From Two Inputs to One
A bilinear form takes two vectors as inputs and returns a scalar. A quadratic form takes one vector as input and returns a scalar. The connection is to put the same vector into both inputs: from a bilinear form \(B\), define \(q(\mathbf{x})=B(\mathbf{x},\mathbf{x})\). When \(B\) is symmetric over \(\mathbb{R}\), the values of \(q\) retain enough information to recover every value of \(B\). The identity that performs this recovery is called polarization.
Throughout this tutorial, the scalar field is \(\mathbb{R}\). Let \(V\) be a finite-dimensional real vector space. A function \(B:V\times V\to\mathbb{R}\) is a bilinear form if it is linear in each input separately: for all \(\mathbf{x},\mathbf{y},\mathbf{z}\in V\) and \(a,b\in\mathbb{R}\), it satisfies \(B(a\mathbf{x}+b\mathbf{y},\mathbf{z})=aB(\mathbf{x},\mathbf{z})+bB(\mathbf{y},\mathbf{z})\) and \(B(\mathbf{x},a\mathbf{y}+b\mathbf{z})=aB(\mathbf{x},\mathbf{y})+bB(\mathbf{x},\mathbf{z})\). It is symmetric if \(B(\mathbf{x},\mathbf{y})=B(\mathbf{y},\mathbf{x})\) for every pair of vectors.
The definition makes \(q(\mathbf{0})=0\), and \(q(a\mathbf{x})=a^2q(\mathbf{x})\). These properties alone are not being used as the definition: the requirement that \(q\) be a homogeneous quadratic polynomial ensures that it has the coordinate structure we will use below.
Matrix Representations
Choose an ordered basis \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) of \(V\). For coordinate columns \(\mathbf{x}=[\mathbf{v}]_{\mathcal{B}}\) and \(\mathbf{y}=[\mathbf{w}]_{\mathcal{B}}\), the matrix of a bilinear form in this basis has entries \(a_{ij}=B(\mathbf{b}_i,\mathbf{b}_j)\). Bilinearity then gives \(B(\mathbf{v},\mathbf{w})=\mathbf{x}^{\top}\mathbf{A}\mathbf{y}\). The matrix is symmetric exactly when the form is symmetric, because \(a_{ij}=B(\mathbf{b}_i,\mathbf{b}_j)=B(\mathbf{b}_j,\mathbf{b}_i)=a_{ji}\).
Proof. Write \(\mathbf{v}=\sum_{i=1}^n x_i\mathbf{b}_i\) and \(\mathbf{w}=\sum_{j=1}^n y_j\mathbf{b}_j\). Applying linearity in each input gives \(B(\mathbf{v},\mathbf{w})=\sum_{i=1}^n\sum_{j=1}^n x_i y_j B(\mathbf{b}_i,\mathbf{b}_j)=\mathbf{x}^{\top}\mathbf{A}\mathbf{y}\). Symmetry of \(B\) gives \(a_{ij}=a_{ji}\), so \(\mathbf{A}^{\top}=\mathbf{A}\). Setting \(\mathbf{w}=\mathbf{v}\) gives the stated formula for \(q\). In the converse direction, \(\mathbf{x}^{\top}\mathbf{A}\mathbf{y}\) is linear in \(\mathbf{x}\) and in \(\mathbf{y}\), so it defines a bilinear form. If \(\mathbf{A}^{\top}=\mathbf{A}\), then \(\mathbf{x}^{\top}\mathbf{A}\mathbf{y}=(\mathbf{x}^{\top}\mathbf{A}\mathbf{y})^{\top}=\mathbf{y}^{\top}\mathbf{A}^{\top}\mathbf{x}=\mathbf{y}^{\top}\mathbf{A}\mathbf{x}\), proving symmetry. \(\square\)
For a symmetric matrix, the diagonal entries contribute square terms, while each off-diagonal entry contributes twice to the corresponding mixed term. Indeed, if \(\mathbf{A}=(a_{ij})\), then
This factor of \(2\) is central when converting a polynomial into a matrix.
Worked Example: Read the Quadratic Form from a Matrix
Let \(B(\mathbf{x},\mathbf{y})=\mathbf{x}^{\top}\mathbf{A}\mathbf{y}\) on \(\mathbb{R}^2\), where \(\mathbf{A}=\begin{bmatrix}2&3\\3&-1\end{bmatrix}\). Find the associated quadratic form and evaluate both \(B\) and its polarization for \(\mathbf{x}=(1,2)^{\top}\) and \(\mathbf{y}=(3,-1)^{\top}\).
The matrix is symmetric. For a general coordinate column \((x_1,x_2)^{\top}\), its quadratic form is
For the specified vectors, first calculate \(\mathbf{A}\mathbf{y}=(3,10)^{\top}\). Thus \(B(\mathbf{x},\mathbf{y})=(1,2)(3,10)^{\top}=3+20=23\). Now \(q(\mathbf{x})=2+12-4=10\), \(q(\mathbf{y})=18-18-1=-1\), and \(\mathbf{x}+\mathbf{y}=(4,1)^{\top}\), so \(q(\mathbf{x}+\mathbf{y})=32+24-1=55\). Polarization gives \(\tfrac12(55-10-(-1))=23\), matching the direct calculation.
Polarization Recovers the Form
Proof. Bilinearity gives \(q(\mathbf{x}+\mathbf{y})=B(\mathbf{x}+\mathbf{y},\mathbf{x}+\mathbf{y})=B(\mathbf{x},\mathbf{x})+B(\mathbf{x},\mathbf{y})+B(\mathbf{y},\mathbf{x})+B(\mathbf{y},\mathbf{y})\). Symmetry makes the two middle terms equal. Subtracting \(q(\mathbf{x})=B(\mathbf{x},\mathbf{x})\) and \(q(\mathbf{y})=B(\mathbf{y},\mathbf{y})\) leaves \(2B(\mathbf{x},\mathbf{y})\). Division by \(2\) is valid over \(\mathbb{R}\), proving the identity. If two symmetric bilinear forms give the same \(q\), the right-hand side has the same value for both at every \(\mathbf{x},\mathbf{y}\); hence the forms are equal. \(\square\)
The identity is also useful in the equivalent form \(B(\mathbf{x},\mathbf{y})=\tfrac14(q(\mathbf{x}+\mathbf{y})-q(\mathbf{x}-\mathbf{y}))\). To obtain it, expand both quadratic-form values: symmetry and bilinearity give \(q(\mathbf{x}+\mathbf{y})=q(\mathbf{x})+2B(\mathbf{x},\mathbf{y})+q(\mathbf{y})\) and \(q(\mathbf{x}-\mathbf{y})=q(\mathbf{x})-2B(\mathbf{x},\mathbf{y})+q(\mathbf{y})\). Subtracting yields \(4B(\mathbf{x},\mathbf{y})\).
There is also a converse: every quadratic form as defined above comes from exactly one symmetric bilinear form. To see existence in coordinates, write its polynomial as \(q(\mathbf{x})=\sum_i d_i x_i^2+\sum_{i<j}c_{ij}x_ix_j\). Form the symmetric matrix with diagonal entries \(d_i\) and entries \(c_{ij}/2\) in positions \((i,j)\) and \((j,i)\). Its product \(\mathbf{x}^{\top}\mathbf{A}\mathbf{x}\) is exactly the given polynomial. The matrix defines a symmetric bilinear form by the preceding theorem, and uniqueness follows from polarization. Thus over \(\mathbb{R}\), symmetric bilinear forms and quadratic forms determine one another.
Worked Example: Build the Matrix from a Polynomial
Find the symmetric matrix and associated bilinear form for \(q(x_1,x_2)=4x_1^2-10x_1x_2+3x_2^2\). Then use polarization to evaluate the form at \(\mathbf{x}=(1,2)^{\top}\) and \(\mathbf{y}=(-2,1)^{\top}\).
The diagonal entries are the coefficients of \(x_1^2\) and \(x_2^2\), namely \(4\) and \(3\). The mixed term has coefficient \(-10\), so each symmetric off-diagonal entry must be \(-5\). Therefore
For a direct check, \(\mathbf{A}\mathbf{y}=(-13,13)^{\top}\), so \(B(\mathbf{x},\mathbf{y})=(1,2)(-13,13)^{\top}=-13+26=13\). The quadratic values are \(q(\mathbf{x})=4-20+12=-4\) and \(q(\mathbf{y})=16+20+3=39\). Since \(\mathbf{x}+\mathbf{y}=(-1,3)^{\top}\), \(q(\mathbf{x}+\mathbf{y})=4+30+27=61\). Polarization gives \(\tfrac12(61-(-4)-39)=13\), in agreement with the matrix calculation.
A Three-Variable Check
Worked Example: Compute Both Representations in R³
Let the symmetric matrix and form be given by
Expanding \(\mathbf{x}^{\top}\mathbf{A}\mathbf{x}\) gives \(q(\mathbf{x})=x_1^2+4x_1x_2-2x_1x_3+6x_2x_3+4x_3^2\). Each mixed coefficient is twice the corresponding off-diagonal entry. For a numerical check, take \(\mathbf{u}=(1,0,1)^{\top}\) and \(\mathbf{v}=(0,1,-1)^{\top}\). Multiplication gives \(\mathbf{A}\mathbf{v}=(3,-3,-1)^{\top}\), so \(B(\mathbf{u},\mathbf{v})=(1,0,1)(3,-3,-1)^{\top}=2\).
Evaluate the quadratic form at \(\mathbf{u}\): \(q(\mathbf{u})=1+4-2=3\). At \(\mathbf{v}\), \(q(\mathbf{v})=4-6=-2\). Their sum is \(\mathbf{u}+\mathbf{v}=(1,1,0)^{\top}\), for which \(q(\mathbf{u}+\mathbf{v})=1+4=5\). Therefore \(\tfrac12(5-3-(-2))=2\), which verifies the direct bilinear-form calculation.
Common Pitfall: The Mixed Coefficient
A frequent error is to copy the coefficient of a mixed term into both off-diagonal matrix positions. For example, for \(q(x_1,x_2)=4x_1^2-10x_1x_2+3x_2^2\), the incorrect matrix \(\begin{bmatrix}4&-10\\-10&3\end{bmatrix}\) gives \(4x_1^2-20x_1x_2+3x_2^2\), not the stated polynomial. The two off-diagonal entries both contribute, so the corrected matrix has \(-5\) in each position and produces \(-10x_1x_2\).
Another distinction is that a quadratic form records only the symmetric part of a matrix. For any real matrix \(\mathbf{M}\), the scalar \(\mathbf{x}^{\top}\mathbf{M}\mathbf{x}\) equals \(\mathbf{x}^{\top}\bigl(\tfrac12(\mathbf{M}+\mathbf{M}^{\top})\bigr)\mathbf{x}\): the skew-symmetric part contributes zero, since its transpose changes the sign of this scalar. The bilinear form \(\mathbf{x}^{\top}\mathbf{M}\mathbf{y}\), however, can depend on that skew-symmetric part. Thus the quadratic form uniquely recovers a symmetric bilinear form, not an arbitrary bilinear form.
These representations are useful because they let a calculation move between two descriptions: the matrix makes evaluation of \(B(\mathbf{x},\mathbf{y})\) direct, while the polynomial makes the dependence of \(q\) on the coordinates visible. Polarization guarantees that, for symmetric forms over \(\mathbb{R}\), no information about the bilinear form is lost by passing to its quadratic form.
Check Your Understanding
For each question, keep track of whether you are working with the bilinear form or with its associated quadratic form.
- Find the quadratic form associated with the symmetric matrix \(\begin{bmatrix}1&-2\\-2&5\end{bmatrix}\).
- Write the symmetric matrix for \(q(x_1,x_2)=7x_1^2+8x_1x_2-2x_2^2\).
- Use polarization to compute \(B(\mathbf{x},\mathbf{y})\) from \(q\), given \(q(\mathbf{x}+\mathbf{y})=12\), \(q(\mathbf{x})=5\), and \(q(\mathbf{y})=3\).
- Explain why a symmetric matrix for \(q(x_1,x_2)=x_1^2+6x_1x_2+2x_2^2\) must have \(3\), not \(6\), in each off-diagonal position.
- State the hypothesis that permits division by \(2\) in the polarization identity.
