In linear algebra, the transpose of a linear map between two vector spaces, defined over the same field, is an induced map between the dual spaces of the two vector spaces. The transpose or algebraic adjoint of a linear map is often used to study the original linear map. This concept is generalised by adjoint functors.

Let ${\displaystyle X^{\#}}$ denote the algebraic dual space of a vector space ⁠${\displaystyle X}$⁠. Let ${\displaystyle X}$ and ${\displaystyle Y}$ be vector spaces over the same field ⁠${\displaystyle {\mathcal {K}}}$⁠. If ${\displaystyle u:X\to Y}$ is a linear map, then its algebraic adjoint or dual, is the map ${\displaystyle {}^{\#\!}u:Y^{\#}\to X^{\#}}$ defined by ⁠${\displaystyle f\mapsto f\circ u}$⁠. The resulting functional ${\displaystyle {}^{\#\!}u(f):=f\circ u}$ is called the pullback of ${\displaystyle f}$ by ⁠${\displaystyle u}$⁠.

The continuous dual space of a topological vector space (TVS) ${\displaystyle X}$ is denoted by ⁠${\displaystyle X^{\prime }}$⁠. If ${\displaystyle X}$ and ${\displaystyle Y}$ are TVSs then a linear map ${\displaystyle u:X\to Y}$ is weakly continuous if and only if ⁠${\displaystyle {}^{\#\!}u\left(Y^{\prime }\right)\subseteq X^{\prime }}$⁠, in which case we let ${\displaystyle {}^{\text{t}}\!u:Y^{\prime }\to X^{\prime }}$ denote the restriction of ${\displaystyle {}^{\#\!}u}$ to ⁠${\displaystyle Y^{\prime }}$⁠. The map ${\displaystyle {}^{\text{t}}\!u}$ is called the transpose or algebraic adjoint of ⁠${\displaystyle u}$⁠. The following identity characterizes the transpose of ⁠${\displaystyle u}$⁠: $${\displaystyle \left\langle {}^{\text{t}}\!u(f),x\right\rangle =\left\langle f,u(x)\right\rangle \quad {\text{ for all }}f\in Y^{\prime }{\text{ and }}x\in X,}$$ where ${\displaystyle \left\langle \cdot ,\cdot \right\rangle }$ is the natural pairing defined by ⁠${\displaystyle {1}}$⁠.

The assignment ${\displaystyle u\mapsto {}^{\text{t}}\!u}$ produces an injective linear map between the space of linear operators from ${\displaystyle X}$ to ${\displaystyle Y}$ and the space of linear operators from ${\displaystyle Y^{\#}}$ to ⁠${\displaystyle X^{\#}}$⁠. If ${\displaystyle X=Y}$ then the space of linear maps is an algebra under composition of maps, and the assignment is then an antihomomorphism of algebras, meaning that ⁠${\displaystyle {1}}$⁠. In the language of category theory, taking the dual of vector spaces and the transpose of linear maps is therefore a contravariant functor from the category of vector spaces over ${\displaystyle {\mathcal {K}}}$ to itself. One can identify ${\displaystyle {}^{\text{t}}\!\!\left({}^{\text{t}}\!u\right)}$ with ${\displaystyle u}$ using the natural injection into the double dual.

$${\displaystyle \|x\|=\sup _{\|x^{\prime }\|\leq 1}\left|x^{\prime }(x)\right|\quad {\text{ for each }}x\in X}$$ and if the linear operator ${\displaystyle u:X\to Y}$ is bounded then the operator norm of ${\displaystyle {}^{\text{t}}\!u}$ is equal to the norm of ${\displaystyle u}$; that is $${\displaystyle \|u\|=\left\|{}^{\text{t}}\!u\right\|,}$$ and moreover, $${\displaystyle \|u\|=\sup \left\{\left|y^{\prime }(ux)\right|:\|x\|\leq 1,\left\|y^{*}\right\|\leq 1{\text{ where }}x\in X,y^{\prime }\in Y^{\prime }\right\}.}$$

Suppose now that ${\displaystyle u:X\to Y}$ is a weakly continuous linear operator between topological vector spaces ${\displaystyle X}$ and ${\displaystyle Y}$ with continuous dual spaces ${\displaystyle X^{\prime }}$ and ⁠${\displaystyle Y^{\prime }}$⁠, respectively. Let ${\displaystyle \langle \cdot ,\cdot \rangle :X\times X^{\prime }\to \mathbb {C} }$ denote the canonical dual system, defined by ${\displaystyle \left\langle x,x^{\prime }\right\rangle =x^{\prime }x}$ where ${\displaystyle x}$ and ${\displaystyle x^{\prime }}$ are said to be orthogonal if ⁠${\displaystyle \left\langle x,x^{\prime }\right\rangle =x^{\prime }x=0}$⁠. For any subsets ${\displaystyle A\subseteq X}$ and ⁠${\displaystyle S^{\prime }\subseteq X^{\prime }}$⁠, let $${\displaystyle A^{\circ }=\left\{x^{\prime }\in X^{\prime }:\sup _{a\in A}\left|x^{\prime }(a)\right|\leq 1\right\}\qquad {\text{ and }}\qquad S^{\circ }=\left\{x\in X:\sup _{s^{\prime }\in S^{\prime }}\left|s^{\prime }(x)\right|\leq 1\right\}}$$ denote the (absolute) polar of ${\displaystyle A}$ in ${\displaystyle X^{\prime }}$ (resp. of ${\displaystyle S^{\prime }}$ in ${\displaystyle X}$).

$${\displaystyle [u(A)]^{\circ }=\left({}^{\text{t}}\!u\right)^{-1}\left(A^{\circ }\right)}$$ and $${\displaystyle u(A)\subseteq B\quad {\text{ implies }}\quad {}^{\text{t}}\!u\left(B^{\circ }\right)\subseteq A^{\circ }.}$$

$${\displaystyle \operatorname {ker} {}^{\text{t}}\!u=\left(\operatorname {Im} u\right)^{\circ }.}$$