Download This lecture: Column Space Basis

In a previous lecture: Basis of the Null Space of a Matrix
This lecture: Column Space Basis
The column space of a matrix is defined in terms of a
spanning set, namely the set of columns of the matrix.
But the columns are not necessarily linearly independent.
In this lecture, we demonstrate a systematic procedure
for obtaining a linearly independent spanning set (i.e. a
basis) for the column space of a matrix.
"
#
Consider the matrix A =
"
and also the matrix J =
1
0
4
1
1
0
0
0
0
1
0
0
4
−1
2
0
3
−1
0
0
−1
1
10
3
0
0
1
0
6
5
−4
0
2
−1
1
0
4
6
2
2
3
−1
1
0
5
6
6
3
#
Notice that the column spaces of A and its ”reduced”
matrix J are NOT the same. It is really easy to find a
basis for the column space of a matrix in reduced echelon form, such as J.
Let the columns of J be called d1 , ...d6 . Then by looking
at the numerical entries in J, the following relationships
can be seen:
0-0
d3 = 3d1 − d2 ,
d5 = 2d1 − d2 + d4 ,
d6 = 3d1 − d2 + d4
That is, d3 , d5 , d6 are linear combinations of d1 , d2 , d4 .
Columns 1, 2 and 4 are linearly independent so they form
a basis of Col(J), the column space of J. Let the columns
of matrix A be called c1 , ...c6 . Now we can check that:
c3 = 3c1 − c2 ,
c5 = 2c1 − c2 + c4 ,
c6 = 3c1 − c2 + c4
" #
" # "
−1
1
10
3
1
0
4
1
=3
−
" #
" # "
4
6
2
2
=2
" #
5
6
6
3
1
0
4
1
−
" #
=3
1
0
4
1
"
−
4
−1
2
0
4
−1
2
0
4
−1
2
0
#
#
"
+
#
"
+
6
5
−4
0
6
5
−4
0
#
#
The reason the numbers add up this way is that the operations of echelon reduction do not affect the dependency
relations between the columns as the matrix A is transformed into J.
0-1
The same result will hold for any matrix A and its row
reduced echelon form matrix J.
Observation If certain columns of the matrix J form a
basis for Col(J), then the corresponding columns of the
matrix A form a basis of Col(A)
Observation If certain columns of the matrix A form a
basis for Col(A), then the corresponding columns in the
matrix J form a basis for Col(J).
So the dimensions of the column spaces of A and J are
equal. The spaces themselves are usually different, but
they do have the same dimension.
Since the dimension of the column space of J is equal
to the number of columns with a leading 1, we have the
following result.
Theorem
If A is an m × n matrix whose reduced
row echelon form J has r leading 1’s, then Col(A) has
dimension r.
0-2
This leads to the matrix version of the famous
Dimension Theorem of Vector Spaces.
Dimension Theorem for Matrices
Theorem If A is an m × n matrix, then
dim Null(A) + dim Col(A) = number of columns of A
Proof Suppose that the reduced row echelon form of A
has r leading 1’s. Then the left hand side of the above
equation equals n − r + r, which equals the number of
columns of A.
Definition The nullity of a matrix A is the dimension
of the Null Space of A.
Definition The rank of a matrix A is the dimension of
the Column Space of A.
Therefore if A is an m × n matrix whose reduced row
echelon form J has r leading 1’s,
nullity = n − r,
rank = r and
rank + nullity = number of columns of the matrix A.
0-3