6Quadratic forms and conics

IA Vectors and Matrices

6.1 Quadrics and conics

6.1.1 Quadrics

Definition

(Quadric)

.

A quadric is an

n

-dimensional surface defined by the

zero of a real quadratic polynomial, i.e.

x

T

Ax + b

T

x + c = 0,

where

A

is a real

n ×n

matrix,

x, b

are

n

-dimensional column vectors and

c

is a

constant scalar.

As noted in example, anti-symmetric matrix has

x

T

Ax

= 0, so for any

A

,

we can split it into symmetric and anti-symmetric parts, and just retain the

symmetric part S = (A + A

T

)/2. So we can have

x

T

Sx + b

T

x + c = 0

with S symmetric.

Since

S

is real and symmetric, we can diagonalize it using

S

=

QDQ

T

with

D diagonal. We write x

0

= Q

T

x and b

0

= Q

T

b. So we have

(x

0

)

T

Dx

0

+ (b

0

)

T

x

0

+ c = 0.

If

S

is invertible, i.e. with no zero eigenvalues, then write

x

00

=

x

0

+

1

2

D

−1

b

0

which shifts the origin to eliminate the linear term (

b

0

)

T

x

0

and finally have

(dropping the prime superfixes)

x

T

Dx = k.

So through two transformations, we have ended up with a simple quadratic form.

6.1.2 Conic sections (n = 2)

From the equation above, we obtain

λ

1

x

2

1

+ λ

2

x

2

2

= k.

We have the following cases:

(i) λ

1

λ

2

>

0: we have ellipses with axes coinciding with eigenvectors of

S

.

(We require

sgn

(

k

) =

sgn

(

λ

1

, λ

2

), or else we would have no solutions at all)

(ii) λ

1

λ

2

< 0: say λ

1

= k/a

2

> 0, λ

2

= −k/b

2

< 0. So we obtain

x

2

1

a

2

−

x

2

2

b

2

= 1,

which is a hyperbola.

(iii) λ

1

λ

2

= 0: Say

λ

2

= 0,

λ

1

6

= 0. Note that in this case, our symmetric

matrix S is not invertible and we cannot shift our origin using as above.

From our initial equation, we have

λ

1

(x

0

1

)

2

+ b

0

1

x

0

1

+ b

0

2

x

0

2

+ c = 0.

We perform the coordinate transform (which is simply completing the

square!)

x

00

1

= x

0

1

+

b

0

1

2λ

1

x

00

2

= x

0

2

+

c

b

0

2

−

(b

0

1

)

2

4λ

1

b

0

2

to remove the x

0

1

and constant term. Dropping the primes, we have

λ

1

x

2

1

+ b

2

x

2

= 0,

which is a parabola.

Note that above we assumed

b

0

2

6

= 0. If

b

0

2

= 0, we have

λ

1

(

x

0

1

)

2

+

b

0

1

x

0

1

+

c

=

0. If we solve this quadratic for

x

0

1

, we obtain 0, 1 or 2 solutions for

x

1

(and x

2

can be any value). So we have 0, 1 or 2 straight lines.

These are known as conic sections. As you will see in IA Dynamics and Relativity,

this are the trajectories of planets under the influence of gravity.