= Transpose of a linear map
{title2=$T^*:W^*\to V^*,\quad T^*(\varphi)=\varphi\circ T$}
{wiki}
= Dual map
{synonym}
For a <linear map> $T:V\to W$, the <dual map> $T^*:W^*\to V^*$ sends a <linear functional> $\varphi$ on the output space to the input measurement $T^*(\varphi)=\varphi\circ T$. It reverses direction because a measurement of the output can be evaluated on an input by first applying $T$:
$$
(T^*\varphi)(v)=\varphi(Tv).
$$
For $T(x,y)=(x+y,2y)$ and $\varphi(u,v)=2u-v$, the input measurement is $T^*\varphi(x,y)=2(x+y)-2y=2x$. The output measurement ignores the part of this transformation contributed by $y$.
If $T$ has <matrix> $A$ in specified <bases>, a row representing $\varphi$ 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. https://www.math.brown.edu/streil/papers/LADW/HTML_2026_04-30/Ch8.html[Treil, Linear Algebra Done Wrong, §8.3: dual transformations and transpose]{ref}.
In finite <dimension>, the <kernel> and image of the <dual map> are described by <annihilators of vector subspaces>:
$$
\ker T^*=(\operatorname{im}T)^\circ,
\qquad
\operatorname{im}T^*=(\ker T)^\circ.
$$
Back to article page
