# Transpose of a linear map

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 of a linear map is often used to study the original linear map. This concept is generalised by adjoint functors.

## Definition

Let V and W be vector spaces over the same field. If f : VW is a linear map, then the transpose[1] (or dual, or adjoint[2]), is defined to be

{\displaystyle {\begin{aligned}{}^{\text{t}}f:W^{*}&\to V^{*}\\\varphi &\mapsto \varphi \circ f.\end{aligned}}}

The resulting functional tf(φ) is called the pullback of φ along f.

The following identity, which characterises the transpose,[3] holds for all φW and vV:

${\displaystyle [{}^{\text{t}}f(\varphi ),\,v]_{V}=[\varphi ,\,f(v)]_{W},}$

where the bracket [·,·]V is the natural pairing of V's dual space with V, and [·,·]W is the same with W.

## Properties

The assignment ftf produces an injective linear map between the space of linear operators from V to W and the space of linear operators from W to V. If V = W then the space of linear maps is an algebra under composition of maps, and the assignment is then an antihomomorphism of algebras, meaning that t(fg) = tg tf. 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 F to itself. Note that one can identify t(tf) with f using the natural injection into the double dual.

• If u : XY and v : YZ are linear maps then t(vu) = tutv.[4]
• If u : XY is a linear map, AX, BY, and A° denotes the polar set of a set then[4]
• [u(A)]° = (tu)−1(A°), and
• u(A) ⊆ B implies tu(B°) ⊆ A°

## Representation as a matrix

If the linear map f is represented by the matrix A with respect to two bases of V and W, then tf is represented by the transpose matrix AT with respect to the dual bases of W and V, hence the name. Alternatively, as f is represented by A acting to the right on column vectors, tf is represented by the same matrix acting to the left on row vectors. These points of view are related by the canonical inner product on Rn, which identifies the space of column vectors with the dual space of row vectors.

## Relation to the Hermitian adjoint

The identity that characterizes the transpose, that is, [f(φ), v] = [φ, f(v)], is formally similar to the definition of the Hermitian adjoint, however, the transpose and the Hermitian adjoint are not the same map. The transpose maps ${\displaystyle W^{*}\to V^{*}}$  and is defined for linear maps between any vector spaces ${\displaystyle V}$  and ${\displaystyle W}$ , without requiring any additional form structure. The Hermitian adjoint maps ${\displaystyle W\to V}$  and is only defined for linear maps between Hilbert spaces, as it is defined in terms of the inner product on the Hilbert space. The Hermitian adjoint therefore requires more mathematical structure than the transpose.

However, the transpose is often used in contexts where the vector spaces are both equipped with a nondegenerate bilinear form such as the Euclidean dot product or another real inner product. In this case, the nondegenerate bilinear form is often used implicitly to map between the vector spaces and their duals, to express the transposed map as a map ${\displaystyle W\to V}$ . For a complex Hilbert space, the inner product is sesquilinear and not bilinear, and these conversions change the transpose into the adjoint map.

More precisely: if X and Y are Hilbert spaces and u : XY is a linear map then the transpose of u and the Hermitian adjoint of u, which we will denote respectively by tu and u, are related. Denote by I : XX and J : YY the canonical antilinear isometries of the Hilbert spaces X and Y onto their duals. Then u is the following composition of maps:[5]

${\displaystyle Y{\overset {J}{\longrightarrow }}Y^{*}{\overset {{}^{\text{t}}u}{\longrightarrow }}X^{*}{\overset {I^{-1}}{\longrightarrow }}X}$

## Applications to functional analysis

Suppose that X and Y are topological vector spaces and that u : XY is a linear map, then many of u's properties are reflected in u.

• If AX and BY are weakly closed, convex sets containing 0, then u(B°) ⊆ A° implies u(A) ⊆ B.[4]
• The null space of tu is the subspace of Y orthogonal to the range u(X) of u.[4]
• tu is injective if and only if the range u(X) of u is weakly closed.[4]

## Notes

1. ^ Treves (1999) p. 240
2. ^ Schaefer (1999) p. 128
3. ^ Halmos (1974, §44)
4. Schaefer (1999), pp. 129–130
5. ^ Treves (1999) p. 488

## References

• Halmos, Paul (1974), Finite-dimensional Vector Spaces, Springer, ISBN 0-387-90093-4
• Schaefer, Helmuth H.; Wolff, M.P. (1999). Topological Vector Spaces. GTM. 3. New York: Springer-Verlag. ISBN 9780387987262.
• Trèves, François (1995). Topological Vector Spaces, Distributions and Kernels. Dover Publications. ISBN 9780486453521.