Bilinear Forms and Their Matrix Representations

Bilinear Forms as Functions of Two Vectors

A linear functional takes one vector as input and returns a scalar. A bilinear form takes two vectors as inputs and is linear in each one while the other is held fixed. Its matrix records how the form acts on pairs of basis vectors. Once that matrix is known, evaluating the form becomes a matrix multiplication involving the coordinate columns of the two vectors.

Throughout this tutorial, \(V\) is a finite-dimensional vector space over \(\mathbb{R}\). The vectors themselves need not be columns until we choose a basis and write their coordinate columns. The order of the basis matters for the coordinates and for the matrix representation.

These conditions imply that \(B(\mathbf{0},\mathbf{v})=B(\mathbf{u},\mathbf{0})=0\). For example, linearity in the first argument gives \(B(\mathbf{0},\mathbf{v})=B(0\mathbf{u},\mathbf{v})=0B(\mathbf{u},\mathbf{v})=0\). Bilinearity does not require the two arguments to be interchangeable: in general, \(B(\mathbf{u},\mathbf{v})\) need not equal \(B(\mathbf{v},\mathbf{u})\).

The Matrix in an Ordered Basis

Let \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) be an ordered basis of \(V\). Each \(\mathbf{u}\in V\) has a unique coordinate column \([\mathbf{u}]_{\mathcal{B}}=(u_1,\dots,u_n)^{\top}\), meaning \(\mathbf{u}=\sum_{i=1}^{n}u_i\mathbf{b}_i\). Define the matrix of \(B\) in this basis by placing \(B(\mathbf{b}_i,\mathbf{b}_j)\) in row \(i\), column \(j\):

The row index corresponds to the first argument and the column index to the second. This convention makes the coordinate formula use the transpose of the first coordinate column, followed by the matrix, followed by the second coordinate column.

Proof. Write \([\mathbf{u}]_{\mathcal{B}}=(u_1,\dots,u_n)^{\top}\) and \([\mathbf{v}]_{\mathcal{B}}=(v_1,\dots,v_n)^{\top}\). Then \(\mathbf{u}=\sum_i u_i\mathbf{b}_i\) and \(\mathbf{v}=\sum_j v_j\mathbf{b}_j\). Applying linearity first in one argument and then in the other gives $$ B(\mathbf{u},\mathbf{v}) =B\left(\sum_i u_i\mathbf{b}_i,\sum_j v_j\mathbf{b}_j\right) =\sum_i\sum_j u_i v_j B(\mathbf{b}_i,\mathbf{b}_j). $$ The entry in row \(i\), column \(j\) of \(\mathbf{M}_{\mathcal{B}}(B)\) is \(B(\mathbf{b}_i,\mathbf{b}_j)\), so the final double sum is exactly the matrix product \([\mathbf{u}]_{\mathcal{B}}^{\top}\mathbf{M}_{\mathcal{B}}(B)[\mathbf{v}]_{\mathcal{B}}\). This proves the formula. Now let \(\mathbf{A}=(a_{ij})\) be any real \(n\times n\) matrix. Define $$ B_{\mathbf{A}}(\mathbf{u},\mathbf{v}) =[\mathbf{u}]_{\mathcal{B}}^{\top}\mathbf{A}[\mathbf{v}]_{\mathcal{B}}. $$ Coordinates depend linearly on the vector, and matrix multiplication is linear in each column input, so \(B_{\mathbf{A}}\) is bilinear. For each pair of basis vectors, \([\mathbf{b}_i]_{\mathcal{B}}=\mathbf{e}_i\) and \([\mathbf{b}_j]_{\mathcal{B}}=\mathbf{e}_j\), hence \(B_{\mathbf{A}}(\mathbf{b}_i,\mathbf{b}_j)=\mathbf{e}_i^{\top}\mathbf{A}\mathbf{e}_j=a_{ij}\). Thus its matrix is \(\mathbf{A}\). Finally, any bilinear form with matrix \(\mathbf{A}\) must have the coordinate formula already proved, so its value on every pair \((\mathbf{u},\mathbf{v})\) is \(B_{\mathbf{A}}(\mathbf{u},\mathbf{v})\). The form is therefore unique. \(\square\)

The theorem has two useful consequences. To construct the matrix of a given form, evaluate it on all ordered pairs \((\mathbf{b}_i,\mathbf{b}_j)\). Conversely, a matrix specifies a bilinear form once a basis is fixed. The matrix is not the form itself: it is the coordinate record of the form relative to that ordered basis.

Evaluating a Form from Its Matrix

The matrix product sums contributions from every pair of coordinates. In indices, it is \(\sum_{i=1}^n\sum_{j=1}^n u_i m_{ij}v_j\). The \(i\)-th coordinate of the first vector multiplies row \(i\), while the \(j\)-th coordinate of the second vector selects column \(j\). This explains why changing the order of the coordinate columns can change the value.

Recovering Values from Basis-Pair Data

The definition of the matrix also works when the only information supplied about \(B\) is its values on basis pairs. Each entry must be recorded in the correct position: \(B(\mathbf{b}_i,\mathbf{b}_j)\) goes in row \(i\), column \(j\), even if that value differs from \(B(\mathbf{b}_j,\mathbf{b}_i)\).

Common Pitfall: Reversing the Arguments

A common error is to calculate \([\mathbf{y}]_{\mathcal{B}}^{\top}\mathbf{M}_{\mathcal{B}}(B)[\mathbf{x}]_{\mathcal{B}}\) when the task asks for \(B(\mathbf{x},\mathbf{y})\). The formula places the first argument on the left and the second on the right. Reversing them computes \(B(\mathbf{y},\mathbf{x})\), which need not be the same number.

For the form and vectors in the first worked example, the correct result is \(B(\mathbf{x},\mathbf{y})=10\). Reversing the columns instead gives

The incorrect value \(-25\) is \(B(\mathbf{y},\mathbf{x})\), not \(B(\mathbf{x},\mathbf{y})\). The correction is to keep the arguments in their stated order: transpose the coordinate column of the first input, then multiply by the form’s matrix, then by the second input’s coordinate column. The matrix in this example is not symmetric, so the reversed calculation gives a different value.

It is useful to keep three pieces of information distinct: the bilinear form \(B\), the chosen ordered basis \(\mathcal{B}\), and the matrix \(\mathbf{M}_{\mathcal{B}}(B)\). The form is defined on vectors; the matrix stores its values on basis pairs; coordinate columns let us evaluate it. A new basis changes the coordinate description, a topic developed through congruence. For now, the essential rule is to identify the basis, place each basis-pair value in the correct row and column, and use the first coordinate column on the left.