Transpose of a linear map ($T^{\ast }:W^{\ast }\to V^{\ast },T^{\ast }(\phi )=\phi \circ T$)

For a linear map $T:V\to W$, the dual map $T^{\ast }:W^{\ast }\to V^{\ast }$ sends a linear functional $\phi$ on the output space to the input measurement $T^{\ast }(\phi )=\phi \circ T$. It reverses direction because a measurement of the output can be evaluated on an input by first applying $T$:For $T(x,y)=(x+y,2y)$ and $\phi (u,v)=2u-v$, the input measurement is $T^{\ast }\phi (x,y)=2(x+y)-2y=2x$. The output measurement ignores the part of this transformation contributed by $y$.

If $T$ has matrix $A$ in specified bases, a row representing $\phi$ changes to $\ell A$. Storing the coefficients of a linear functional as a column in the corresponding dual basis gives $A^T{\ell }^T$, explaining why the transpose of a linear map is represented by the matrix transpose. This algebraic operation uses no inner product; it agrees with a matrix transpose even over the complex numbers, rather than automatically introducing complex conjugation. The adjoint operator is defined using an inner product and is a different construction. ^[ref].

In finite dimension, the kernel and image of the dual map are described by annihilators of vector subspaces:

For a linear map between finite-dimensional vector spaces, the image of its dual map is the annihilator of a vector subspace given by its kernel. Inclusion follows by composition. Conversely a linear functional vanishing on the kernel descends to the image and extends to the target by basis extension. The identity and the rank-nullity theorem show that a linear map and its dual have equal rank.

For the map represented by a matrix $A$, column rank is the rank of the map and row rank is the rank of its dual. Since the dual map has kernel equal to the annihilator of the original image, rank-nullity gives equal ranks.

If the columns $v_1,\dots ,v_m$ of an $n$ by $m$ matrix are independent, some $m$ by $m$ row minor is nonzero. A nonzero term of that minor's determinant selects distinct nonzero coordinates $f(i)$, and replacing the corresponding standard basis vectors $e_{f(i)}$ by the $v_i$ leaves a basis.

For a finite-dimensional vector space and $U\le V$, dualizing the quotient and inclusion maps gives