The fourth subspace:
Column space (span of columns, Ax for all possible x)
Row space (span of rows, yTA for all possible y)
Null space (solutions of Ax=0, all vectors orthogonal to the rows, all dependencies among
columns)
Left null space (solutions of yTA=0, all vectors orthogonal to the columns, all dependencies
among the rows, nullspace of AT).
B =
1
-2
4
-3
-1
0
2
3
1
8
2
3
1
1
1
3
2
0
3
-1
7
1
1
1
2
1
-2
2
0
1
5
5
0
13
0
8
>> rref(B)
ans =
1.0000
0
0
0.2000
0
0.2000
0
1.0000
0
2.4000
0
1.4000
0
0
1.0000
0.4000
0
0.4000
0
0
0
0
1.0000
-1.0000
0
0
0
0
0
0
0
0
0
0
0
0
Rank(B)=4=dim row space = dim col space
Basis of col space:
B(:,[1 2 3 5])
ans =
1
-2
4
-1
2
3
1
2
1
1
1
2
3
-1
7
1
2
1
-2
0
5
5
0
0
Basis of row space:
1.0000
0
0
0.2000
0
0.2000
0
1.0000
0
2.4000
0
1.4000
0
0
1.0000
0.4000
0
0.4000
0
0
0
0
1.0000
-1.0000
Null space basis (worked out by hand from above):
[ -.2
-2.4
[ -.2 -1.4
-.4
1
0
0]
-.4
0
1
1]
>> null(B,'r')
%let matlab do it
ans =
-0.2000
-0.2000
-2.4000
-1.4000
-0.4000
-0.4000
1.0000
0
0
1.0000
0
1.0000
Left null space basis
>> null(B','r')
ans =
-1.4444
-0.4444
-0.3333
-2.3333
-0.8889
2.1111
1.0000
0
0
-1.0000
0
1.0000
>> [B(:,[1 2 3 5]) null(B','r')]
ans =
1.0000
-2.0000
4.0000
-1.0000
-1.4444
-0.4444
2.0000
3.0000
1.0000
2.0000
-0.3333
-2.3333
1.0000
1.0000
1.0000
2.0000
-0.8889
2.1111
3.0000
-1.0000
7.0000
1.0000
1.0000
0
2.0000
1.0000
-2.0000
0
0
-1.0000
5.0000
5.0000
0
0
0
1.0000
First four columns are basis of column space
Last two columns are basis of left null space
Together they are a basis for R6
Characterization of column space: Ax=b has a solution (b is in the
column space) if and only if yTb=0 for any y in left null space.
Definition of inverse: If A is square, we define the inverse of A (if
it exists) as the matrix A-1 that satisfies AA-1=I and A-1A=I.
Theorem: A-1 exists (A has an inverse) if and only if A~I (or the rank
of A is n or dim col space=n , Ax=0 only if x=0 )
How to calculate A-1
Think of solving for B that satisfies AB=I – this is, column by column
in B, a system of linear equations, i.e. A*(col j of B)=(col j of I)
and we solve all these equations at once by putting all the columns of
I into a super-augmented matrix: [A | I] . Then we reduce as usual,
and assuming that A~I we would obtain
[A | I] ~ [I | A-1]
because each column on the right is one of the columns of B that we
seek.
(If you discover that the rank of A is less than n, then A-1 doesn’t
exist, e.g. note that if A-1 exists then A(A-1b)=b for any b, which
shows that Ax=b always has a solution. If the rank of A is less than
n, this can’t be true and so A-1 can’t exist)
4 subspaces in one reduction: We already know how to get a basis of
the column space, row space and null space of the matrix A from the
reduced row echelon form of A. Now consider the reduction
[A | I] ~ [R | B] where R is the (reduced) row echelon form of A. R
will have m-r rows of zeros at the bottom, where r is the rank of A.
Then the bottom m-r rows of B are a basis for the left null space.
Here’s the reason: Every row operation can be carried out by a leftmultiplication so [A | I] ~ [R | B] means that C[A | I]=[R | B] for
some C. But then we must have CA=R and CI=B; from the latter we see
that C=B so that BA=R. Now the bottom m-r rows of R are equal to the
corresponding m-r rows of B times A and the result is zero for each
such row. Thus each of the bottom m-r rows of B is in the left null
space. The dimension of the left null space is m-r and those rows of B
are linearly independent because B is row equivalent to I. Thus those
rows are a basis of the left null space.
Here is our previous example of the matrix B of rank 4. We find the
left null space using the procedure outlined above.
>> [B eye(6)]
ans =
1
-2
4
-3
-1
0
1
0
0
0
0
0
2
3
1
8
2
3
0
1
0
0
0
0
1
1
1
3
2
0
0
0
1
0
0
0
3
-1
7
1
1
1
0
0
0
1
0
0
2
1
-2
2
0
1
0
0
0
0
1
0
5
5
0
13
0
8
0
0
0
0
0
1
>> rref([B eye(6)])
ans =
1
0
0
0.2000
0
0.2000
0
0
-0.0690
0.1379
0.4483
-0.0483
0
1
0
2.4000
0
1.4000
0
0
0.0690
-0.1379
-0.4483
0.2483
0
0
1
0.4000
0
0.4000
0
0
-0.0345
0.0690
-0.2759
0.0759
0
0
0
0
1
-1
0
0
0.5172
-0.0345
0.1379
-0.1379
0
0
0
0
0
0
1
0
0.8621
-0.7241
-0.1034
0.1034
0
0
0
0
0
0
0
1
-1.0690
0.1379
0.4483
-0.4483
The rank of B is 4, the last two rows in the reduced form of B are
zeroes. The basis for the left null space is then the last two rows in
the right-hand side of the augmented matrix, namely
[1
0
0.8621
-0.7241
-0.1034
0.1034]
[0
1
-1.0690
0.1379
0.4483
-0.4483]
Of course, it’s not clear that this is any easier than simply finding
the null space of the transpose of the matrix in question, which is
the more direct approach.