The Unique Symmetric Matrix of a Quadratic Form

Why a Quadratic Form Does Not Determine Every Matrix Entry

A quadratic form records the value of an expression \(\mathbf{x}^{\top}\mathbf{A}\mathbf{x}\), not every entry of the matrix \(\mathbf{A}\) separately. In fact, different real matrices can give the same value for every vector \(\mathbf{x}\). Nevertheless, each real quadratic form has exactly one symmetric matrix that represents it. The key is that a quadratic expression combines the two entries on opposite sides of the diagonal: the entries \(a_{ij}\) and \(a_{ji}\) contribute together.

Throughout, \(\mathbf{x}\in\mathbb{R}^n\) is a column vector. A matrix \(\mathbf{S}\in\mathbb{R}^{n\times n}\) is symmetric if \(\mathbf{S}^{\top}=\mathbf{S}\), and a matrix \(\mathbf{K}\in\mathbb{R}^{n\times n}\) is skew-symmetric if \(\mathbf{K}^{\top}=-\mathbf{K}\). A real quadratic form is a function \(q:\mathbb{R}^n\to\mathbb{R}\) that is a homogeneous polynomial of degree two. Thus it can be written using terms \(x_i^2\) and \(x_ix_j\), with real coefficients.

Expanding the matrix product explains where the ambiguity occurs:

For each \(i<j\), the coefficient of \(x_ix_j\) is the sum \(a_{ij}+a_{ji}\). The quadratic form therefore cannot distinguish those entries individually. A symmetric representation assigns half of that combined coefficient to each of the two positions.

The Symmetric Part Preserves the Form

Every square real matrix can be separated into a symmetric part and a skew-symmetric part. The formula for the symmetric part is important because it works for any matrix that represents a quadratic form, even if the original matrix is not symmetric.

Proof. Taking transposes gives \(\mathbf{S}^{\top}=\tfrac12(\mathbf{A}^{\top}+\mathbf{A})=\mathbf{S}\), and \(\mathbf{K}^{\top}=\tfrac12(\mathbf{A}^{\top}-\mathbf{A})=-\mathbf{K}\). Adding their definitions gives \(\mathbf{S}+\mathbf{K}=\mathbf{A}\). It remains to show that \(\mathbf{x}^{\top}\mathbf{K}\mathbf{x}=0\). This expression is a scalar, so it equals its transpose. Therefore \((\mathbf{x}^{\top}\mathbf{K}\mathbf{x})^{\top}=\mathbf{x}^{\top}\mathbf{K}^{\top}\mathbf{x}=-\mathbf{x}^{\top}\mathbf{K}\mathbf{x}\). A real scalar equal to its own negative must be zero, so \(\mathbf{x}^{\top}\mathbf{K}\mathbf{x}=0\). Consequently \(\mathbf{x}^{\top}\mathbf{A}\mathbf{x}=\mathbf{x}^{\top}(\mathbf{S}+\mathbf{K})\mathbf{x}=\mathbf{x}^{\top}\mathbf{S}\mathbf{x}\) for every \(\mathbf{x}\). \(\square\)

This argument includes the case \(\mathbf{x}=\mathbf{0}\), for which both quadratic expressions are zero directly. It also shows precisely why the skew-symmetric part is invisible to a quadratic form: its contribution vanishes for every input vector, not merely for selected vectors.

Existence and Uniqueness of the Symmetric Representation

The symmetrization theorem shows that whenever a form is already expressed as \(\mathbf{x}^{\top}\mathbf{A}\mathbf{x}\), replacing \(\mathbf{A}\) by its symmetric part preserves its values. To establish the full result, we must also show that every quadratic form has a matrix representation and that no two different symmetric matrices can represent the same form.

Proof. Write the quadratic form by collecting its terms: \(q(\mathbf{x})=\sum_{i=1}^n c_{ii}x_i^2+\sum_{1\le i<j\le n}c_{ij}x_ix_j\). Define the entries of \(\mathbf{S}\) by \(s_{ii}=c_{ii}\) and \(s_{ij}=s_{ji}=c_{ij}/2\) when \(i<j\). These assignments make \(\mathbf{S}\) symmetric. In \(\mathbf{x}^{\top}\mathbf{S}\mathbf{x}\), the diagonal entries give \(\sum_i c_{ii}x_i^2\), and each pair of off-diagonal entries gives \(s_{ij}x_ix_j+s_{ji}x_jx_i=(c_{ij}/2)x_ix_j+(c_{ij}/2)x_ix_j=c_{ij}x_ix_j\). Thus \(\mathbf{x}^{\top}\mathbf{S}\mathbf{x}=q(\mathbf{x})\) for every \(\mathbf{x}\), proving existence.

For uniqueness, suppose symmetric matrices \(\mathbf{S}\) and \(\mathbf{T}\) represent the same form. Let \(\mathbf{e}_i\) be the vector with entry \(1\) in position \(i\) and \(0\) elsewhere. Evaluating at \(\mathbf{e}_i\) gives \(s_{ii}=q(\mathbf{e}_i)=t_{ii}\), so the diagonal entries agree. For \(i<j\), evaluate at \(\mathbf{e}_i+\mathbf{e}_j\). Symmetry gives \(q(\mathbf{e}_i+\mathbf{e}_j)=s_{ii}+2s_{ij}+s_{jj}\) and also \(q(\mathbf{e}_i+\mathbf{e}_j)=t_{ii}+2t_{ij}+t_{jj}\). Since the diagonal entries already agree, these equalities imply \(2s_{ij}=2t_{ij}\), hence \(s_{ij}=t_{ij}\). Symmetry then gives \(s_{ji}=t_{ji}\). Every entry agrees, so \(\mathbf{S}=\mathbf{T}\). This proves uniqueness. \(\square\)

The proof also gives a direct way to recover the entries of the unique symmetric matrix from values of \(q\): its diagonal entries are \(q(\mathbf{e}_i)\), and its off-diagonal entries are determined by the values at sums of two standard basis vectors.

Indeed, for \(i\ne j\), symmetry gives \(q(\mathbf{e}_i+\mathbf{e}_j)=s_{ii}+2s_{ij}+s_{jj}\), while \(q(\mathbf{e}_i)=s_{ii}\) and \(q(\mathbf{e}_j)=s_{jj}\). Subtracting the last two equations and dividing by \(2\) yields the formula. This is also the coordinate form of the polarization identity: the symmetric expression \(\tfrac12(q(\mathbf{x}+\mathbf{y})-q(\mathbf{x})-q(\mathbf{y}))\) equals \(\mathbf{x}^{\top}\mathbf{S}\mathbf{y}\). To check it, expand \(q(\mathbf{x}+\mathbf{y})=(\mathbf{x}+\mathbf{y})^{\top}\mathbf{S}(\mathbf{x}+\mathbf{y})\); the two pure terms cancel on subtraction, and symmetry makes the two remaining terms equal.

Different Matrices, the Same Quadratic Form

Uniqueness applies only among symmetric matrices. An arbitrary representing matrix need not be unique: adding any skew-symmetric matrix changes the matrix but leaves the form unchanged. Conversely, the symmetrization theorem shows that every matrix representing a given form has the same symmetric part, namely the unique symmetric matrix of that form.

Common Pitfall: Doubling a Cross-Term Coefficient

A frequent error is to copy the entire coefficient of a cross term into both symmetric positions. For \(q(\mathbf{x})=6x_1x_2\), that mistake produces \(\begin{bmatrix}0&6\\6&0\end{bmatrix}\). But multiplying out gives \(6x_1x_2+6x_2x_1=12x_1x_2\), which is twice the required form. The correction is to split the coefficient: the unique symmetric matrix is \(\begin{bmatrix}0&3\\3&0\end{bmatrix}\), and its two cross-term contributions add to \(3x_1x_2+3x_2x_1=6x_1x_2\).

The rule is always to halve each cross-term coefficient when assigning it to a symmetric pair of entries. Diagonal coefficients are not halved: \(x_i^2\) comes from the single diagonal entry \(s_{ii}\), not from a pair of entries.