Dual Bases and Coordinate Extraction

Why a Dual Basis Extracts Coordinates

A basis represents each vector using a unique list of scalar coefficients. A dual basis provides a direct way to retrieve those coefficients: its functionals act on a vector and return one coordinate at a time. This is useful both as a conceptual link between vectors and linear functionals and as a practical computation. In coordinates, the dual functionals can be read from the inverse of the matrix whose columns are the basis vectors.

Let \(V\) be a finite-dimensional vector space over a field \(\mathbb{F}\), where \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\). A linear functional on \(V\) is a linear map \(f:V\to\mathbb{F}\). The dual space \(V^*\) is the vector space of all such functionals, with addition and scalar multiplication defined pointwise. In the complex case, these are linear functionals, not conjugate-linear functionals.

The order matters: \(f_i\) returns the coefficient of \(\mathbf{b}_i\), so matching the functional to the corresponding basis vector is essential. If \(\mathbf{v}=c_1\mathbf{b}_1+\cdots+c_n\mathbf{b}_n\), then linearity and the defining values give \(f_i(\mathbf{v})=c_i\). The next result proves this for every vector and also establishes that the required functionals exist.

Proof. Since \(\mathcal{B}\) is a basis, every \(\mathbf{v}\in V\) has a unique expression \(\mathbf{v}=\sum_{j=1}^n c_j\mathbf{b}_j\). Define \(f_i(\mathbf{v})=c_i\). This defines a function on all of \(V\) because the coefficients are unique. If \(\mathbf{v}=\sum_j c_j\mathbf{b}_j\) and \(\mathbf{w}=\sum_j d_j\mathbf{b}_j\), then \(a\mathbf{v}+b\mathbf{w}=\sum_j(ac_j+bd_j)\mathbf{b}_j\); its \(i\)-th coefficient is \(ac_i+bd_i\). Thus \(f_i(a\mathbf{v}+b\mathbf{w})=af_i(\mathbf{v})+bf_i(\mathbf{w})\), so \(f_i\) is linear. The \(i\)-th coefficient of \(\mathbf{b}_j\) is \(\delta_{ij}\), giving \(f_i(\mathbf{b}_j)=\delta_{ij}\). Any functional \(g_i\) with these values must satisfy \(g_i(\mathbf{v})=\sum_j c_jg_i(\mathbf{b}_j)=c_i\), proving uniqueness. The same computation proves coordinate extraction. To show the functionals form a basis of \(V^*\), take any \(g\in V^*\). For every \(\mathbf{v}=\sum_j c_j\mathbf{b}_j\), linearity gives \(g(\mathbf{v})=\sum_j c_jg(\mathbf{b}_j)=\sum_j g(\mathbf{b}_j)f_j(\mathbf{v})\). Hence \(g=\sum_j g(\mathbf{b}_j)f_j\), so the functionals span \(V^*\). If \(\sum_i a_if_i\) is the zero functional, evaluating at \(\mathbf{b}_j\) gives \(a_j=0\) for every \(j\). They are linearly independent and therefore form a basis. \(\square\)

In particular, coordinate extraction can be written as a vector identity: \(\mathbf{v}=\sum_{i=1}^n f_i(\mathbf{v})\mathbf{b}_i\). The functionals return the coefficients; the basis vectors then assemble those coefficients back into the vector. This identity is valid for the zero vector as well: every \(f_i(\mathbf{0})=0\), so the reconstruction gives \(\mathbf{0}\).

Computing the Dual Basis with a Matrix Inverse

For \(V=\mathbb{F}^n\), let \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) be an ordered basis of column vectors, and form the basis matrix \(\mathbf{P}=[\mathbf{b}_1\ \cdots\ \mathbf{b}_n]\). Its columns form a basis, so \(\mathbf{P}\) is invertible by the Invertible Matrix Theorem. If \(\mathbf{x}\) is a vector and \(\mathbf{c}\) is its coordinate column relative to \(\mathcal{B}\), then \(\mathbf{x}=\mathbf{P}\mathbf{c}\), hence \(\mathbf{c}=\mathbf{P}^{-1}\mathbf{x}\). Each row of \(\mathbf{P}^{-1}\) therefore represents one dual functional.

Proof. By the definition of an inverse, \(\mathbf{P}^{-1}\mathbf{P}=\mathbf{I}_n\). The entry in row \(i\), column \(j\) of this product is \(\mathbf{r}_i\mathbf{b}_j\), so it equals \(\delta_{ij}\). Matrix multiplication also shows that each function \(f_i(\mathbf{x})=\mathbf{r}_i\mathbf{x}\) is linear. Thus these functionals have the defining values of the dual basis, which is unique by the dual-basis theorem. For any \(\mathbf{x}\), the coordinate column is \(\mathbf{P}^{-1}\mathbf{x}\), whose \(i\)-th entry is \(\mathbf{r}_i\mathbf{x}=f_i(\mathbf{x})\). \(\square\)

This row formula uses ordinary matrix multiplication, including when the scalar field is \(\mathbb{C}\). No conjugation is applied to the rows: the objects being represented are linear functionals.

Dual Bases on a Subspace

The inverse-matrix formula applies directly when the basis matrix is square and its columns form a basis of the whole coordinate space. For a subspace, the same coordinate-extraction idea works, but the functionals have that subspace as their domain. A rectangular matrix of basis columns does not have an ordinary matrix inverse; instead, one can express a vector in the given basis and identify its coefficients.

Common Pitfall: Using the Wrong Matrix

For coordinate extraction, the correct operation is \(\mathbf{P}^{-1}\mathbf{x}\), not \(\mathbf{P}^{\top}\mathbf{x}\). The transpose generally does not undo the action of the basis matrix. For the basis in the preceding two-dimensional worked example, \(\mathbf{P}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\) and \(\mathbf{x}=(3,2)^{\top}\). The incorrect calculation gives

which cannot be the coordinates, since \(3\mathbf{b}_1+5\mathbf{b}_2=(8,5)\ne(3,2)\). The correction is to use the inverse: \(\mathbf{P}^{-1}\mathbf{x}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\begin{bmatrix}3\\2\end{bmatrix}=(1,2)^{\top}\), and \(1\mathbf{b}_1+2\mathbf{b}_2=(3,2)\). A transpose can occur in other settings, but it is not the general coordinate-extraction rule for an arbitrary basis.

There is also no need for an inner product to define a dual basis. A dual functional is specified by its linearity and its values on the basis vectors. In particular, for a complex vector space, using conjugated rows would represent conjugate-linear expressions rather than the linear functionals defined here.

Why Dual Coordinates Are Useful

The dual basis turns the abstract statement “a vector has unique basis coordinates” into a collection of scalar-valued measurements. Each \(f_i\) isolates exactly one coefficient, while the identity \(\mathbf{v}=\sum_i f_i(\mathbf{v})\mathbf{b}_i\) rebuilds the vector from those measurements. In a coordinate space, the rows of the inverse basis matrix perform these measurements simultaneously. The same viewpoint applies to any finite-dimensional vector space, even when its vectors are not initially presented as columns.