4Matrices and linear equations

IA Vectors and Matrices

4.5 Homogeneous problem Ax = 0

We restrict our attention to the square case, i.e. number of unknowns = number

of equations. Here A is an n × n matrix. We want to solve Ax = 0.

First of all, if

det A 6

= 0, then

A

−1

exists and

x

−1

=

A

−1

0

=

0

, which is the

unique solution. Hence if Ax = 0 with x 6= 0, then det A = 0.

4.5.1 Geometrical interpretation

We consider a 3 × 3 matrix

A =

r

T

1

r

T

2

r

T

3

Ax

=

0

means that

r

i

· x

= 0 for all

i

. Each equation

r

i

· x

= 0 represents a

plane through the origin. So the solution is the intersection of the three planes.

There are three possibilities:

(i)

If

det A

= [

r

1

, r

2

, r

3

]

6

= 0, span

{r

1

, r

2

, r

3

}

=

R

3

and thus

r

(

A

) = 3. By

the rank-nullity theorem,

n

(

A

) = 0 and the kernel is

{0}

. So

x

=

0

is the

unique solution.

(ii) If det A = 0, then dim(span{r

1

, r

2

, r

3

}) = 1 or 2.

(a)

If rank = 2, wlog assume

r

1

, r

2

are linearly independent. So

x

lies

on the intersection of two planes

x · r

1

= 0 and

x · r

2

= 0, which is

the line

{x ∈ R

3

:

x

=

λr

1

× r

2

}

(Since

x

lies on the intersection of

the two planes, it has to be normal to the normals of both planes).

All such points on this line also satisfy

x · r

3

= 0 since

r

3

is a linear

combination of r

1

and r

2

. The kernel is a line, n(A) = 1.

(b)

If rank = 1, then

r

1

, r

2

, r

3

are parallel. So

x·r

1

= 0

⇒ x·r

2

=

x·r

3

= 0.

So all

x

that satisfy

x ·r

1

= 0 are in the kernel, and the kernel now is

a plane. n(A) = 2.

(We also have the trivial case where

r

(

A

) = 0, we have the zero mapping and

the kernel is R

3

)

4.5.2 Linear mapping view of Ax = 0

In the general case, consider a linear map

α

:

R

n

→ R

n

x 7→ x

0

=

Ax

. The

kernel k(A) = {x ∈ R

n

: Ax = 0} has dimension n(A).

(i)

If

n

(

A

) = 0, then

A

(

e

1

)

, A

(

e

2

)

, ··· , A

(

e

n

) is a linearly independent set,

and r(A) = n.

(ii)

If

n

(

A

)

>

0, then the image is not the whole of

R

n

. Let

{u

i

}, i

=

1

, ··· , n

(

A

) be a basis of the kernel, i.e. so given any solution to

Ax

=

0

,

x

=

n(A)

X

i=1

λ

i

u

i

for some

λ

i

. Extend

{u

i

}

to be a basis of

R

n

by introducing

extra vectors

u

i

for

i

=

n

(

A

) + 1

, ··· , n

. The vectors

A

(

u

i

) for

i

=

n(A) + 1, ··· , n form a basis of the image.