Bilinear Forms and Congruence

Definition Let $V$ be a vector space over $\mathbb{R}$. A bilinear form on $V$ is a function $b \colon V \times V \to \mathbb{R}$ that is linear in each argument when the other is held fixed. For all vectors and all scalars $a, c \in \mathbb{R}$,

The form is symmetric if $b(x, y) = b(y, x)$ for all $x, y \in V$, and skew-symmetric if $b(x, y) = -b(y, x)$ for all $x, y \in V$. For each fixed $x$ the map $y \mapsto b(x, y)$ is a linear functional, an element of $V^{\vee}$. The same definition makes sense over $\mathbb{C}$, but a complex inner product is conjugate-linear in its first argument, so on a nonzero space it is not a bilinear form.

Definition Let $V$ be a real vector space of finite dimension $n$, let $\mathcal{B} = (v_1, \dots, v_n)$ be an ordered basis and let $b$ be a bilinear form on $V$. The matrix of $b$ in $\mathcal{B}$ is $M_{\mathcal{B}} \in \mathbb{R}^{n \times n}$ with entries $m_{ij} = b(v_i, v_j)$. If $\xi = [x]_{\mathcal{B}}$ and $\eta = [y]_{\mathcal{B}}$, expanding $x = \sum_i \xi_i v_i$ in the first slot and $y = \sum_j \eta_j v_j$ in the second gives

So the values on pairs of basis vectors determine $b$, and $M_{\mathcal{B}}$ is the only matrix with this property. The form is symmetric exactly when $M_{\mathcal{B}}^{\mathsf{T}} = M_{\mathcal{B}}$. A vector $y$ satisfies $b(x, y) = 0$ for every $x$ exactly when $M_{\mathcal{B}}\eta = 0$, and a vector $x$ satisfies $b(x, y) = 0$ for every $y$ exactly when $M_{\mathcal{B}}^{\mathsf{T}}\xi = 0$. The form is nondegenerate when neither kind of nonzero vector exists, which for the square matrix $M_{\mathcal{B}}$ happens exactly when it is invertible.

Definition Let $b$ be a bilinear form on a real $n$-dimensional space $V$ with ordered bases $\mathcal{B}$ and $\mathcal{C}$. Let $S = P_{\mathcal{B} \leftarrow \mathcal{C}}$, the matrix whose columns are the $\mathcal{C}$ vectors in $\mathcal{B}$ coordinates, so that $[x]_{\mathcal{B}} = S[x]_{\mathcal{C}}$. Substituting into both slots with $\xi = [x]_{\mathcal{C}}$ and $\eta = [y]_{\mathcal{C}}$ gives

so

With $P = P_{\mathcal{C} \leftarrow \mathcal{B}} = S^{-1}$ the same rule reads $M_{\mathcal{C}} = (P^{-1})^{\mathsf{T}} M_{\mathcal{B}}\, P^{-1}$. Real matrices $M, N \in \mathbb{R}^{n \times n}$ are congruent if $N = S^{\mathsf{T}} M S$ for some invertible $S \in \mathbb{R}^{n \times n}$, so congruent matrices are exactly the matrices of one bilinear form in two bases. Congruence preserves symmetry, rank and nondegeneracy, and

so the sign of the determinant survives but its value need not. The eigenvalues and the trace can change. When $S$ is orthogonal, $S^{\mathsf{T}} = S^{-1}$ and congruence gives the same matrix as similarity.

Definition The quadratic form of a bilinear form $b$ on a real space $V$ is $q(v) = b(v, v)$. A function $q \colon V \to \mathbb{R}$ is called a quadratic form if it arises this way from some bilinear form. When $V$ is finite-dimensional with ordered basis $\mathcal{B}$ and $\xi = [v]_{\mathcal{B}}$,

and $q(cv) = c^2 q(v)$ for every scalar $c$. When $\dim V \ge 2$, different bilinear forms can have the same quadratic form, because for $i \ne j$ only the sum $m_{ij} + m_{ji}$ affects the coefficient of $\xi_i \xi_j$. Exactly one of them is symmetric, the one whose matrix splits each cross coefficient equally between $m_{ij}$ and $m_{ji}$.

Definition Let $b$ be a symmetric bilinear form on a real vector space $V$ with quadratic form $q$. Then for all $x, y \in V$

To prove it, expand by bilinearity,

and use symmetry to combine the two middle terms. Hence a symmetric bilinear form is determined by its quadratic form, and symmetric bilinear forms on $V$ correspond one to one with quadratic forms on $V$. Without symmetry the right side of the identity equals $\tfrac{1}{2}\bigl(b(x, y) + b(y, x)\bigr)$, the symmetric part of $b$, not $b$ itself.

Definition Every $M \in \mathbb{R}^{n \times n}$ splits uniquely as $M = H + K$ with $H^{\mathsf{T}} = H$ and $K^{\mathsf{T}} = -K$, namely

The split is unique because a matrix that is both symmetric and skew-symmetric is zero. A bilinear form splits the same way into a symmetric part and a skew-symmetric part. For every real column $x$ the scalar $x^{\mathsf{T}}Kx$ equals its own transpose $x^{\mathsf{T}}K^{\mathsf{T}}x = -x^{\mathsf{T}}Kx$, so it is $0$ and

A real bilinear form satisfies $b(v, v) = 0$ for every $v$ exactly when it is skew-symmetric, and such a form is called alternating. One direction is the computation above, and for the other, expanding $b(x + y, x + y) = 0$ gives $b(x, y) = -b(y, x)$.