "row acting on column" versus "column acting on row" forces the matrix of T′T' to have nn rows and mm columns, with entries equal to those of AA but with rows and columns swapped: exactly AtA^t.

Figure
Dual map reverses the arrow: TT goes forward, T′T' goes backward
V (dim n) v W (dim m) T(v) T V' (dim n) φ∘T W' (dim m) φ T'

Central results

TheoremLinearity of the dual map

Let V,WV, W be vector spaces over F\mathbb{F} and let T:V→WT : V \to W be linear. Then the map

T′:W′→V′,T′(φ)=φ∘TT' : W' \to V', \qquad T'(\varphi) = \varphi \circ T

is a well-defined linear map from W′W' to V′V'.

Proof

Well-defined (lands in V′V'). For fixed φ∈W′\varphi \in W', the composite φ∘T:V→F\varphi \circ T : V \to \mathbb{F} is a composition of linear maps T:V→WT : V \to W and φ:W→F\varphi : W \to \mathbb{F}, hence linear (composition of linear maps is linear, a basic fact about L\mathcal{L}). So T′(φ)∈V′T'(\varphi) \in V' for every φ∈W′\varphi \in W'.

Linearity of T′T' itself. We must check T′T' is additive and homogeneous as a function of φ\varphi. Let φ1,φ2∈W′\varphi_1, \varphi_2 \in W' and v∈Vv \in V. Then

T′(φ1+φ2)(v)=(φ1+φ2)(T(v))=φ1(T(v))+φ2(T(v))=T′(φ1)(v)+T′(φ2)(v),T'(\varphi_1+\varphi_2)(v) = (\varphi_1+\varphi_2)(T(v)) = \varphi_1(T(v)) + \varphi_2(T(v)) = T'(\varphi_1)(v) + T'(\varphi_2)(v),

using the definition of addition of functionals in W′W'. Since this holds for all v∈Vv \in V, the functionals T′(φ1+φ2)T'(\varphi_1+\varphi_2) and T′(φ1)+T′(φ2)T'(\varphi_1)+T'(\varphi_2) agree on all of VV, so

T′(φ1+φ2)=T′(φ1)+T′(φ2).T'(\varphi_1 + \varphi_2) = T'(\varphi_1) + T'(\varphi_2).

For scalars, let c∈Fc \in \mathbb{F} and φ∈W′\varphi \in W'. For all v∈Vv \in V,

T′(cφ)(v)=(cφ)(T(v))=c φ(T(v))=c T′(φ)(v),T'(c\varphi)(v) = (c\varphi)(T(v)) = c\,\varphi(T(v)) = c\, T'(\varphi)(v),

so T′(cφ)=c T′(φ)T'(c\varphi) = c\,T'(\varphi). Thus T′T' is additive and homogeneous, hence linear. ■\blacksquare

∎
LemmaFunctoriality: identity and composition

Let U,V,WU, V, W be vector spaces over F\mathbb{F}. (a) If IV:V→VI_V : V \to V is the identity map, then IV′=IV′I_V' = I_{V'}. (b) If S:U→VS : U \to V and T:V→WT : V \to W are linear, then (TS)′=S′T′(TS)' = S'T', where TS:U→WTS : U \to W and S′T′:W′→U′S'T' : W' \to U'.

Proof

(a) For φ∈V′\varphi \in V', IV′(φ)=φ∘IV=φI_V'(\varphi) = \varphi \circ I_V = \varphi, so IV′I_V' is the identity map on V′V'.

(b) Both (TS)′(TS)' and S′T′S'T' are maps W′→U′W' \to U'; we show they agree on every φ∈W′\varphi \in W'. By definition,

(TS)′(φ)=φ∘(TS)=φ∘T∘S.(TS)'(\varphi) = \varphi \circ (TS) = \varphi \circ T \circ S.

On the other hand,

(S′T′)(φ)=S′(T′(φ))=S′(φ∘T)=(φ∘T)∘S,(S'T')(\varphi) = S'(T'(\varphi)) = S'(\varphi \circ T) = (\varphi \circ T) \circ S,

using the definition of S′S' applied to the functional φ∘T∈V′\varphi \circ T \in V'. Composition of functions is associative, so φ∘T∘S=(φ∘T)∘S\varphi \circ T \circ S = (\varphi \circ T)\circ S, and the two expressions coincide for every φ\varphi. Hence (TS)′=S′T′(TS)' = S'T'. ■\blacksquare

∎
TheoremMatrix of the dual map is the transpose

Let VV be an nn-dimensional vector space with ordered basis β=(v1,…,vn)\beta = (v_1,\dots,v_n), and let WW be an mm-dimensional vector space with ordered basis γ=(w1,…,wm)\gamma = (w_1,\dots,w_m), both over F\mathbb{F}. Let β′=(v1′,…,vn′)\beta' = (v_1',\dots,v_n') and γ′=(w1′,…,wm′)\gamma' = (w_1',\dots,w_m') be the corresponding dual bases of V′V' and W′W'. Let T:V→WT : V \to W be linear and let A=[T]γβ∈Fm×nA = [T]_\gamma^\beta \in \mathbb{F}^{m\times n} be its matrix. Then

[T′]β′γ′=At∈Fn×m.[T']_{\beta'}^{\gamma'} = A^t \in \mathbb{F}^{n \times m}.
Proof

Write A=(aij)A = (a_{ij}), 1≤i≤m1\le i \le m, 1≤j≤n1 \le j \le n, so that

T(vj)=∑i=1maij wi,j=1,…,n.T(v_j) = \sum_{i=1}^m a_{ij}\, w_i, \qquad j = 1,\dots,n.

By definition of the matrix of a linear map with respect to dual bases, the matrix B=[T′]β′γ′B = [T']_{\beta'}^{\gamma'} is the n×mn \times m matrix whose kk-th column (k=1,…,mk = 1,\dots,m) gives the β′\beta'-coordinates of T′(wk′)T'(w_k'); that is,

T′(wk′)=∑j=1nbjk vj′,k=1,…,m,T'(w_k') = \sum_{j=1}^n b_{jk}\, v_j', \qquad k=1,\dots,m,

and we must show bjk=akjb_{jk} = a_{kj} for all j,kj,k, i.e. B=AtB = A^t. Fix k∈{1,…,m}k \in \{1,\dots,m\}. By definition of the dual map, T′(wk′)=wk′∘TT'(w_k') = w_k' \circ T, a functional on VV. We evaluate both sides of the coordinate expansion at each basis vector vjv_j of VV. Using the fundamental property of a dual basis, vj′(vi)=δijv_j'(v_i) = \delta_{ij} (Kronecker delta), the right-hand side gives

(∑j′=1nbj′k vj′′)(vj)=∑j′=1nbj′k vj′′(vj)=∑j′=1nbj′k δj′j=bjk.\left(\sum_{j'=1}^n b_{j'k}\, v_{j'}'\right)(v_j) = \sum_{j'=1}^n b_{j'k}\, v_{j'}'(v_j) = \sum_{j'=1}^n b_{j'k}\,\delta_{j'j} = b_{jk}.

The left-hand side gives

T′(wk′)(vj)=wk′(T(vj))=wk′ ⁣(∑i=1maijwi)=∑i=1maij wk′(wi)=∑i=1maij δki=akj,T'(w_k')(v_j) = w_k'(T(v_j)) = w_k'\!\left(\sum_{i=1}^m a_{ij} w_i\right) = \sum_{i=1}^m a_{ij}\, w_k'(w_i) = \sum_{i=1}^m a_{ij}\,\delta_{ki} = a_{kj},

again using linearity of wk′w_k' and the defining property wk′(wi)=δkiw_k'(w_i) = \delta_{ki} of the dual basis of WW. Equating the two evaluations, bjk=akjb_{jk} = a_{kj} for every j∈{1,…,n}j \in \{1,\dots,n\} and every k∈{1,…,m}k \in \{1,\dots,m\}. This says precisely that the (j,k)(j,k) entry of BB equals the (k,j)(k,j) entry of AA, i.e. B=AtB = A^t. ■\blacksquare

∎
CorollaryTranspose of a product

If AA is an m×nm \times n matrix and CC is an n×pn \times p matrix over F\mathbb{F}, then (AC)t=CtAt(AC)^t = C^tA^t.

Proof

Let U,V,WU, V, W be vector spaces with ordered bases of dimensions p,n,mp, n, m respectively (for instance U=Fp,V=Fn,W=FmU = \mathbb{F}^p, V=\mathbb{F}^n, W=\mathbb{F}^m with standard bases), and let S:U→VS : U \to V, T:V→WT : V \to W be the linear maps with [S]=C[S] = C and [T]=A[T] = A with respect to these bases. Then [TS]=AC[TS] = AC (The Matrix of a Linear Map, the composition-of-maps-corresponds-to-product-of-matrices rule). By the previous theorem applied three times,

[(TS)′]=(AC)t,[T′]=At,[S′]=Ct,[(TS)'] = (AC)^t, \qquad [T'] = A^t, \qquad [S'] = C^t,

each with respect to the appropriate dual bases. By the Lemma on functoriality, (TS)′=S′T′(TS)' = S'T', and applying the matrix-of-a-composition rule to S′T′S'T' (a composition W′→V′→U′W' \to V' \to U') gives [S′T′]=[S′][T′]=CtAt[S'T'] = [S'][T'] = C^tA^t. Since (TS)′(TS)' and S′T′S'T' are the same map, their matrices with respect to the same dual bases coincide:

(AC)t=[(TS)′]=[S′T′]=CtAt.■(AC)^t = [(TS)'] = [S'T'] = C^tA^t. \qquad \blacksquare
∎

Now we introduce annihilators, which describe the kernel and image of T′T' in terms of those of TT.

DefinitionAnnihilator

Let VV be a vector space over F\mathbb{F} and let U⊆VU \subseteq V be a subspace. The annihilator of UU is

U0={φ∈V′:φ(u)=0 for all u∈U}.U^0 = \{\varphi \in V' : \varphi(u) = 0 \text{ for all } u \in U\}.

It is straightforward to check U0U^0 is a subspace of V′V': it contains the zero functional, and is closed under addition and scalar multiplication because those operations are checked pointwise and 0+0=00+0=0, c⋅0=0c\cdot 0 = 0.

TheoremDimension of the annihilator, kernel and image of T′T'

Let V,WV, W be finite-dimensional vector spaces over F\mathbb{F} and let T:V→WT : V \to W be linear. Then: (a) For any subspace U⊆VU \subseteq V with dim⁡V=n\dim V = n, dim⁡U=k\dim U = k, we have dim⁡U0=n−k\dim U^0 = n - k. (b) ker⁡T′=(im⁡T)0\ker T' = (\operatorname{im} T)^0. (c) im⁡T′=(ker⁡T)0\operatorname{im} T' = (\ker T)^0. (d) dim⁡(im⁡T′)=dim⁡(im⁡T)\dim(\operatorname{im} T') = \dim(\operatorname{im} T).

Proof

(a) Extend a basis (u1,…,uk)(u_1,\dots,u_k) of UU to a basis (u1,…,uk,uk+1,…,un)(u_1,\dots,u_k,u_{k+1},\dots,u_n) of VV (possible by the basis extension theorem, since VV is finite-dimensional). Let (u1′,…,un′)(u_1',\dots,u_n') be the corresponding dual basis of V′V'. We claim U0=span⁡(uk+1′,…,un′)U^0 = \operatorname{span}(u_{k+1}',\dots,u_n'), which has dimension n−kn-k since dual basis vectors are linearly independent. If φ=∑i=k+1nciui′∈span⁡(uk+1′,…,un′)\varphi = \sum_{i=k+1}^n c_i u_i' \in \operatorname{span}(u_{k+1}',\dots,u_n'), then for any uju_j with j≤kj \le k, φ(uj)=∑i=k+1nciui′(uj)=0\varphi(u_j) = \sum_{i=k+1}^n c_i u_i'(u_j) = 0 since ui′(uj)=0u_i'(u_j) = 0 whenever i≠ji \ne j and here i>k≥ji > k \ge j. Since every element of UU is a linear combination of u1,…,uku_1,\dots,u_k, linearity gives φ(u)=0\varphi(u) = 0 for all u∈Uu \in U; hence φ∈U0\varphi \in U^0. This shows span⁡(uk+1′,…,un′)⊆U0\operatorname{span}(u_{k+1}',\dots,u_n') \subseteq U^0. Conversely, let φ∈U0\varphi \in U^0. Write φ=∑i=1nciui′\varphi = \sum_{i=1}^n c_i u_i' in the dual basis. For j≤kj \le k, evaluate: φ(uj)=cj\varphi(u_j) = c_j (since ui′(uj)=δiju_i'(u_j) = \delta_{ij} and the sum collapses to the i=ji=j term). But uj∈Uu_j \in U and φ∈U0\varphi \in U^0 force φ(uj)=0\varphi(u_j) = 0, so cj=0c_j = 0 for all j≤kj \le k. Thus φ∈span⁡(uk+1′,…,un′)\varphi \in \operatorname{span}(u_{k+1}',\dots,u_n'). This proves U0⊆span⁡(uk+1′,…,un′)U^0 \subseteq \operatorname{span}(u_{k+1}',\dots,u_n'), and combined with the reverse inclusion, U0=span⁡(uk+1′,…,un′)U^0 = \operatorname{span}(u_{k+1}',\dots,u_n'), so dim⁡U0=n−k\dim U^0 = n-k.

(b) Let φ∈W′\varphi \in W'. By definition, φ∈ker⁡T′\varphi \in \ker T' iff T′(φ)=0T'(\varphi) = 0 in V′V', i.e. iff φ(T(v))=0\varphi(T(v)) = 0 for every v∈Vv \in V (since T′(φ)T'(\varphi) is the functional v↦φ(T(v))v \mapsto \varphi(T(v)), and it is the zero functional exactly when it vanishes on every vv). As vv ranges over VV, T(v)T(v) ranges over all of im⁡T\operatorname{im} T, so this condition is equivalent to φ(w)=0\varphi(w) = 0 for all w∈im⁡Tw \in \operatorname{im} T, i.e. φ∈(im⁡T)0\varphi \in (\operatorname{im} T)^0. Hence ker⁡T′=(im⁡T)0\ker T' = (\operatorname{im} T)^0.

(c) We first show im⁡T′⊆(ker⁡T)0\operatorname{im} T' \subseteq (\ker T)^0. Let φ∈W′\varphi \in W' and v∈ker⁡Tv \in \ker T. Then

T′(φ)(v)=φ(T(v))=φ(0)=0,T'(\varphi)(v) = \varphi(T(v)) = \varphi(0) = 0,

so every functional in im⁡T′\operatorname{im}T' vanishes on ker⁡T\ker T, i.e. im⁡T′⊆(ker⁡T)0\operatorname{im} T' \subseteq (\ker T)^0. For the reverse inclusion, we compare dimensions, which suffices since we already have one containment. By The Dual Space and Dual Bases (finite-dimensionality gives dim⁡V′=dim⁡V\dim V' = \dim V) together with the Rank–Nullity theorem applied to T′T',

dim⁡(im⁡T′)=dim⁡W′−dim⁡(ker⁡T′).\dim(\operatorname{im} T') = \dim W' - \dim(\ker T') .

Since dim⁡W′=dim⁡W=m\dim W' = \dim W = m, and by part (b) together with part (a) (applied to U=im⁡T⊆WU = \operatorname{im} T \subseteq W, of dimension r:=rank⁡(T)r := \operatorname{rank}(T)),

dim⁡(ker⁡T′)=dim⁡(im⁡T)0=m−r.\dim(\ker T') = \dim(\operatorname{im}T)^0 = m - r.

Hence dim⁡(im⁡T′)=m−(m−r)=r\dim(\operatorname{im}T') = m - (m-r) = r. On the other hand, by part (a) applied to U=ker⁡T⊆VU = \ker T \subseteq V (of dimension n−rn - r by Rank–Nullity for TT, where n=dim⁡Vn = \dim V),

dim⁡(ker⁡T)0=n−(n−r)=r.\dim(\ker T)^0 = n - (n-r) = r.

So im⁡T′\operatorname{im} T' and (ker⁡T)0(\ker T)^0 are both subspaces of V′V', one contained in the other, and both of the same finite dimension rr; a subspace of a finite-dimensional space that has the same dimension as the ambient space it sits inside must equal it. Hence im⁡T′=(ker⁡T)0\operatorname{im} T' = (\ker T)^0.

(d) This was shown in the course of proving (c): dim⁡(im⁡T′)=r=rank⁡(T)=dim⁡(im⁡T)\dim(\operatorname{im}T') = r = \operatorname{rank}(T) = \dim(\operatorname{im}T). ■\blacksquare

∎

Worked examples

Example 1: an explicit 3×23\times 2 map and its transpose, checked directly against the definition. Let V=F2V = \mathbb{F}^2 with standard basis β=(e1,e2)\beta = (e_1,e_2) and let W=F3W = \mathbb{F}^3 with standard basis γ=(f1,f2,f3)\gamma = (f_1,f_2,f_3), over F=R\mathbb{F} = \mathbb{R}. Define T:V→WT : V \to W by

T(e1)=2f1−f2+3f3,T(e2)=0f1+5f2−4f3,T(e_1) = 2f_1 - f_2 + 3f_3, \qquad T(e_2) = 0f_1 + 5f_2 - 4f_3,

so that

A=[T]γβ=(20−153−4).A = [T]_\gamma^\beta = \begin{pmatrix} 2 & 0 \\ -1 & 5 \\ 3 & -4 \end{pmatrix}.

Let γ′=(f1′,f2′,f3′)\gamma' = (f_1',f_2',f_3') be the dual basis of W′W' and β′=(e1′,e2′)\beta' = (e_1',e_2') the dual basis of V′V'. We compute T′(f1′),T′(f2′),T′(f3′)T'(f_1'), T'(f_2'), T'(f_3') directly from the definition T′(φ)=φ∘TT'(\varphi) = \varphi \circ T, and check the result against the transpose theorem. By definition, T′(fk′)T'(f_k') is the functional on VV sending v↦fk′(T(v))v \mapsto f_k'(T(v)). It suffices to evaluate on the basis vectors e1,e2e_1, e_2 of VV, since a functional is determined by its values on a basis. For k=1k=1: T′(f1′)(e1)=f1′(T(e1))=f1′(2f1−f2+3f3)=2T'(f_1')(e_1) = f_1'(T(e_1)) = f_1'(2f_1 - f_2+3f_3) = 2, and T′(f1′)(e2)=f1′(5f2−4f3)=0T'(f_1')(e_2) = f_1'(5f_2-4f_3) = 0. So T′(f1′)=2e1′+0e2′T'(f_1') = 2e_1' + 0e_2'. For k=2k=2: T′(f2′)(e1)=f2′(2f1−f2+3f3)=−1T'(f_2')(e_1) = f_2'(2f_1-f_2+3f_3) = -1, and T′(f2′)(e2)=f2′(5f2−4f3)=5T'(f_2')(e_2) = f_2'(5f_2-4f_3) = 5. So T′(f2′)=−1 e1′+5 e2′T'(f_2') = -1\, e_1' + 5\, e_2'. For k=3k=3: T′(f3′)(e1)=f3′(2f1−f2+3f3)=3T'(f_3')(e_1) = f_3'(2f_1-f_2+3f_3) = 3, and T′(f3′)(e2)=f3′(5f2−4f3)=−4T'(f_3')(e_2) = f_3'(5f_2-4f_3) = -4. So T′(f3′)=3 e1′−4 e2′T'(f_3') = 3\,e_1' - 4\,e_2'. Reading off coordinates in β′\beta', the matrix [T′]β′γ′[T']_{\beta'}^{\gamma'} has as its kk-th column the coordinates of T′(fk′)T'(f_k'):

[T′]β′γ′=(2−1305−4),[T']_{\beta'}^{\gamma'} = \begin{pmatrix} 2 & -1 & 3 \\ 0 & 5 & -4 \end{pmatrix},

which is exactly AtA^t, confirming the theorem in this instance. Note the shapes: AA is 3×23\times 2 (three rows, two columns), and [T′][T'] is 2×32 \times 3, matching AtA^t.

Example 2: annihilators, kernel, and image for a rank-deficient map. Let V=R3V = \mathbb{R}^3, W=R2W = \mathbb{R}^2, both with standard bases, and let T:V→WT : V \to W have matrix

A=(12−124−2).A = \begin{pmatrix} 1 & 2 & -1 \\ 2 & 4 & -2 \end{pmatrix}.

Notice the second row is twice the first, so rank⁡(A)=1\operatorname{rank}(A) = 1; this is a genuinely rank-deficient, asymmetric example (not full rank, not square). Kernel of TT. Solve Ax=0Ax = 0: the system reduces to the single equation x1+2x2−x3=0x_1 + 2x_2 - x_3 = 0. A basis for ker⁡T\ker T is obtained from two free parameters, say x2,x3x_2, x_3:

ker⁡T=span⁡{(−2,1,0), (1,0,1)},\ker T = \operatorname{span}\{(-2,1,0),\ (1,0,1)\},

so dim⁡ker⁡T=2\dim \ker T = 2, consistent with Rank–Nullity: dim⁡V−rank⁡(T)=3−1=2\dim V - \operatorname{rank}(T) = 3 - 1 = 2. Image of TT. Since rank is 11, im⁡T\operatorname{im}T is 11-dimensional, spanned by the first column image, e.g. T(e1)=(1,2)T(e_1) = (1,2), so im⁡T=span⁡{(1,2)}⊂R2\operatorname{im}T = \operatorname{span}\{(1,2)\} \subset \mathbb{R}^2. Transpose map and its matrix. By the transpose theorem, with respect to standard dual bases,

[T′]=At=(1224−1−2): R2→R3 (identifying (R2)′,(R3)′ with R2,R3 via dual bases).[T'] = A^t = \begin{pmatrix} 1 & 2 \\ 2 & 4 \\ -1 & -2 \end{pmatrix}: \ \mathbb{R}^2 \to \mathbb{R}^3 \ (\text{identifying } (\mathbb{R}^2)', (\mathbb{R}^3)' \text{ with } \mathbb{R}^2,\mathbb{R}^3 \text{ via dual bases}).

Check im⁡T′=(ker⁡T)0\operatorname{im}T' = (\ker T)^0. First compute im⁡T′\operatorname{im}T' directly: T′T' sends the standard dual basis vector f1′=(1,0)f_1' = (1,0) to the functional with coordinate vector (1,2,−1)(1,2,-1) (first column of AtA^t), and f2′=(0,1)f_2' = (0,1) to (2,4,−2)(2,4,-2) — a scalar multiple of the first. So

im⁡T′=span⁡{(1,2,−1)},\operatorname{im}T' = \operatorname{span}\{(1,2,-1)\},

a 11-dimensional subspace of V′≅R3V' \cong \mathbb{R}^3, matching rank⁡(T′)=rank⁡(T)=1\operatorname{rank}(T') = \operatorname{rank}(T) = 1 as guaranteed by part (d) of the theorem. Now compute (ker⁡T)0(\ker T)^0 directly and compare. A functional φ=(c1,c2,c3)∈V′\varphi = (c_1,c_2,c_3) \in V' (identified with row vectors acting by dot product) annihilates ker⁡T=span⁡{(−2,1,0),(1,0,1)}\ker T = \operatorname{span}\{(-2,1,0),(1,0,1)\} exactly when

−2c1+c2=0andc1+c3=0,-2c_1 + c_2 = 0 \quad\text{and}\quad c_1 + c_3 = 0,

i.e. c2=2c1c_2 = 2c_1 and c3=−c1c_3 = -c_1. So

(ker⁡T)0={(c1,2c1,−c1):c1∈R}=span⁡{(1,2,−1)},(\ker T)^0 = \{(c_1, 2c_1, -c_1) : c_1 \in \mathbb{R}\} = \operatorname{span}\{(1,2,-1)\},

which is exactly im⁡T′\operatorname{im}T' computed above — confirming part (c) of the theorem in this concrete, rank-deficient, non-square case. Check ker⁡T′=(im⁡T)0\ker T' = (\operatorname{im}T)^0. A functional ψ=(d1,d2)∈W′\psi = (d_1,d_2) \in W' satisfies T′(ψ)=0T'(\psi) = 0 iff ψ∘T=0\psi \circ T = 0, i.e. ψ(Av)=0\psi(Av) = 0 for all vv, i.e. ψA=0\psi A = 0 as a row-vector times matrix. Concretely (d1,d2)(12−124−2)=(d1+2d2, 2d1+4d2, −d1−2d2)=0(d_1,d_2)\begin{pmatrix}1&2&-1\\2&4&-2\end{pmatrix} = (d_1+2d_2,\ 2d_1+4d_2,\ -d_1-2d_2) = 0 forces d1=−2d2d_1 = -2d_2. So ker⁡T′=span⁡{(−2,1)}\ker T' = \operatorname{span}\{(-2,1)\}, a 11-dimensional subspace of W′≅R2W' \cong \mathbb{R}^2. Independently, (im⁡T)0={(d1,d2):d1⋅1+d2⋅2=0}={(d1,d2):d1=−2d2}=span⁡{(−2,1)}(\operatorname{im}T)^0 = \{(d_1,d_2) : d_1\cdot 1 + d_2 \cdot 2 = 0\} = \{(d_1,d_2) : d_1 = -2d_2\} = \operatorname{span}\{(-2,1)\}, matching exactly.

Figure
Annihilator picture for Example 2: im⁡T⊂R2\operatorname{im}T \subset \mathbb{R}^2 and its annihilator line in W′W'
w₁ w₂ im T = span(1,2) W = ℝ² f₁' f₂' (im T)⁰ = span(-2,1) = ker T' W' = ℝ²
Counterexamplewhy finite-dimensionality is needed for the annihilator dimension formula

Consider V=F[x]V = \mathbb{F}[x], the (infinite-dimensional) vector space of all polynomials over F\mathbb{F}, and let U={p∈V:p(0)=0}U = \{p \in V : p(0) = 0\}, the subspace of polynomials with zero constant term. This UU has codimension 11 in VV (a complement is the span of the constant polynomial 11), so one might hope for an analogue of "dim⁡U0=dim⁡V−dim⁡U\dim U^0 = \dim V - \dim U" reading "U0U^0 has dimension equal to the codimension of UU", i.e. dimension 11. But V′=L(V,F)V' = \mathcal{L}(V,\mathbb{F}) for infinite-dimensional VV is vastly larger than VV itself, and the annihilator U0U^0 is not 11-dimensional. Indeed, the "evaluation at 00" functional ε0(p)=p(0)\varepsilon_0(p) = p(0) certainly lies outside U0U^0 (it does not vanish on all of UU's complement, in fact it is nonzero on the complement), but there are many functionals other than scalar multiples of ε0\varepsilon_0 that vanish on all of UU. For example, one can extend a basis of UU (any basis of the polynomials-vanishing-at-00 subspace, e.g. {x,x2,x3,… }\{x, x^2, x^3, \dots\}) together with the constant 11 to a Hamel basis of VV, then define a functional sending 1↦01 \mapsto 0 and every basis vector of UU to 00 as well — that is only the zero functional, so instead consider that VV actually has uncountably many linearly independent functionals vanishing on UU once UU itself is infinite-dimensional and admits many linear functionals on a complement-free description; concretely, because dim⁡V/U=1\dim V/U = 1 but dim⁡U\dim U is already infinite, the space of functionals on VV restricting to zero on UU is naturally identified with (V/U)′(V/U)' , and while V/UV/U is 11-dimensional (so (V/U)′(V/U)' is also 11-dimensional), the subtlety is that U0≅(V/U)′U^0 \cong (V/U)' still holds by the general isomorphism theorem for annihilators — so in this particular example U0U^0 is 11-dimensional. To get a genuine failure of the counting formula, instead take U={0}U = \{0\}, the zero subspace of the infinite-dimensional space V=F[x]V = \mathbb{F}[x]. The formula "dim⁡U0=dim⁡V−dim⁡U\dim U^0 = \dim V - \dim U" would predict dim⁡U0=dim⁡V−0=dim⁡V\dim U^0 = \dim V - 0 = \dim V (an appropriate infinite cardinal), which is at least formally consistent, but the proof technique itself breaks down: the proof of part (a) crucially used basis extension of a finite list to a basis, followed by reading off dual-basis coordinates of an arbitrary φ∈U0\varphi \in U^0 using only finitely many nonzero coordinates. In infinite dimensions, an element of V′V' need not be expressible as a finite linear combination of "dual basis vectors" at all — the dual space of an infinite-dimensional space is not spanned by any naive analogue of a dual basis, because V′V' is strictly larger than the span of the functionals vi′v_i' dual to a Hamel basis (vi)i∈I(v_i)_{i \in I} of VV (a functional can be nonzero on infinitely many basis vectors simultaneously, which no finite linear combination of the vi′v_i' can achieve). This is precisely why the theorem is stated only for finite-dimensional VV: the proof genuinely requires representing every functional in U0U^0 as a finite linear combination of dual-basis vectors, a step that fails outside finite dimensions. Dropping finite-dimensionality breaks the proof, and indeed the clean dimension count dim⁡U0=dim⁡V−dim⁡U\dim U^0 = \dim V - \dim U fails to hold in the naive cardinal-arithmetic sense expected, since dim⁡V′\dim V' for infinite-dimensional VV always strictly exceeds dim⁡V\dim V itself (a cardinality fact), making any such formula for U={0}U=\{0\}, where U0=V′U^0 = V', already inconsistent with dim⁡V′=dim⁡V\dim V' = \dim V.

Scope, edge cases & common traps

SituationWhat happensWhy
V,WV, W finite-dimensionalT′T' well-defined, matrix is AtA^t, annihilator identities hold exactlyTheorem above, full proof given
VV or WW infinite-dimensionalT′T' still well-defined and linear (Theorem on linearity needs no finite-dim hypothesis)Definition and linearity proof used no dimension count
Dimension-counting corollaries (annihilator dimension formula, dim⁡im⁡T′=dim⁡im⁡T\dim\operatorname{im}T' = \dim\operatorname{im}T)Can fail or need reinterpretationProof uses finite bases and Rank–Nullity, both finite-dimensional tools
Non-square AA (m×nm \times n, m≠nm \ne n)AtA^t is n×mn \times m; T′T' genuinely maps between spaces of different dimensionsMatches shapes: T:V→WT:V\to W vs T′:W′→V′T':W'\to V'
Composition order(ST)′=T′S′(ST)' = T'S', reversedPrecomposition reverses order, proved via associativity of function composition

The mathematician behind it

Origins · 1 minHermann Grassmann and how this was discovered
HG
Hermann Grassmann
1809–1877 · German mathematician, linguist, and physicist working largely in isolation from mainstream mathematics of his time

Hermann Grassmann developed the algebra of extension ("Ausdehnungslehre", 1844), introducing what we would now recognize as vector spaces, exterior algebra, and — implicitly — dual pairings between spaces of extended magnitudes, decades before the language of linear algebra was standardized. His work was so far ahead of the notational and conceptual conventions of his era that it was largely ignored by contemporaries and only fully appreciated after his death, once mathematicians such as Giuseppe Peano and later the Bourbaki school built the modern axiomatic framework of vector spaces and dual spaces in which Grassmann's ideas found their natural home. The formal notion of a dual map and its transpose, in the abstract coordinate-free form used in this article, crystallized in the early twentieth century as functional analysis and abstract algebra matured, but its conceptual seed — treating "measuring devices" (functionals) on a space as objects with their own algebraic structure, transformable by maps on the underlying space — traces back to this pioneering and long-underappreciated work.

History
- 1844: Grassmann publishes the Ausdehnungslehre, introducing extension algebra with structures anticipating dual spaces.
- 1888: Peano gives the first fully modern axiomatic definition of a vector space, building on Grassmann's ideas.
- Early 1900s: Functional analysis (Fréchet, Hilbert, Riesz) develops the general theory of linear functionals on infinite-dimensional spaces.
- 1930s: Abstract linear algebra, including dual spaces and adjoint/transpose maps, is systematized in textbooks and by the Bourbaki collective.
- Mid-20th century: The transpose-as-dual-map viewpoint becomes standard in linear algebra pedagogy, unifying matrix transpose with abstract duality.
Why is the dual map defined by precomposition rather than postcomposition?

Precomposition is the only option that type-checks. Given T:V→WT : V \to W and a functional φ∈W′\varphi \in W' (so φ:W→F\varphi : W \to \mathbb{F}), the composite φ∘T\varphi \circ T is well-defined because the output of TT, living in WW, matches the input of φ\varphi. Postcomposing would require applying something to the output of φ\varphi, which is already a scalar, so there is no natural map F→V\mathbb{F} \to V to postcompose with in general. This forces the dual map to run backward: T′T' takes functionals on WW to functionals on VV, giving T′:W′→V′T' : W' \to V', the reverse of T:V→WT : V \to W.

Does the transpose theorem depend on the choice of dual bases, or is it basis-independent?

The map T′T' itself is completely basis-independent — it is defined purely by T′(φ)=φ∘TT'(\varphi) = \varphi \circ T, with no reference to any basis. What depends on a choice of bases is the matrix representing T′T'. The theorem states that if you use the specific dual bases associated to chosen bases of VV and WW (not arbitrary bases of V′V' and W′W'), then the matrix of T′T' works out to be exactly AtA^t. Using different bases of V′V' or W′W' not arising as duals of bases of V,WV, W would generally produce a different-looking (though similar) matrix.

Is T′′T'', the dual of the dual map, the same as TT?

For finite-dimensional VV and WW, yes, in a precise sense: there is a canonical isomorphism V≅V′′V \cong V'' sending vv to the "evaluation functional" v^(φ)=φ(v)\hat v(\varphi) = \varphi(v), and under this identification (and the analogous one for WW), T′′:V′′→W′′T'' : V'' \to W'' corresponds exactly to T:V→WT : V \to W. This is why duality is often called an involution on finite-dimensional spaces. In infinite dimensions the canonical map V→V′′V \to V'' is still injective but typically not surjective, so T′′T'' genuinely lives on a larger space V′′V'' than VV, and the identification T′′↔TT'' \leftrightarrow T breaks down without further hypotheses (e.g. reflexivity in Banach space theory).

How does this explain why row rank equals column rank of a matrix?

The column rank of A=[T]γβA = [T]_\gamma^\beta is dim⁡(im⁡T)=rank⁡(T)\dim(\operatorname{im}T) = \operatorname{rank}(T). By part (d) of the theorem, this equals dim⁡(im⁡T′)=rank⁡(T′)\dim(\operatorname{im}T') = \operatorname{rank}(T'). But by the transpose theorem, T′T' has matrix AtA^t, so rank⁡(T′)=rank⁡(At)\operatorname{rank}(T') = \operatorname{rank}(A^t), which is the column rank of AtA^t — exactly the row rank of AA. Chaining these equalities gives column rank of AA equals row rank of AA, with no appeal to row reduction at all. This is the cornerstone argument developed further in Rank of a Matrix.

Can the dual map be an isomorphism even when TT is not injective or not surjective?

No — for finite-dimensional spaces, T′T' is injective if and only if TT is surjective, and T′T' is surjective if and only if TT is injective. This follows directly from the annihilator identities: T′T' injective means ker⁡T′={0}\ker T' = \{0\}, i.e. (im⁡T)0={0}(\operatorname{im}T)^0 = \{0\}, which by the dimension formula for annihilators forces im⁡T=W\operatorname{im}T = W, i.e. TT surjective (and conversely). The symmetric statement for im⁡T′=(ker⁡T)0\operatorname{im}T' = (\ker T)^0 gives the other equivalence. In particular T′T' is an isomorphism exactly when TT is.

What goes wrong if I try to apply the transpose theorem using two different, unrelated bases of V′V' and W′W' instead of dual bases?

The theorem's proof relies crucially on the defining relations vj′(vi)=δijv_j'(v_i) = \delta_{ij} and wk′(wi)=δkiw_k'(w_i) = \delta_{ki}, which hold only because β′\beta', γ′\gamma' are the dual bases to β,γ\beta,\gamma. If you instead pick arbitrary bases of V′V' and W′W' with no relation to bases of VV and WW, the matrix of T′T' with respect to those bases will differ from AtA^t by change-of-basis matrices on each side; it need not literally equal AtA^t, though it will always be similar in structure (related by conjugation). The clean identity "[T′]=At[T'] = A^t" is a special feature of using dual bases specifically, not a basis-free fact about arbitrary representations.

Is the annihilator U0U^0 the same as the orthogonal complement U⊥U^\perp from an inner product space?

They are closely related but conceptually distinct. The annihilator U0⊆V′U^0 \subseteq V' lives in the dual space and consists of functionals vanishing on UU; it makes sense for any vector space, with no inner product required. The orthogonal complement U⊥⊆VU^\perp \subseteq V lives in VV itself and requires an inner product to make sense of "perpendicular." When VV is a finite-dimensional inner product space, the Riesz map v↦⟨⋅,v⟩v \mapsto \langle \cdot, v \rangle identifies VV with V′V' and carries U⊥U^\perp exactly onto U0U^0 under this identification, which is why the two notions are often conflated in practice — but the annihilator is the more primitive, purely linear-algebraic notion.

Before this