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Row Space, Column Space, and Nullspace
Linear Algebra
MATH 2010
- Terminology: Let A be the 2x4 matrix
A =
[
]
The row vectors of A are [ 2 3 − 1 0
]
[
]
(the rows of A) in <
4 .
The column vectors of A are [ 2
4
]
[
]
[
]
[
]
(the columns of A) in <
2
- Definition: Let A be a mxn matrix (recall m is the number of rows and n is the number of columns),
then
- The row space of A is the subspace of <
n spanned by the row vectors of A
- The column space of A is the subspace of <
m spanned by the column vectors of A.
- Theorem: If a mxn matrix A is row-equivalent to a mxn matrix B, then the row space of A is equal
to the row space of B. (NOT true for the column space)
- Theorem: If a matrix A is row-equivalent to a matrix B in row-echelon form, then the nonzero row
vectors of B form a basis for the row space of A.
- Example - Finding a Basis for Row Space Let
A =
Find a basis for the row space of A.
We must reduce A: 1 1 4 1 2
0 1 2 1 1
0 0 0 1 2
1 − 1 0 0 2
2 1 6 0 1
Then
w 1 = [1, 1 , 4 , 1 , 2] w 2 = [0, 1 , 2 , 1 , 1] w 3 = [0, 0 , 0 , 1 , 2]
form a basis for the row space of A.
- Example- Finding a basis spanned by a set S: Let S = {v 1 , v 2 , v 3 , v 4 } where
v 1 = [1, − 2 , 0 , 3 , −4]
v 2 = [3, 2 , 8 , 1 , 4]
v 3 = [2, 3 , 7 , 2 , 3]
v 4 = [− 1 , 2 , 0 , 4 , −3]
Find a basis for the subspace of < 5 spanned by S.
If we look the matrix
A =
then a basis for the row space of A gives a basis for the subspace of <
5 spanned by S.
Then w 1 = [1, − 2 , 0 , 3 , −4]
w 2 = [0, 1 , 1 , − 1 , 2]
w 3 = [0, 0 , 0 , 1 , −1]
form a basis for the subspace spanned by S. The dimension of the row space is 3.
- Example - Finding a basis for the column space of A: There are two ways to find a basis for
the column space:
- Find the row-echelon form of A:
1 1 4 1 2
0 1 2 1 1
0 0 0 1 2
1 − 1 0 0 2
2 1 6 0 1
Then find the column space of A:
1 3 2 − 1
− 2 2 3 2
0 8 7 0
3 1 2 4
− 4 4 3 − 3
There are leading ones in columns 1,2, and 3, so columns 1,2, and 3 from the original matrix form a
basis for the column space of A. This corresponds to the vectors
v 1 = [1, − 2 , 0 , 3 , −4]
v 2 = [3, 2 , 8 , 1 , 4]
v 3 = [2, 3 , 7 , 2 , 3]
- Theorem: If A is an mxn matrix, then the row space and column space of A have the same dimension.
- Definition: The dimension of the row (or column) space of a matrix A is called the rank of A; denoted
rank(A).
A =
Then (^)
Therefore, rank(A) = 2 because there are 2 leading ones.
- Definition: If A is a mxn matrix, then the set of all solutions of the homogeneous system of linear
equations
Ax = 0
is a subspace of <
n called the nullspace of A and denoted N (A).
N (A) = {x ∈ <
n : Ax = 0}
The dimension of N (A) is called the nullity of A.
- Example: Finding a basis for the nullspace of A: Let
A =
We need to solve the system Ax = 0:
1 1 4 1 2 | 0
0 1 2 1 1 | 0
0 0 0 1 2 | 0
1 − 1 0 0 2 | 0
2 1 6 0 1 | 0
Therefore, x 3 = s and x 5 = t are free parameters. The solution to the system is given by
x =
− 2 s − t
− 2 s + t
s
− 2 t
t
s +
t
Therefore, the basis for N (A) is given by
We have dim(N (A)) = 2, i.e, nullity(A) = 2. Note that dim(rowspace(A)) = 3 and nullity(A) = 2.
3+2 = 5 = n (the number of columns of A).
- Theorem: If A is a mxn matrix of rank A (r), then the dimension of the solution space of Ax = 0 is
n − r, i.e.
n = rank(A) + nullity(A)
A =
Find
- basis for row space of A
- basis for column space of A that is a subset of the column vectors of A
- basis for nullspace of A
- rank(A)
- nullity(A)
Answers:
- basis for row space of A: {[1, 2 , − 1 , −1], [0, 0 , 1 , 4]}
- basis for column space of A that is a subset of the column vectors of A:
- basis for nullspace of A:
- rank(A): 2
- nullity(A): 2
- Solutions of Systems of Linear Equations: The solution x to
Ax = b
can be written as
x = xp + xh
where xp is called a particular solution of Ax = b and xh is called the homogeneous solution of Ax = 0.
