# 4.6 Null Space, Column Space, Row Space

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4.6. NULL SPACE, COLUMN SPACE, ROW SPACE

147

4.6 Null Space, Column Space, Row Space
In applications of linear algebra, subspaces of Rn typically arise in one of two situations: 1) as the set of solutions of a linear homogeneous system or 2) as the set of all linear combinations of a given set of vectors. In this section, we will study, compare and contrast these two situations. We will …nish the section with an introduction to linear transformations.

4.6.1 The Null Space of a Matrix
De…nitions and Elementary Remarks and Examples
In previous section, we have already seen that the set of solutions of a homogeneous linear system formed a vector space (theorem 271). This space has a name.

De…nition 342 The null space of an m n matrix A, denoted N ull A, is the set of all solutions to the homogeneous equation Ax = 0. Written in set notation, we have

N ull A = fx : x 2 Rn and Ax = 0g Remark 343 As noted earlier, this is a subspace of Rn. In particular, the elements of N ull A are vectors in Rn if we are working with an m n matrix.

Remark 344 We know that N ull A 6= ; since it always contain 0, the trivial solution. The question is can it have nontrivial solutions?

Example 345 The elements of N ull A if A is 3 Example 346 The elements of N ull A if A is 2 Example 347 The elements of N ull A if A is 3 Example 348 The elements of N ull A if A is 5

5 are vectors of R5. 2 are vectors in R2. 2 are vectors in R2. 2 are vectors in R2.

Remark 349 The kind of elements N ull A contains (which vector space they belong to) depends only on the number of columns of A.

We now look at speci…c examples and how to …nd the null space of a matrix.

Examples
Usually, when one is trying to …nd the null space of a matrix, one tries to …nd a basis for it. So, when asked to "…nd the null space" of a matrix, one is asked to …nd a basis for it. The examples below illustrate how to do this.

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CHAPTER 4. VECTOR SPACES

2 36

3 11 7

Example 350 Find N ull A for A = 4 1 2 2 3 1 5

2 458 4

First, let us remark that the elements of N ull A will be elements of R5. We

a2re …nding the solution set of Ax3 = 0. The augmented matrix of th2e system is

3

6 3 6 1 1 7 ... 0 7

6 1 2 0 1 3 ... 0 7

66 4

1

223

1

...

0

77, the reduced row-echelon form is: 5

66 4

0

0

1

2

2

...

0

77. 5

2 4 5 8 4 ... 0 8

0 0 0 0 0 ... 0

>>>>< x1 =x22r =+ rs 3t

The solutions of the system are >>>

x3 = 2s + 2t x =s

>:

4

which can be written as

2 3 2 3 2 3 2 x53= t

x1

2

1

3

66 x2 77 66 1 77 66 0 77 66 0 77

66 x3 77 = r 66 0 77 + s 66 2 77 + t 66 2 77. N ull A is the subspace spanned

4 x4 5 4 0 5 4 1 5 4 0 5

x5

0

20 3

21

2

66 1 77

66

by fu; v; wg where u = 66 0 77, v = 66

405

4

3

2

1

0 77

66

2 77 and w = 66

15

4

3 3 0 77 2 77. 05

It should

0

0

1

be clear that this set is also linearly independent. So, it is a basis for N ull

8A.2 He3nce2, N ull3A2has d3im9ension 3 and it is the subspace of R5 with basis

>>>><6

2 1

7

6

1 0

76

3 0

7>>>>=

666 0 777 ; 666 2 777 ; 666 2 777 . >>>>:4 0 5 4 1 5 4 0 5>>>>;

0

0

1

2

3

2 2 10 1

Example 351 Find N ull A for A = 664 11 11 22 03 11 775

00111

First, let us remark that the elements of N ull A will be elemen2ts of R5. To …nd 2 2 10
N ull A, we solve the system Ax = 0. The augmented matrix is 664 11 11 22 03

2

3

0011

110010

Its reduced row-echelon form is 664 00 00 10 01 10 00 775. So, the solutions are

000000

3 10 11 00 775. 10

4.6. NULL SPACE, COLUMN SPACE, ROW SPACE

149

8 < x1 + x2 + x5 = 0 : x3 x+4 x=5 0= 0

2

3

2

x1 = t s

66 x2 = t 77

66

that is 66 x3 = s 77which can be written as x = s 66

4 x4 = 0 5

4

3 1 0 77 1 77+ 05

23

x5 = s

82 31 2 39

1 617

>>>><6

1 0 76

1 1

7>>>>=

t 666 0 777. Thus N ull A is a subspace of R5, of dimension 2 with basis 666 1 777 ; 666 0 777

405

>>>>:4 0 5 4 0 5>>>>;

.0

1

0

If should be clear to the reader that if A is invertible then N ull A = f0g. Indeed, if A is invertible, then Ax = 0 only has the trivial solution. We state it as a theorem.

Theorem 352 If A is invertible then N ull A = f0g.

In earlier chapters, we developed the technique of elementary row transformations to solve a system. In particular, we saw that performing elementary row operations did not change the solutions of linear systems. We state this result as a theorem.

Theorem 353 Elementary row operations on a matrix A do not change N ull A.

De…nition 354 The nullity of a matrix A, denoted nullity (A) is the dimension of its null space.

2 2 2 10
Example 355 From the previous examples, we see that if A = 664 11 11 22 03

2 36

11

then nullity (A) = 2 and if B = 4 1 2 2 3

3

0011

7

1 5 then nullity (B) =

2 458 4

3.

3 1 1 77 15 1

4.6.2 Column Space and Row Space of a Matrix
De…nitions and Elementary Concepts Let us start by formalizing some terminology we have already used.

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2 a11 a12 a13
66 a21 a22 a23 De…nition 356 Consider an m n matrix 66 a31 a32 a33
64 ... ... ... am1 am2 am3

3 a1n a2n 77 a3n 77.
... 75 amn

1. The vectors r1 = a11; a12; a13;

; a1n , r2 = a21; a22; a23;

:::, rm = am1; am2; am3;

; amn in Rn obtained from the rows

of A are called the row vectors of A.

2

3

2

3

2

3

a11

a12

a1n

66 a21 77

66 a22 77

66 a2n 77

2. The vectors c1 = 66 a31 77, c2 = 66 a32 77, :::, cn = 66 a3n 77 in Rm

64 ... 75

64 ... 75

64 ... 75

am1

am2

amn

obtained from the columns of A are called the column vectors of A.

We now turn to the main de…nitions of this section.

De…nition 357 Let A be an m n matrix.

1. The subspace of Rn spanned by the row vectors of A is called the row space of A.

2. The subspace of Rm spanned by the column vectors of A is called the column space of A.

In this subsection, we will try to answer the following questions:

1. How does one …nd the row space of a matrix A?

2. How does one …nd the column space of a matrix A?

3. Is there a relationship between them?

4. Is there a relationship between them and the null space of a matrix A?

5. Since matrices are closely related to solving systems of linear equations, is there a relationship between the row space, column space, null space of a matrix A and the linear system Ax = b? 23 x1 66 x2 77
With the notation already introduced, and letting x = 64 ... 75 we see that xn
we can write Ax = x1c1 + x2c2 + ::: + xncn
Thus, solving Ax = b amounts to solving x1c1 +x2c2 +:::+xncn = b. So, we see that for the system to have a solution, b must be in the span of fc1; c2; :::; cng. We state this a a theorem.

; a2n ,

4.6. NULL SPACE, COLUMN SPACE, ROW SPACE

151

Theorem 358 A system of linear equations Ax = b is consistent if and only if b is in the column space of A.
We now look at some important results about the column space and the row space of a matrix.
Theoretical Results
First, we state and prove a result similar to one we already derived for the null space.
Theorem 359 Elementary row operations do not change the row space of a matrix A. Proof. Suppose that A is m n with rows r1; rn; :::; rm. Let B be obtained from A by one of the elementary row operations. Suppose that the rows of B are r01; r0n; :::; r0m. We divide the proof according to the kind of elementary row transformation being applied. We do the proof by showing that every vector in the row space of B is in the row space of A and vice-versa. In other words, we prove inclusion both ways; a standard technique to show two sets are equal.
Row interchange. This case is easy. A and B will still have the same row space since they will have the same rows.
Replacing a row by a multiple of another or by itself plus a multiple of another. This can be generalized by saying that one or more of r0i are linear combinations of the rjs. Thus the vectors r01; r0n; :::; r0m lie in the row space of A. Since a vector space is closed under linear combinations, any linear combination of r01; r0n; :::; r0m will also be in the row space of A. But that’s precisely what the row space of B is, linear combinations of r01; r0n; :::; r0m. Thus the row space of B is in the row space of A.
Since every elementary row transformation has an inverse transformation, we can transform B into A. Using the same argument as above, we will prove that the row space of A in in the row space of B.

Remark 360 This result, unfortunately, does not apply to the column space of A. Only its row space is preserved under elementary row operations. However, we do have the following results: The next two theorems are given without proof.
Theorem 361 Let A and B be two matrices which are row equivalent (one is obtained from the other with elementary row operations).
1. A given set of column vectors of A is linearly independent if and only if the corresponding column vectors of B are linearly independent.
2. A given set of column vectors of A form a basis for the column space of A if and only if the corresponding column vectors of B form a basis for the column space of B.

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CHAPTER 4. VECTOR SPACES

The following theorem makes it easy to …nd a basis for the row and column space of a matrix. We will use it in the examples.

Theorem 362 If a matrix R is in row-echelon form then:

1. The row vectors with the leading 10s form a basis for the row space of R.

2. The column vectors with the leading 10s of the row vectors form a basis for the column space of R.

Remark 363 Combining theorems 361 and 362, we see that to …nd a basis for the row space of a matrix of a matrix, we put it in row-echelon form and extract the row vectors with a leading 1.

Remark 364 Since row elementary row operations do not preserve the column

space, to …nd the column space of a matrix will be a little more di¢ cult. But

not too much. Let A be a matrix and B be the row-echelon form of A. If we call

c1; c2; :::; cn

the

columns

of

A

and

c

0 1

;

c

0 2

;

:::;

c

0 n

the

columns

of

B.

We

look

at

the

columns of B which have a leading 1 of the rows of B. A basis for the column

space of A will be the vectors ci for the values of i for which c0i has a leading 1

of the rows of B. This sounds complicated but it is not. We will illustrate this

with examples.

Finally, we give a theorem which relates the null space of A and solutions of Ax = b.

Theorem 365 If x0 is any solution of Ax = b and v1; v2; :::; vr form a basis for the null space of A, then any solution of Ax = b can be written as

x = x0 + c1v1 + c2v2 + ::: + crvr

(4.1)

and conversely, any vector written this way for any scalars c1; c2; :::; cr is a solution of Ax = b. Proof. We prove both ways.

Assume x0 is any solution of Ax = b and v1; v2; :::; vr form a basis for the null space of A. We must prove that any solution of Ax = b can be written
as x = x0 + c1v1 + c2v2 + ::: + crvr. If x is any solution of Ax = b then we have Ax0 = b and Ax = b that is Ax0 = Ax or A (x x0) = 0. Thus, x x0 is in the null space of A, that is there exists scalars c1; c2; :::; cr such that x x0 = c1v1 + c2v2 + ::: + crvr or x = x0 + c1v1 + c2v2 + ::: + crvr.

Conversely, suppose that for any scalars c1; c2; :::; cr, x = x0 + c1v1 + c2v2 + ::: + crvr, where x0 is a solution of Ax = b we must show that x is a solution of Ax = b.

Ax = (x0 + c1v1 + c2v2 + ::: + crvr) = Ax0 + c1Av1 + c2Av2 + ::: + crAvr = b + 0 + 0 + ::: + 0

since the vi0s are a basis for N ull A. Thus we see that Ax = b.

4.6. NULL SPACE, COLUMN SPACE, ROW SPACE

153

We now see how these results help us …nding the column space and the row space of a matrix.

Examples

We begin with very easy examples. 2 1
Example 366 Consider the matrix in row-echelon form A = 664 00

3 2503 10 30 01 00 775.

0 0 000

Find a basis for the row space and column space.

Since the matrix is in row-echelon form, we can apply theorem 362 directly.

Row Space: A basis is the set of row vectors with a leading 1, that is

1 2 5 0 3 ; 0 1 3 0 0 ; 0 0 0 1 0 . Hence

the row space has dimension 3.

82 3 2

>><6

1 0

7

6

Column Space: A basis is >>64 0 75 ; 64

:0

3 2 39

2 1

7

6

0 0

7>>=

0 75 ; 64 1 75>>. Hence the column

0

0;

space has dimension 3. 2 1
Example 367 Find bases for the row space and column space of A = 664 22

34 69 69

25 18 19

3 4 27 775.

2

1 334 2

1 34 2 5 4

We begin by …nding the row-echelon form of A. It is 664 00 00 10 30 12 56 775.

54

0000 0 0

Row Space: A basis is the set of row vectors with a leading 1, that is 1 34 254;0013 2 6;000015 .
Hence the row space has dimension 3.

Column Space: The columns with a leading 1 from the row vectors are 1,

3, and 5. Hence, a basis for the8c2olumn 3spa2ce are3co2lumns 31,93, and 5

>><6

1 2

76

4 9

76

5 8

7>>=

from the original matrix that is >>64 2 75 ; 64 9 75 ; 64 9 75>>. Hence

:1

4

5;

the column space has dimension 3.

Remark 368 In the two examples above, we see that the row space and column space have the same dimension. This was not an accident. We will see that indeed they do have the same dimension.

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4.6.3 Application to Linear Transformations
A transformation is the linear algebra term of a concept studied in calculus which was then called a function. Another di¤erence is that a transformation works on vectors. Its input values are vectors, its output values are also vectors. The input and output values may not be from the same vector space. Like functions, a transformation assigns to each input value a unique output value. In this section, we only look at a special kind of transformations called linear transformations. We now give a precise mathematical de…nition.
De…nition 369 (Linear Transformation) A linear transformation T from a vector space V to another vector space W is a rule which assigns to each vector x in V a unique vector T (x) in W , such that:
1. T (u + v) = T (u) + T (v) for every u and v in V .
2. T (cu) = cT (u) for every u in V and every scalar c.
Remark 370 The two conditions which have to be satis…ed is what make the transformation a linear transformation.
Remark 371 If T is a transformation from V to W , we often write T : V ! W.
Let us look at some linear transformations we already know.
Example 372 Let A be an m n matrix. De…ne T : Rn ! Rm by T (x) = Ax. It is easy to see that T is a linear transformation from the properties of matrices.
T (x + x0) = A (x + x0) = Ax + Ax0 = T (x) + T (x0)
and
T (cx) = A (cx) = cAx = cT (x)
Example 373 Let V be the vector space of real valued functions with continuous …rst derivatives de…ned on an interval [a; b] and let W be the space of continuous functions on [a; b]. Consider the transformation D : V ! W de…ned by D (f ) = f 0. It is easy to see that D is indeed a linear transformation.
D (f + g) = (f + g)0 = f 0 + g0 (sum rule for di¤ erentiation) = D (f ) + D (g)

4.6. NULL SPACE, COLUMN SPACE, ROW SPACE

155

and
D (cf ) = (cf )0 = cf 0 = cD (f )

When dealing with linear transformation, there are two concepts of importance, de…ned below.

De…nition 374 Let T : V ! W be a linear transformation.

1. The kernel of T , denoted ker (T ) is the set of vectors v in V such that T (v) = 0.

2. The range of T is the set of all vectors in W of the form T (x) for some x in V .

Example 375 If the linear transformation is as in example 373, then ker (D) is the set of functions f such that D (f ) = 0 that is f 0 = 0. So, it is the set of
constant functions.

Remark 376 If T is as in example 372, then the kernel of T is simply the null

space of A and the range of T is simply the column space of A.

2

1

Example 377 Let T : R6 ! R4 be de…ned by T (x) = Ax where A = 664

2 2

34 69 69

25 18 19

3 4 27 775.

13 42 5 4

Find the kernel and the range of T .

Kernel of T : As noticed in the remark, …nding the kernel of T amounts to

…nding the null space of A. So, we2need to solve the system Ax = 0. T3he

1 34 25 40

augmented matrix of the system is 664 22

69 69

18 19

27 00 775.

2 1
Its reduced row-echelon form is 664 00

13 30 01 00

42 14 0 30 01

5 34 0 37 0 45 00 775. Thus,

8

000 0 0 0 0

< x1 3x2 14x4 37x6 = 0

the solutions are :

x3 + 3x4 + 4x6 = 0

, or written in paramet-

8

x5 + 5x6 =20 3

23 2 3

>>>>>

x1

= 3s + 14t + 37u x =s

x1 6x 7

3 617

14 607

>< x3 = 2 3t 4u

666

2
x3

777

666 0 777 666 3 777

ric form:>>>

x4 = t

or

66

x4

77

=

s6 6

0

7 + t6 76

1

7+ 7

>>>:

x5 = 5u

4 x5 5

405 4 0 5

x6 = u

x6

0

0

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CHAPTER 4. VECTOR SPACES

23 37
66 0 77 u 6666 04 7777. So, we see that the dimension of ker (T ) is 3.
4 55 1

Range of T : As noticed in the remark, it is the column space of A. We

already computed it in a8p2reviou3s e2xample3an2d foun3d9that it was the sub-

>><6

1 2

76

4 9

76

5 8

7>>=

space of R4 with basis >>64 2 75 ; 64 9 75 ; 64 9 75>>. Hence the range

:1

4

5;

of T has dimension 3.

4.6.4 Problems
1. Do # 1, 2, 3a, 3b, 3c, 6, 7, 8, 9, 11, 14, 16 on pages 277, 278 2. Suppose that T : V ! W is a linear transformation.
(a) Prove that ker (T ) is a subspace of V . (b) Prove that the range of T is a subspace of W .
Column SpaceRow SpaceMatrixSpaceBasis