Basic algebraic concepts

Definition 2.1   Let $V$ be a vector space over a field ${\Bbb F}$. A function

$$B : V \times V \longrightarrow {\Bbb F}$$

is said to be a bilinear form if

$$B(\lambda v + v' , w) = \lambda B(v,w) + B(v', w)$$

$$B(v, \lambda w + w') = \lambda B(v, w) + B(v, w')$$

for any vectors $v,v',w,w' \in V$ and scalar $\lambda \in {\Bbb F}$.

Definition 2.2   A bilinear form $B$ is said to be symmetric (resp. skew-symmetric) if $B(v,w) = B(w,v)$ (resp. $B(v,w) = -B(w,v)$) for any vectors $v,w \in V$. The kernel of a bilinear form $B$ is the set of vectors $v$ for which $B(v,w) = 0$ for all $w \in V$. The dimension of the kernel is called the nullity of $B$. If the nullity of a bilinear form is zero, the form is said to be nondegenerate.

Exercise 2.1 (10)   Let $W$ be a vector space over ${\Bbb F}$, let $W^*$ be its dual, and let $V$ denote the product $W \times W^*$. Show that the function

$$\omega : V \times V \longrightarrow {\Bbb F}$$

defined by $\omega((v,f),(v',f')) := f'(v) - f(v')$ is a skew-symmetric, nondegenerate bilinear form. This is the canonical symplectic form on $V$.

Exercise 2.2 (*05)   Let $V$ be a vector space of dimension $n$ over a field with $p$ elements. How many different bilinear forms are there on $V$?

Exercise 2.3 (10)   Let $V$ be an $n$-dimensional vector space over a field ${\Bbb F}$. Show that if we fix a basis of $V$, bilinear forms are in one-to-one correspondence with $n \times n$ matrices with entries in ${\Bbb F}$. Moreover, the form is symmetric if and only if the matrix is symmetric.

Definition 2.3   Let $V$ be a vector space over a field ${\Bbb F}$. A quadratic form on $V$ is a function $Q : V \rightarrow {\Bbb F}$ such that there exists a bilinear form $B$ with $Q(v) = B(v,v)$ for all $v \in V$

Exercise 2.4 (15)   Show that if $Q$ is a quadratic form, there exists a unique symmetric bilinear form $B$ such that $Q(v) = B(v,v)$ for all $v \in V$. Give a formula for $B$ in terms of $Q$.

Definition 2.4   A quadratic form $Q$ on ${\Bbb R}^n$ is said to be positive definite (resp. negative definite) if $Q(v) > 0$ (resp. $Q(v) < 0$ ) for all nonzero vectors $v \in V$.

Remark. If you did the previous exercises, you know that a quadratic form on ${\Bbb R}^n$ is of the form

$$Q(x_{1},\dots,x_{n}) := \begin{pmatrix} x_{1} & \cdots & x_{n} \e... ...n{pmatrix} x_{1} \\ \cdot \\ \cdot \\ \cdot \\ x_{n} \end{pmatrix} \ ,$$

with $a_{ij} = a_{ji}$.

Let $V$ be a vector space and let $P(V)$ be its projectivization. If $Q$ is a quadratic form on $V$ and $v$ is a nonzero vector such that $Q(v) = 0$, then $Q(\lambda v) = 0$ for any scalar $\lambda$. This suggests the following definition:

Definition 2.5   Let $V$ be a vector space and let $P(V)$ be its projectivization. If $Q$ is a quadratic form on $V$, the projective quadric associated to $Q$ is the set

$${\cal Q} := \{[v] \in P(V) : v \in V \setminus \{0\}, Q(v) = 0 \} .$$

A projective quadric on ${\Bbb R}P^2$ is called a conic. If the symmetric bilinear form associated to $Q$ is nondegenerate, we shall say that ${\cal Q}$ is nondegenerate or proper.

It may happen that the projective quadric associated to a quadratic form be empty. For example if $Q(x_{1},x_{2},x_{3}) := x_{1}^2 + x_{2}^2 + x_{3}^2$, then ${\cal Q} \subset {\Bbb R}P^2$ is the empty set.

We end the section by proving the following important result:

Theorem 2.1   If ${\cal Q} \subset {\Bbb R}P^n$ is a nondegenerate quadric, the set of all hyperplanes tangent to ${\cal Q}$ is a nondegenerate quadric in the dual space ${\Bbb R}P^{n*}$.

Definition 2.6   The projective quadric in ${\Bbb R}P^{n*}$ formed by the set of hyperplanes tangent to a nondegenderate quadric ${\cal Q} \subset {\Bbb R}P^n$ is called the dual quadric of ${\cal Q}$ and will be denoted by ${\cal Q}^*$.

Let $Q : {\Bbb R}^{n+1} \rightarrow {\Bbb R}$ be the nondegenerate quadratic form associated to ${\cal Q}$ and let $B$ be the symmetric bilinear form satisfying $Q(v) = B(v,v)$. Define the map

$$dQ : {\Bbb R}^{n+1} \rightarrow {\Bbb R}^{n+1 *}$$

by the equation $dQ(v)(w) = 2B(v,w)$.

Exercise 2.5 (00)   Show that $dQ$ is an invertible linear map.

Exercise 2.6 (10)   Let $v \in {\Bbb R}^{n+1}$ be a nonzero vector lying on the cone $Q = 0$. Show that

$$\Pi_{v} := \{w \in {\Bbb R}^{n+1}: dQ(v)(w) = 0 \}$$

is a hyperplane in ${\Bbb R}^{n+1}$ that passes through the origin and is tangent to the cone $Q = 0$ at the point $v$. Show that if $\lambda$ is a nonzero scalar, $\Pi_{\lambda v} = \Pi_{v}$.

Exercise 2.7 (05)   Prove that the dual quadric ${\cal Q}^*$ is the image of ${\cal Q}$ under the projective transformation induced by $dQ$. Use this to prove theorem 2.1.

Exercise 2.8 (10)   Study the duality of degenerate quadrics.