Transpose of a linear map
In linear algebra, the transpose of a linear map, also referred to as the dual map or pullback, is a canonical construction that assigns to each linear transformation T: [V](/page/V.) \to [W](/page/W) between vector spaces over a field [K](/page/K) a corresponding linear transformation T^t: W^* \to V^* between their dual spaces, defined by (T^t \phi)(v) = \phi(T v) for all \phi \in W^* and v \in V, where V^* and W^* denote the spaces of linear functionals on V and W, respectively.[1][2][3] This definition ensures that T^t preserves the algebraic structure of linear maps, as it is itself linear: for any \phi_1, \phi_2 \in W^* and scalars c \in K, T^t(\phi_1 + c \phi_2) = T^t \phi_1 + c T^t \phi_2.[2][3] Key properties of the transpose highlight its role in preserving dimensions and relating kernels and images across dual spaces. Specifically, the null space of T^t is the annihilator of the range of T, denoted \ker T^t = (\operatorname{im} T)^0, and the range of T^t is the annihilator of the kernel of T, so \operatorname{im} T^t = (\ker T)^0.[2][3] As a consequence, the rank of T equals the rank of T^t, i.e., \operatorname{rank} T = \operatorname{rank} T^t, which extends the well-known equality of row and column ranks for matrices.[1][3] Additionally, the transpose reverses composition: for linear maps S: U \to V and T: V \to W, (S T)^t = T^t S^t, and it commutes with addition and scalar multiplication in an analogous manner.[1] When V and W are finite-dimensional, the transpose corresponds directly to the matrix transpose under choice of bases. If T has matrix representation A with respect to bases for V and W, then T^t has matrix A^t (the transpose of A) with respect to the dual bases for W^* and V^*.[3] This connection is fundamental in applications such as solving systems of linear equations, where the transpose arises in least-squares problems and orthogonal projections, and in more abstract settings like tensor algebra and representation theory.[1] In the context of inner product spaces, the transpose coincides with the adjoint operator when identifying a space with its dual via the inner product, though the pure transpose does not require an inner product structure.[2]Definition and Foundations
Dual Spaces
The dual space of a vector space V over a field F, denoted V^*, is the set of all linear functionals on V, that is, all linear maps \phi: V \to F.[4] These linear functionals, also called covectors, assign to each vector in V an element of the scalar field F in a linear manner.[5] The dual space V^* itself forms a vector space over F, with pointwise addition of functionals defined by (\phi + \psi)(v) = \phi(v) + \psi(v) for \phi, \psi \in V^* and v \in V, and scalar multiplication by (\alpha \phi)(v) = \alpha \phi(v) for \alpha \in F.[4] The zero element in V^* is the zero functional that maps every vector in V to $0 \in F.[4] The dual space V^* corepresents the space of linear maps from V to F, meaning it parametrizes all such maps. A key aspect is the evaluation map, which associates to each v \in V the functional \mathrm{ev}_v \in (V^*)^* defined by \mathrm{ev}_v(\phi) = \phi(v) for \phi \in V^*; for finite-dimensional V, this induces a natural isomorphism V \cong (V^*)^*.[4] While the algebraic dual V^* is defined for any vector space without additional structure, the topological dual consists of continuous linear functionals and forms a subspace of V^* when V is equipped with a topology; the algebraic dual is the focus here unless otherwise specified.[5] For example, if V = F^n, then V^* is isomorphic to the space of row vectors in F^n, where each functional acts via the dot product: \phi(x_1, \dots, x_n) = a_1 x_1 + \dots + a_n x_n for coefficients a_i \in F.[6]Definition of the Transpose Map
Let V and W be vector spaces over a field F, and let T: V \to W be a linear map.[4] The transpose (or dual map) of T, denoted T^*: W^* \to V^*, is the linear map between the dual spaces defined by (T^* \phi)(v) = \phi(T v) for all \phi \in W^* and v \in V, where V^* = \mathrm{Hom}_F(V, F) and W^* = \mathrm{Hom}_F(W, F).[7] To verify that T^* is linear, consider \phi, \psi \in W^* and \alpha \in F. For any v \in V, (T^*(\phi + \psi))(v) = (\phi + \psi)(T v) = \phi(T v) + \psi(T v) = (T^* \phi)(v) + (T^* \psi)(v), and (T^*(\alpha \phi))(v) = (\alpha \phi)(T v) = \alpha (\phi(T v)) = \alpha (T^* \phi)(v). Thus, T^* preserves addition and scalar multiplication, confirming its linearity.[4] The notation T^* is standard in many treatments, though alternatives such as T^t or T^\vee appear in some sources.[4][2] This construction generalizes to the setting of modules over a commutative ring R: for an R-linear map T: V \to W between R-modules, the transpose T^\vee: W^\vee \to V^\vee is defined by pre-composition, T^\vee(\phi) = \phi \circ T for \phi \in W^\vee = \mathrm{Hom}_R(W, R).[8] However, the primary focus here is on vector spaces. As an example, consider finite-dimensional spaces V = F^m and W = F^n, where elements of V^* and W^* may be identified with row vectors in F^{1 \times m} and F^{1 \times n}, respectively. In this identification, T^* maps row vectors via pre-composition with T.[7]Algebraic Properties
Linearity and Composition
The transpose of a linear map inherits linearity from the original map. Suppose T: V \to W is a linear map between vector spaces over the field \mathbb{F}, and T^*: W^* \to V^* denotes its transpose, defined by (T^* \psi)(v) = \psi(T v) for all \psi \in W^* and v \in V. To verify linearity, consider the sum of functionals: for \psi_1, \psi_2 \in W^*, ((T^* (\psi_1 + \psi_2))(v) = (\psi_1 + \psi_2)(T v) = \psi_1(T v) + \psi_2(T v) = (T^* \psi_1)(v) + (T^* \psi_2)(v) = (T^* \psi_1 + T^* \psi_2)(v), which holds for all v \in V, so T^*(\psi_1 + \psi_2) = T^* \psi_1 + T^* \psi_2. Similarly, for scalar multiplication and c \in \mathbb{F}, (T^* (c \psi))(v) = (c \psi)(T v) = c \, \psi(T v) = c (T^* \psi)(v) = (c T^* \psi)(v), confirming T^* is linear.[9] The transpose reverses the order under composition. Let S: W \to U and T: V \to W be linear maps, with transposes S^*: U^* \to W^* and T^*: W^* \to V^*. The composition rule states (S \circ T)^* = T^* \circ S^*. To prove this, evaluate the defining action on an arbitrary \phi \in U^* and v \in V: (((S \circ T)^* \phi)(v) = \phi((S \circ T) v) = \phi(S (T v)) = (S^* \phi)(T v) = (T^* (S^* \phi))(v). Thus, (S \circ T)^* \phi = T^* (S^* \phi) for all \phi, establishing the equality of maps. This reversal arises naturally from the contravariant nature of the dual space functor.[9] The identity map's transpose is itself the identity on the dual space. For the identity \mathrm{Id}_V: V \to V, its transpose satisfies (\mathrm{Id}_V^* \psi)(v) = \psi(\mathrm{Id}_V v) = \psi(v) for all \psi \in V^* and v \in V, so \mathrm{Id}_V^* = \mathrm{Id}_{V^*}. This follows directly from the definition and holds over any field.[9] Invertibility is preserved under transposition. If T: V \to W is invertible (i.e., bijective), then T^*: W^* \to V^* is also invertible, with inverse (T^*)^{-1} = (T^{-1})^*. To see surjectivity of T^*, take any \psi \in V^* and define \phi = \psi \circ T^{-1} \in W^*; then T^* \phi (v) = \phi(T v) = \psi(T^{-1} (T v)) = \psi(v). For injectivity, if T^* \phi = 0, then \phi(T v) = 0 for all v \in V, so \phi(w) = 0 for all w \in W (since T is surjective), hence \phi = 0. The inverse relation follows by composing the definitions: (T^{-1})^* T^* \psi (v) = T^* \psi (T^{-1} v) = \psi(T (T^{-1} v)) = \psi(v), and similarly for the other order. This isomorphism property underscores the duality between V and V^* for finite-dimensional spaces.[9]Dimension and Rank Relations
In finite-dimensional vector spaces V and W over a field K, the dual spaces V^* and W^* are isomorphic to V and W, respectively, so \dim V^* = \dim V and \dim W^* = \dim W. For a linear map T: V \to W, the transpose T^*: W^* \to V^* satisfies \dim \operatorname{Im} T^* = \dim \operatorname{Im} T and \dim \ker T^* = \dim W - \dim \operatorname{Im} T. This follows from the rank-nullity theorem applied to both T and T^*: \dim V = \rank T + \nullity T and \dim W^* = \rank T^* + \nullity T^*, combined with the duality of dimensions and \rank T^* = \rank T, yielding \nullity T^* = \dim W - \rank T.[2][10] The equality of ranks arises from explicit relations between the kernels and images via annihilators. The annihilator of a subspace S \subset U is the subspace S^0 = \{ f \in U^* \mid f(s) = 0 \ \forall s \in S \}. For the transpose, \ker T^* = (\operatorname{Im} T)^0 \subset W^* and \operatorname{Im} T^* = (\ker T)^0 \subset V^*. In finite dimensions, \dim (\operatorname{Im} T)^0 = \dim W - \dim \operatorname{Im} T, so \dim \ker T^* = \dim W - \dim \operatorname{Im} T; similarly, \dim (\ker T)^0 = \dim V - \dim \ker T = \rank T, so \dim \operatorname{Im} T^* = \rank T. These annihilator identifications hold in arbitrary dimensions, providing structural relations even when dimensions are infinite, though the numerical equalities rely on finite-dimensionality.[2] A related relation involves the cokernel, defined for T: V \to W as \coker T = W / \operatorname{Im} T. Dually, \coker T^* = V^* / \operatorname{Im} T^* = V^* / (\ker T)^0. By the isomorphism theorems for dual spaces, the natural map V^* \to (V / \ker T)^* has kernel (\ker T)^0, yielding V^* / (\ker T)^0 \cong (V / \ker T)^*. Since T induces V / \ker T \cong \operatorname{Im} T, it follows that \coker T^* \cong (\operatorname{Im} T)^*. This duality preserves the structure of the original map's image in the dual setting and holds generally, without assuming finite dimensions.[2] In infinite dimensions, the rank equality \rank T^* = \rank T (understood as the dimension of the image) follows from the above isomorphisms: \operatorname{Im} T^* \cong (V / \ker T)^* \cong (\operatorname{Im} T)^*, so the image of T^* is dual to the image of T. While dimensions may differ (e.g., \dim (\operatorname{Im} T)^* \geq \dim \operatorname{Im} T with strict inequality possible for infinite \dim \operatorname{Im} T), the structural equivalence via dualization maintains the rank relation in this generalized sense. As an illustrative example, consider a nilpotent linear map T: V \to V on a finite-dimensional space, meaning T^k = 0 for some minimal index k but T^{k-1} \neq 0. The transpose T^* is also nilpotent with the same index k, since the Jordan canonical form of T^* consists of the transposed Jordan blocks of T, preserving the sizes of the nilpotent blocks and thus the nilpotency index.[11]Geometric and Set-Theoretic Properties
Polars
In the context of dual vector spaces, the polar of a subset A \subseteq V of a vector space V over a field F is defined as the setA^0 = \{ \phi \in V^* \mid \phi(a) = 0 \ \forall \, a \in A \} \subseteq V^*,
where V^* denotes the algebraic dual space of V, consisting of all linear functionals from V to F.[12] This construction provides a dual analog to the orthogonal complement, capturing the functionals that vanish on every element of A. The polar A^0 is always a subspace of V^*, and it depends only on the span of A, since A^0 = (\operatorname{span} A)^0.[13] The bipolar of A, denoted (A^0)^0, is the polar of A^0 taken in the double dual (V^*)^*. Algebraically, (A^0)^0 always contains \operatorname{span} A, as any element of \operatorname{span} A annihilates precisely the functionals in A^0. In more general settings, such as topological vector spaces, (A^0)^0 coincides with the closure of \operatorname{span} A under suitable topologies on V. However, when V is finite-dimensional, reflexivity ensures exact equality: (A^0)^0 = \operatorname{span} A. This follows from the natural isomorphism between V and its double dual V^{**}, which identifies elements of V with evaluation functionals on V^*.[13] For a linear map T: V \to W between vector spaces, the polar of the image \operatorname{Im} T relates directly to the transpose (or dual map) T^*: W^* \to V^*, defined by (T^* \psi)(v) = \psi(T v) for \psi \in W^* and v \in V. Specifically, the kernel of T^* is the polar of \operatorname{Im} T:
\ker T^* = (\operatorname{Im} T)^0 = \{ \psi \in W^* \mid \psi(T v) = 0 \ \forall \, v \in V \}.
This identity highlights how the transpose encodes information about the functionals vanishing on the range of T.[12] In the finite-dimensional case over \mathbb{R}, with V = \mathbb{R}^n equipped with its standard dual V^* \cong \mathbb{R}^n via the pairing \langle [x, y](/page/X&Y) \rangle = x^T y (identifying functionals with row vectors), the polar of a subspace U \subseteq \mathbb{R}^n is precisely its orthogonal complement U^\perp = \{ y \in \mathbb{R}^n \mid x^T y = 0 \ \forall \, x \in U \}. For instance, if U is the xy-plane in \mathbb{R}^3, spanned by (1,0,0) and (0,1,0), then U^0 = U^\perp is the z-axis, spanned by (0,0,1). This identification bridges algebraic duality with Euclidean geometry.[14]