4 Vector Spaces

4.2
NULL SPACES, COLUMN
SPACES, AND LINEAR
TRANSFORMATIONS

© 2016 Pearson Education, Ltd.

NULL SPACE OF A MATRIX
 Definition: The null space of an m × n matrix A,
written as NulA, is the set of all solutions of the
homogeneous equation Ax = 0. In set notation,
Nul A {x = : x is in ℝ
 𝑛𝑛n and Ax 0}.
 Theorem 2: The null space of an m × n matrix A is
a subspace of ℝ𝑛𝑛 . Equivalently, the set of all
solutions to a system Ax = 0 of m homogeneous
linear equations in n unknowns is a subspace of ℝ𝑛𝑛 .
 Proof:NulA is a subset of ℝ𝑛𝑛 because A has n
columns.
 We need to show that NulA satisfies the three
properties of a subspace.
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NULL SPACE OF A MATRIX

 0 is in Null A.
 Next, let u and v represent any two vectors in Nul A.
 Then
Au = 0 and Av = 0
 To show that u + v is in Nul A, we must show that
A(u + v) = 0.
 Using a property of matrix multiplication, compute
A(u + v) = Au + Av = 0 + 0 = 0
 Thus u + v is in Nul A, and Nul A is closed under
vector addition.
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NULL SPACE OF A MATRIX

 Finally, if c is any scalar, then

A(c=
u) c( A=
u) c=
(0) 0
which shows that cu is in NulA.
 Thus NulA is a subspace of ℝ𝑛𝑛 .

 An Explicit Description of NulA

 There is no obvious relation between vectors in NulA
and the entries in A.
 We say that NulA is defined implicitly, because it is
defined by a condition that must be checked.
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NULL SPACE OF A MATRIX
 No explicit list or description of the elements in Nul
A is given.
 Solving the equation Ax = 0 amounts to producing an
explicit description of Nul A.

 Example 3: Find a spanning set for the null space of

the matrix
 −3 6 −1 1 −7 
 1 −2 2 3 −1
A= .
 
 2 −4 5 8 −4 
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NULL SPACE OF A MATRIX
 Solution: The first step is to find the general solution
of Ax = 0 in terms of free variables.

 Row reduce the augmented matrix [ A 0] to reduce

echelon form in order to write the basic variables in
terms of the free variables:

 1 −2 0 −1 3 0  x1 − 2 x2 − x4 + 3 x5 =
0
0 0 1 2 −2 0 
 , x3 + 2 x4 − 2 x5 =
0
0 0 0 0 0 0  0=0
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NULL SPACE OF A MATRIX
 The general solution is x1 = 2 x2 + x4 − 3 x5,
x3 =−2 x4 + 2 x5 , with x2, x4, and x5 free.
 Next, decompose the vector giving the general
solution into a linear combination of vectors where
the weights are the free variables. That is,
 x1   2 x2 + x4 − 3 x5  2  1  −3
x   x  1   0  0
 2  2
      
 x3  = −2 x4 + 2 x5  =x2  0  + x4  −2  + x5  2 
x   x  0  1  0
 4  4       
 x5   x5   0   0   1

© 2016 Pearson Education, Ltd.

u v w Slide 4.2- 7
NULL SPACE OF A MATRIX
= x2 u + x4 v + x5 w (3)
 Every linear combination of u, v, and w is an
element of Nul A.
 Thus {u, v, w} is a spanning set for Nul A.

1. The spanning set produced by the method in

Example (3) is automatically linearly
independent because the free variables are the
weights on the spanning vectors.
2. When Nul A contains nonzero vectors, the
number of vectors in the spanning set for Nul
A equals the number of free variables in the
equation Ax = 0 .
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COLUMN SPACE OF A MATRIX

 Definition: The column space of an m × n matrix A,

written as Col A, is the set of all linear combinations
of the columns of A. If A = [ a1  a n ], then
Col A = Span{a1 ,...,a n }.

 Theorem 3: The column space of an m × n matrix A

is a subspace of ℝ𝑚𝑚 .
 A typical vector in Col A can be written as Ax for
some x because the notation Ax stands for a linear
combination of the columns of A. That is,
Col
= A {b = : b Ax for some x in ℝ }. n
𝑛𝑛

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 COLUMN SPACE OF A MATRIX
 The notation Ax for vectors in Col A also shows that Col
A is the range of the linear transformation x  Ax .
 The column space of an m × n matrix A is all of ℝ𝑚𝑚 if
and only if the equation Ax = b has a solution for each
b in ℝ𝑚𝑚 .  3
 2 4 −2 1  −2 
 2 −5 7 3 , u =  
 Example 7: Let A =−
   −1
 3  3 7 −8 6 
   0
and v = −1 .  
 
 3
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COLUMN SPACE OF A MATRIX

a. Determine if u is in Nul A. Could u be in Col A?

b. Determine if v is in Col A. Could v be in Nul A?
 Solution:
a. An explicit description of Nul A is not needed
here. Simply compute the product Au.

 3
 2 4 −2 1    0  0 
  −2    
Au = −2 −5 7 3   = −3 ≠ 0
   −1    
 3 7 −8 6     3 0 
 0
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COLUMN SPACE OF A MATRIX

 u is not a solution of Ax = 0 , so u is not in NulA.

 Also, with four entries, u could not possibly be in
Col A, since Col A is a subspace of ℝ3 .
b. Reduce [ A v ] to an echelon form.
 2 4 −2 1 3  2 4 −2 1 3
 2 −5 7 3 −1 ~  0 1 −5 −4 −2 
[ A v ] 
=−
  
 3 7 −8 6 3  0 0 0 17 1

 The equation Ax = v is consistent, so v is in Col A.

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 KERNEL AND RANGE OF A LINEAR
TRANSFORMATION

 With only three entries, v could not possibly be

in NulA, since NulA is a subspace of ℝ4 .
 Subspaces of vector spaces other than ℝ𝑛𝑛 are often
described in terms of a linear transformation instead of
a matrix.
 Definition: A linear transformationT 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) for all u, v in V, and
ii. T (cu) = cT (u) for all u in V and all scalars c.

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KERNEL AND RANGE OF A LINEAR
TRANSFORMATION

 The kernel (or null space) of such a T is the set of all

u in V such that T (u) = 0 (the zero vector in W ).

 The range of T is the set of all vectors in W of the

form T (x) for some x in V.

 The kernel of T is a subspace of V.

 The range of T is a subspace of W.

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CONTRAST BETWEEN NUL A AND COL A FOR AN
m × n MATRIX A

NulA Col A
1. Nul A is a subspace of 1. Col A is a subspace of
ℝ𝑛𝑛 . ℝ𝑚𝑚 .
2. NulA is implicitly 2. Col A is explicitly
defined; i.e., you are defined; i.e., you are
given only a condition told how to build
( Ax = 0) that vectors in vectors in Col A.
NulA must satisfy.

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CONTRAST BETWEEN NUL A AND COL A FOR AN
m × n MATRIX A

3. It takes time to find 3. It is easy to find vectors

vectors in Nul A. Row in Col A. The columns
operations on [ A 0] of a are displayed;
are required. others are formed from
them.
4. There is no obvious 4. There is an obvious
relation between Nul A relation between Col A
and the entries in A. and the entries in A,
since each column of A
is in Col A.

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 CONTRAST BETWEEN NUL A AND COL A FOR AN
m × n MATRIX A

5. A typical vector v in Nul 5. A typical vector v in Col

A has the property that A has the property that
Av = 0. the equation Ax = v is
consistent.
6. Given a specific vector v, 6. Given a specific vector
it is easy to tell if v is in v, it may take time to tell
Nul A. Just compare Av. if v is in Col A. Row
operations on [ A v ] are
required.

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 CONTRAST BETWEEN NUL A AND COL A FOR AN
m × n MATRIX A

7. Nul A = {0}if and only if 7. Col 𝐴𝐴 = ℝ𝑚𝑚 if and only

the equation Ax = 0has if the equation Ax = b
only the trivial solution. has a solution for every b
in ℝ𝑚𝑚 .
8. Nul A = {0} if and only if 8. Col 𝐴𝐴 = ℝ𝑚𝑚 if and only
the linear transformation if the linear
x  Ax
is one-to-one. transformation x  Ax
maps ℝ𝑛𝑛 onto ℝ𝑚𝑚 .

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