4.2 Null Spaces, Column Spaces, and Linear
Transformations
Two useful subspaces of a vector space that
turn up in applications are the null space and
the column space of a matrix A representing a
linear transformation.
The Null Space of A
Definition: The null space of an m x n matrix A
is the set (Nul A) of all solutions to the
homogeneous equation Ax = 0.
So, Nul A = {x: x  Rn and Ax = 0}
Or, Nul A is the set of all x in Rn that get
mapped into the 0 in Rm under x  Ax.
Theorem 4.2: The null space of and an m x n
matrix A is a subspace of Rn. Equivalently,
the set of all solutions to a system Ax = 0 of m
homogeneous linear equations in n unknowns
is a subspace of Rn.
Proof: Nul A is a subset of Rn, so we show:
a. 0 is in Nul A,
b. that Nul A is closed under vector addition,
c. Nul A is closed under scalar multiplication.
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a. A0=0, so 0 is in NulA
b. Let u, v be in Nul A.
We need to show that A(u + v) = 0
A(u + v)
= Au + Av By the definition of the null space
of A, Au = 0 & Av = 0, so
Au + Av
=0+0
=0
So, Nul A is closed under vector addition.
c. Let u be in Nul A and c a scalar.
Then, by the definition of the null space of A,
Au = 0.
Acu = cAu
= c0
=0
So, Nul A is closed under scalar
multiplication.
Since properties a, b, and c hold, Nul A is a
subspace of Rn. QED
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►Given a matrix A, we can solve the
equation Ax = 0 to get a specific description
of Null A.
Example: Let
3 6 6 3 9
A

6 12 13 0 3
Row reduce the augmented matrix
corresponding to Ax = 0 to get
1 2 0 13 33
0 0 1  6  15

0
0
The free variables are x2 , x4 , x5
The solution is:
 x1   2 x2  13x4  33x5 
x  

x2
 2 

 x3   

6 x4  15 x5
  

x
x
4
 4 

 x5  

x5
So, any vector that satisfies Ax = 0 is a linear
combination of three vectors:
3
 2
 13
 33
1
 0 
 0 
 




x2  0   x5  6   x5  15 
 




0
 1 
 0 
 0 
 0 
 1 
If we name these three vectors u, v, w
respectively, Nul A = Span{u, v, w}
Note: The set {u, v, w} found using this
method is automatically linearly independent.
  2
 13
 33 0
1
 0 
 0  0 
 



  
c1  0   c2  6   c3  15   0
 



  
0
 1 
 0  0 
 0 
 0 
 1  0
Look at the second, fourth, and fifth entries to
see that c1 = c2 = c3 = 0
Note: If Nul A ≠ {0}, the number of vectors in
the spanning set of Nul A equals the number
of free variables in Ax = 0.
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The Column Space of A
Definition: The column space of an m x n
matrix A is the set (Col A) of all linear
combinations of the columns of A.
That is, if A = [a1 … an],
Col A = Span{a1, …, an}.
OR!
Col A = {b: b = Ax for some x in Rn}.
From this definition, Col A is a subset of Rm.
Col A is also a subspace of Rm.
Theorem 4.3: The column space of an m x n
matrix A is a subspace of Rm.
Proof: Col A = Span{a1, …, an}. By Th. 4.1,
Span{a1, …, an} is a subspace of the space
that contains {a1, …, an}, and {a1, …, an}  Rm.
Thus, Col A is a subspace of Rm.
QED
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Example: Find a matrix A such that
W = Col A where
 x  2 y 




W    3 y  : x, y  R 
 x  y 




Write W as the set of all linear combinations
of vectors.
1
  2
x 0  y  3 
1
 1 
Let these vectors be called u and v
respectively, then W = Span{u, v}, so a matrix
whose column space is W is:
1  2
A  0 3 
1 1 
Note: By Th. 1.4, The column space of an
m x n matrix A is all of Rm iff the equation
Ax = b has a solution for every b in Rm.
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Differences Between Nul A and Col A:
Example: Let
1
2
A
3

0
2 3
4 7 
6 10

0 1
►The column space of A is a subspace of R?.
The columns of A have 4 entries, so they and
linear combinations’s of them are in R4.
► The null space of A is a subspace of R?.
Null A contains vectors that can be multiplied
by this matrix, so they must be in R3.
►Find a non-zero vector in Col A.
Any linear combination of the columns will
do:
1 
 2
3
 2
 4
7
c1    c2    c3  
 3
6 
10
 
 
 
0 
0 
1
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► Find a non-zero vector in Nul A.
Solve the equ. Ax = 0, and pick a solution.
The augmented matrix of Ax = 0 is
1
2

3

0
2 3 0
4 7 0
6 10 0

0 1 0
The reduced row echelon form is:
1
0

0

0
2 0 0

0 1 0
0 0 0

0 0 0
Solution:
x1  2 x2
x2 is free
x3  0
Since x2 is free, we can choose any value for
x2, say x2 = 1:
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 x1   1




x   x2    1 
 x3   0 
Be sure to read page 232 for more contrasts!
Summing Up:
A subspace H of a vector space V is a subset
of V that has three properties:
a. The zero vector of V is in H.
b.  u and v  H, u + v  H.
c.  u  H, and each scalar c  R, cu  H.
Theorems 4.1, 4.2, and 4.3 are ways of
proving that sets are subspaces without
resorting to the definition.
The null space of an m x n matrix A is a
subspace of Rn.
The column space of an m x n matrix A is a
subspace of Rm.
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Kernel and Range of a Linear Transformation
Definition: A linear transformation T from a
vector space V into a vector space W is a rule
that assigns to each vector x in V a unique
vector T(x) in W such that
i T(u + v) = T(u) + T(v)
ii T(cu) = cT(u)
for all u, v in V and all scalars c  R.
Definition: the kernel (or null space) of T is the
set of all vectors u in V such that T(u) = 0.
Definition: The range of T is the set of all
vectors in W of the form T(u) where u is in V.
So, if T(x) = Ax, Col A is the range of T.
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