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.