Quadratic Forms

Similar to endomorphisms, we can restrict bilinear forms on a single vector space and choose the same basis for both sides.

Proposition. Suppose that $V$ is a finite-dimensional vector space over $\mathbf{F}$ and $\phi: V \times V \to \mathbf{F}$ is a bilinear form and $(e_1, …, e_n)$ and $(f_1, …, f_n)$ are two bases of $V$ such that $f_i = \sum P_{ki} e_k$. If $A$ represents $\phi$ with respect to $(e_i)$ and $B$ represents $\phi$ with respect to $(f_i)$ then

\[B = P^\intercal AP\]

Definition. The square matrices $A$ and $B$ are congruent if there is an invertible matrix $P$ such that $B = P^\intercal AP$.

Definition. A bilinear form $\phi: V \times V \to \mathbf{F}$ is symmetric if $\phi(v_1, v_2) = \phi(v_2, v_1)$ for all $v_1, v_2 \in V$.

Proposition. Suppose that $\phi: V \times V \to \mathbf{F}$ is a bilinear form and $(e_1, …, e_n)$ is a basis for $V$ and $M$ is the matrix representing $\phi$ with respect to this basis. Then $\phi$ is symmetric iff $M$ is symmetric, i.e. $M^\intercal = M$.

Proof.

($\Rightarrow$) If $\phi$ is symmetric then $M_{ij} = \phi(e_i, e_j) = \phi(e_j, e_i) = M_{ji}$ so $M$ is symmetric.

($\Leftarrow$) If $M$ is symmetric then

\[\phi(x, y) = \sum_{i,j=1}^n x_i M_{ij} y_j = \sum_{i,j=1}^n j_i M_{ji} x_i = \phi(y, x)\]

Definition. A quadratic form on $V$ is the map $q: V \to \mathbf{F}, q(v) = \phi(v, v)$ where $\phi: V \times V \to \mathbf{F}$ is a bilinear form.

For example, if $V = \mathbf{R^2}$ and $\phi$ is represented by $A$ with respect to the standard basis then

\[q(v) = \begin{pmatrix} x & y \end{pmatrix} A \begin{pmatrix} x \\ y \end{pmatrix} = A_{11} x^2 + (A_{12} + A_{21}) xy + A_{22} y^2\]

Note that quadratic form is not linear.

Proposition. [Polarization Identity] If $q: V \to \mathbf{F}$ is a quadratic form then there exists a unique symmetric bilinear form $\phi: V \times V \to \mathbf{F}$ such that $q(v) = \phi(v, v)$ for all $v \in V$.

Proof.

Let $\psi: V \times V \to \mathbf{F}$ be the bilinear form such that $\phi(v, v) = q(v)$. Then

\[\phi(v, w) = {1 \over 2} (\psi(v, w) + \psi(w, v))\]

is a symmetric bilinear form such that $\phi(v, v) = q(v)$.

Suppose that $\phi$ is symmetric bilinear form such that $\phi(v, v) = q(v)$, then

\[q(v + w) = \phi(v + w, v + w) = \phi(v, v + w) + \phi(w, v + w) = \phi(v, v) + 2 \phi(v, w) + \phi(w, w)\]

and hence

\[\phi(v, w) = {1 \over 2} (q(v + w) - q(v) - q(w))\]

is unique.

To conclude, there is a one-to-one mapping between symmetric bilinear form and quadratic form in which the above formula can be used to determine the matrix of the corresponding bilinear form given a quadratic form.

Matrix Representation

Proposition. [Diagonalization of symmetric bilinear forms] If $\phi: V \times V \to \mathbf{F}$ is a symmetric bilinear form of a finite dimensional vector space $V$ then there is a basis $(e_1, …, e_n)$ for $V$ such that $\phi$ is represented by a diagonal matrix.

Proof.

By induction on $n = \dim V$. When $n = 0, 1$, every matrix is diagonal matrix. Assume it is true for all spaces of dimension strictly smaller than $n$. If $\phi(v, v) = 0$ for all $v \in V$ then by polarization identity $\phi(v, w) = 0$ so $\phi$ is represented by the zero matrix when is diagonal. Otherwise, we can choose $e_1$ such that $\phi(e_1, e_1) \not= 0$. Consider the subspace

\[U = \Set{ u \in V : \phi(e_1, u) = 0 } = \ker \phi(e_1, -): V \to \mathbf{F}\]

Since the codomain $\mathbf{F}$ is one-dimensional, by rank-nullity theorem, $\dim U = n - 1$. Also, $e_1 \not= U$ so $V = \langle e_1 \rangle \oplus U$.

The bilinear map $\phi_{U \times U} : U \times U \to \mathbf{F}$ is also symmetric so by the induction hypothesis, there exists basis $\Set{e_2, …, e_n}$ such that the matrix representing $\phi_{U \times U}$ is diagonal. Hence, the basis $\Set{e_1, …, e_n}$ satisfies $\phi(e_i, e_j) = 0$ for $i \not= j$ as required.

Corollary. Let $\phi$ be a symmetric bilinear form on a finite dimensional $\mathbf{C}$-vector space $V$. Then there is basis $(v_1, …, v_n)$ for $V$ such that $\phi$ is represented by a matrix of the form

\[\begin{pmatrix} I_r & 0 \\ 0 & 0 \\ \end{pmatrix}\]

with $r = r(\phi)$ or equivalently such that the corresponding quadratic form $q$ is given by

\[q(v) = q \left( \sum_{i=1}^n a_i v_i \right) = \sum_{i=1}^r a_i^2\]

Proof.

From the above, there is basis $(e_1, …, e_n)$ such that $\phi(e_i, e_j) = \delta_{ij} \lambda_i$ for $\lambda_i \in \mathbf{C}$. By reordering the $e_i$’s we can assume

\[\begin{cases} \lambda_i \not= 0 &\qquad 1 \le i \le r \\ \lambda_i = 0 &\qquad i > r \\ \end{cases}\]

Since we are working with $\mathbf{C}$, we can define

\[v_i = \begin{cases} {1 \over \sqrt{\lambda_i}} e_i &\qquad 1 \le i \le r \\ e_i &\qquad i > r \\ \end{cases}\]

so that $\phi(v_i, v_j) = 0$ for $i \not= j$ or $i = j > r$ and $\phi(v_i, v_i) = 1$ for $1 \le i \le r$ as required.

Corollary. Every symmetric matrix $S \in \text{Mat}_n(\mathbf{C})$ is congruent to a unique matrix of the form

\[\begin{pmatrix} I_r & 0 \\ 0 & 0 \\ \end{pmatrix}\]

Corollary. Let $\phi$ be a symmetric bilinear form on a finite dimensional $\mathbf{R}$-vector space $V$. Then there is basis $(v_1, …, v_n)$ for $V$ such that $\phi$ is represented by a matrix of the form

\[\begin{pmatrix} I_p & 0 & 0 \\ 0 & -I_q & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}\]

with $p, q \ge 0$ and $p + q = r(\phi)$ or equivalently such that the corresponding quadratic form $q$ is given by

\[q(v) = q \left( \sum_{i=1}^n a_i v_i \right) = \sum_{i=1}^p a_i^2 - \sum_{i=p+1}^{p+q} a_i^2\]

Proof.

Similarily, there is basis $(e_1, …, e_n)$ such that $\phi(e_i, e_j) = \delta_{ij} \lambda_i$ for $\lambda_i \in \mathbf{R}$. By reordering the $e_i$’s we can assume

\[\begin{cases} \lambda_i > 0 &\qquad 1 \le i \le p \\ \lambda_i < 0 &\qquad p+1 < i < p+q = r(\phi) \\ \lambda_i = 0 &\qquad i > p+q \\ \end{cases}\]

for some $p, q \ge 0$. Since we are working with $\mathbf{R}$, we can define

\[v_i = \begin{cases} {1 \over \sqrt{\lambda_i}} e_i &\qquad 1 \le i \le p \\ {1 \over \sqrt{-\lambda_i}} e_i &\qquad p+1 < i < p+q \\ e_i &\qquad i > p+q \\ \end{cases}\]

so that the matrix representing $\phi$ with respect to $(v_1, …, v_n)$ is of the required form.

Definition. A symmetric bilinear form $\phi$ on a $\mathbf{R}$-vector space $V$ is

  • positive definite if $\phi(v, v) > 0$ for all $v \in V \setminus 0$;

  • positive semi-definite if $\phi(v, v) \ge 0$ for all $v \in V$;

  • negative definite if $\phi(v, v) < 0$ for all $v \in V \setminus 0$;

  • negative semi-definite if $\phi(v, v) \le 0$ for all $v \in V$.

From the above we can see that it depends on the value of $p$ and $q$, e.g. $\phi$ is positive definite if $p = n$ and $q = 0$.

Theorem. [Sylvester’s Law of Inertia] Let $\phi$ be a symmetric bilinear form on a finite dimensional $\mathbf{R}$-vector space $V$. Then there are unique integers $p, q$ such that $V$ has basis $(v_1, …, v_n)$ such that the matrix representing $\phi$ with respect to it is of the form

\[\begin{pmatrix} I_p & 0 & 0 \\ 0 & -I_q & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}\]

Proof.

We have proved the existence of such a basis and $p + q = r(\phi)$ is unique. Suppose that we have found a certain $p$ such that the matrix of the corresponding basis $(v_1, …, v_n)$ is of the form shown. Then $\phi$ is positive definite on the space spanned by $v_1, …, v_p$.

Let $P$ be any subspace of $V$ such that $\phi_{P \times P}$ is positive definite and let $Q = \langle v_{p+1}, … v_n \rangle$. $\phi_{Q \times Q}$ is negative semi-definite so $P \cap Q = 0$ and therefore $\dim P + \dim Q = \dim(P + Q) \le n$. Thus, $\dim P \le p$. It means that if both $\phi_{P \times P}$ and $\phi_{P’ \times P’}$ are positive definite, then we have $p \le p’$ and $p’ \le p$ so $p = p’$ is unique.

Definition. The signature of the symmetric bilinear form is defined to be $p - q$.

Corollary. Every symmetric matrix $S \in \text{Mat}_n(\mathbf{R})$ is congruent to a unique matrix of the form

\[\begin{pmatrix} I_p & 0 & 0 \\ 0 & -I_q & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}\]

To conclude, congruence of real matrices preserves the signature, i.e. the number of positive, negative and zero eigenvalues is constant. One way we can determine the congruence of symmetric matrices by computing their eigenvalues. If we know the quadratic form we can find the completing the squares, i.e. making it of the form

\[q(v) = a_1 y_1^2 + a_2 y_2^2 + ... + a_n y_n^2\]

where each $y_i$ is in terms of the original $x_i$. This process is precisely changing basis and the $a_i$ are the entries in the resulting diagonal matrix. Therefore, we can find the rank ($p + q$) and signature ($p - q$) by counting the number of postiive, negative and zero coefficients.

Reference