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**Definition 2.1**
Let

be a vector space over a field

. A function

is said to be a

*bilinear form* if

for any vectors

and scalar

.

**Definition 2.2**
A bilinear form

is said to be

*symmetric*
(resp.

*skew-symmetric*) if

(resp.

) for any vectors

.
The

*kernel* of a bilinear form

is the set of vectors

for which

for all

. The dimension of the kernel
is called the

*nullity* of

. If the nullity of a bilinear form
is zero, the form is said to be

*nondegenerate.*

**Exercise 2.1** (10)
Let

be a vector space over

, let

be its dual, and
let

denote the product

. Show that the function

defined by

is a skew-symmetric,
nondegenerate bilinear form. This is the

*canonical symplectic form* on

.

**Exercise 2.2** (*05)
Let

be a vector space of dimension

over a field with

elements. How many different bilinear forms are there on

?

**Exercise 2.3** (10)
Let

be an

-dimensional vector space over a field

. Show that if
we fix a basis of

, bilinear forms are in one-to-one correspondence with

matrices with entries in

. Moreover, the form is symmetric
if and only if the matrix is symmetric.

**Definition 2.3**
Let

be a vector space over a field

. A

*quadratic form* on

is a function

such that there exists a
bilinear form

with

for all

**Exercise 2.4** (15)
Show that if

is a quadratic form, there exists a unique symmetric
bilinear form

such that

for all

. Give
a formula for

in terms of

.

**Definition 2.4**
A quadratic form

on

is said to be

*positive definite*
(resp.

*negative definite*) if

(resp.

)
for all nonzero vectors

.

*Remark.*
If you did the previous exercises, you know that a quadratic form on
is of the form

with
.

Let be a vector space and let be its projectivization. If
is a quadratic form on and is a nonzero vector such that
, then
for any scalar . This
suggests the following definition:

**Definition 2.5**
Let

be a vector space and let

be its projectivization. If

is a quadratic form on

, the

*projective quadric*
associated to

is the set

A projective quadric on

is called a

*conic.*
If the symmetric bilinear form associated to

is nondegenerate, we
shall say that

is

*nondegenerate* or

*proper.*
It may happen that the projective quadric associated to a quadratic form
be empty. For example if
, then
is the empty set.

We end the section by proving the following important result:

**Theorem 2.1**
If

is a nondegenerate quadric,
the set of all hyperplanes tangent to

is a
nondegenerate quadric in the dual space

.

**Definition 2.6**
The projective quadric in

formed by the set of hyperplanes
tangent to a nondegenderate quadric

is called
the

*dual quadric* of

and will be denoted by

.

Let
be the nondegenerate quadratic
form associated to and let be the symmetric bilinear
form satisfying
. Define the map

by the equation
.

**Exercise 2.5** (00)
Show that

is an invertible linear map.

**Exercise 2.6** (10)
Let

be a nonzero vector lying on the cone

.
Show that

is a hyperplane in

that passes through the origin and is tangent
to the cone

at the point

. Show that if

is a nonzero
scalar,

.

**Exercise 2.7** (05)
Prove that the dual quadric

is the image of

under the projective transformation induced by

. Use this to prove
theorem

2.1.

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

** Next:** Action of on the
** Up:** The Projective Geometry of
** Previous:** Motivation
Juan Carlos Alvarez
2001-01-30