Why the Dual Map Has the Transposed Matrix

Why a Transpose Appears

A linear map sends vectors forward, while its dual map acts on linear functionals by composing them with that map. The transpose appears because the coordinates of a functional record its values on basis vectors: the value of a functional on the image of one basis vector is assembled from the corresponding column of the original map’s matrix. Tracking those values carefully explains both the transpose rule and why dual maps reverse the order of composition.

Let \(\mathbb{F}\) be \(\mathbb{R}\) or \(\mathbb{C}\), and let \(V\) and \(W\) be finite-dimensional vector spaces over \(\mathbb{F}\). The dual space \(V^*\) is the space of linear functionals \(f:V\to\mathbb{F}\). For a linear map \(T:V\to W\), its dual map is \(T^*:W^*\to V^*\), defined by \(T^*(g)=g\circ T\). Thus \(T^*(g)\) takes an input from \(V\), sends it through \(T\), and then evaluates \(g\) on the result. This is the algebraic dual map: over \(\mathbb{C}\), the functionals in \(V^*\) are linear, not conjugate-linear.

Choose ordered bases \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) of \(V\) and \(\mathcal{C}=(\mathbf{c}_1,\dots,\mathbf{c}_m)\) of \(W\). Their dual bases are \(\mathcal{B}^*=(\beta_1,\dots,\beta_n)\) and \(\mathcal{C}^*=(\gamma_1,\dots,\gamma_m)\), where \(\beta_j(\mathbf{b}_k)=1\) if \(j=k\) and \(0\) otherwise, and likewise \(\gamma_i(\mathbf{c}_k)=1\) if \(i=k\) and \(0\) otherwise. In these bases, a functional’s coordinate vector lists its values on the corresponding basis, in order.

Proof. Write \(\mathbf{A}=(a_{ij})\), so its \(j\)th column records the \(\mathcal{C}\)-coordinates of \(T(\mathbf{b}_j)\). In particular, \(T(\mathbf{b}_j)=\sum_{i=1}^{m}a_{ij}\mathbf{c}_i\). For each \(i\), apply the dual map to the \(i\)th dual basis functional of \(W\). For each \(j\), linearity and the dual-basis property give \[ (T^*\gamma_i)(\mathbf{b}_j) =\gamma_i(T(\mathbf{b}_j)) =\gamma_i\left(\sum_{k=1}^{m}a_{kj}\mathbf{c}_k\right) =a_{ij}. \] A functional’s coordinate in position \(j\) relative to \(\mathcal{B}^*\) is its value on \(\mathbf{b}_j\). Therefore the \(j\)th coordinate of \(T^*\gamma_i\) is \(a_{ij}\). The \(i\)th column of the matrix of \(T^*\) is consequently \((a_{i1},\dots,a_{in})^\top\), the \(i\)th row of \(\mathbf{A}\) written as a column. This holds for every \(i=1,\dots,m\), so the matrix of \(T^*\) is \(\mathbf{A}^\top\). \(\square\)

The sizes help keep the rule straight. The matrix of \(T\) has \(m\) rows and \(n\) columns because \(T\) maps from an \(n\)-dimensional space to an \(m\)-dimensional space. The dual map goes from \(W^*\), with basis \(\mathcal{C}^*\), to \(V^*\), with basis \(\mathcal{B}^*\). Its matrix must therefore have \(n\) rows and \(m\) columns, just like \(\mathbf{A}^\top\).

Applying the Rule in Standard Coordinates

Why the Rule Works in Any Bases

The transpose rule is not limited to standard coordinates. Its meaning depends on using the dual bases in the correct order: the basis \(\mathcal{C}^*\) for the input functionals and \(\mathcal{B}^*\) for the output functionals. Here is an example where neither vector-space basis is the standard basis.

The calculation also shows why basis labels matter. A matrix is not just a list of numbers: its rows and columns refer to specific ordered bases. The transpose rule pairs the original input basis with the output dual basis, and the original output basis with the input dual basis. Using the transpose while keeping the original bases in place would generally describe a different map.

Dual Maps Reverse Composition

Suppose \(S:U\to V\) and \(T:V\to W\) are linear maps. The composite \(T\circ S\) first applies \(S\), then \(T\). Its dual map starts with a functional on \(W\), so it must first pull that functional back through \(T\), and then pull the result back through \(S\). This reverses the order.

Proof. Take any \(g\in W^*\). Both \((T\circ S)^*(g)\) and \(S^*(T^*(g))\) are functionals on \(U\). To show they are equal, evaluate them at an arbitrary \(\mathbf{u}\in U\). By the definition of the dual map, \[ \big((T\circ S)^*(g)\big)(\mathbf{u}) =g\big((T\circ S)(\mathbf{u})\big) =g\big(T(S(\mathbf{u}))\big). \] On the other hand, \[ \big(S^*(T^*(g))\big)(\mathbf{u}) =(T^*(g))(S(\mathbf{u})) =g\big(T(S(\mathbf{u}))\big). \] The values agree for every \(\mathbf{u}\in U\), so the functionals are equal. This holds for every \(g\in W^*\), proving the stated equality of maps. \(\square\)

In coordinates, the same reversal is the familiar rule for transposing a product. If \(\mathbf{A}\) is the matrix of \(S\) and \(\mathbf{B}\) is the matrix of \(T\), using compatible bases, then \(T\circ S\) has matrix \(\mathbf{B}\mathbf{A}\). The dual maps have matrices \(\mathbf{A}^\top\) and \(\mathbf{B}^\top\), in the corresponding dual bases, so the composite \(S^*\circ T^*\) has matrix \(\mathbf{A}^\top\mathbf{B}^\top=(\mathbf{B}\mathbf{A})^\top\).

A Common Mistake: Confusing Transpose and Adjoint

Over \(\mathbb{C}\), the algebraic dual map uses the transpose, not the conjugate transpose. The conjugate transpose belongs to the inner-product adjoint, which is a different construction. For a concrete check, take the complex-linear map on \(\mathbb{C}^2\) with matrix \[ \mathbf{A}=\begin{bmatrix}i&1\\0&1\end{bmatrix} \] in the standard basis, and let \(g(y_1,y_2)=y_1\). The coordinate vector of \(g\) in the dual basis is \((1,0)^\top\). The dual-map rule gives \[ \mathbf{A}^\top\begin{bmatrix}1\\0\end{bmatrix} =\begin{bmatrix}i&0\\1&1\end{bmatrix} \begin{bmatrix}1\\0\end{bmatrix} =\begin{bmatrix}i\\1\end{bmatrix}. \] Indeed, \(g(T(x_1,x_2))=ix_1+x_2\), so those are the correct functional coefficients.

A wrong calculation would replace \(\mathbf{A}^\top\) with \(\mathbf{A}^*=\overline{\mathbf{A}}^\top\). It would produce \[ \mathbf{A}^*\begin{bmatrix}1\\0\end{bmatrix} =\begin{bmatrix}-i&0\\1&1\end{bmatrix} \begin{bmatrix}1\\0\end{bmatrix} =\begin{bmatrix}-i\\1\end{bmatrix}, \] which represents \(g(T(x_1,x_2))=-ix_1+x_2\), not \(ix_1+x_2\). The conjugation changes the result and fails the defining evaluation \(T^*(g)=g\circ T\). Over \(\mathbb{R}\), transpose and conjugate transpose coincide numerically, but the algebraic definition is still composition with \(T\).

Why These Rules Are Useful

The transpose rule converts questions about the action of a map on functionals into ordinary matrix multiplication. The reversal-of-composition rule also prevents an easy order error: when maps are composed forward, their duals are composed in reverse. This order is not a convention added to the matrix calculation; it follows directly from the definition of composition and evaluation.

When using these rules, check three items before multiplying:

• Identify the domain and codomain of the original map and the dual map.
• Write the bases and dual bases in their correct order.
• For a composite, reverse the map order before composing the duals.

In particular, the dual-map matrix is not usually the inverse of the original matrix, and it is not generally the conjugate transpose. It is the transpose relative to the specified dual bases. The examples show that the decisive check is always evaluation: a proposed formula must give the same value as applying the functional to the image under the original map.