Basis of the Columns pace of a Matrix What about a basis for the column

space of a matrix A? It is remarkable that the same row reduction that gives the basis for the rowspace of A also indicates how to find a basis for the columnspace of A.

However, the method is more subtle and requires a little more attention.

Again, let B be the reduced row echelon form of A. Recall that the whole point of the method of row reduction is that a vector x satisfies Ax =

0

if and only if it satisfies Bx =

0.

That is, if we let i11, ^{• • • ,} i111 denote the columns of A and

b1,

• • • ,

b11

denote the columns of B, then

if and only if

So, any statement about the linear dependence of the columns of A is true if and only if the same statement is true for the corresponding columns of B.

Suppose that B is the reduced row echelon form of A. Then the columns of A that correspond to the columns of B with leading ls form a basis of the columnspace of A. Hence, the dimension of the columnspace equals the rank of A.

Proof: For any m x n matrix B ⁼

[E1 b11]

in reduced row echelon form, the set of columns containing leading ls is linearly independent as they are standard basis vectors from JR"'. Moreover, if

b;

is a column of B that does not contain a leading 1, then, by definition of the reduced row echelon form, it can be written as a linear combi­

nation of the columns containing leading 1 s. Therefore, from our argument above, the corresponding column a; in A can be written as a linear combination of the columns in A that correspond to the columns containing leading l s in B. Thus, we can remove this column from the spanning set without changing the set it spans by Theorem 1.2.3.

We continue to do this until we have removed all the columns of A that correspond to columns of B that do not have leading l s, and we get a basis for the columnspace

�A. •

EXAMPLE 13

EXERCISE 7

EXERCISE 8

Definition

Nullity

Section 3.4 Special Subspaces for Systems and Mappings: Rank Theorem 159

2

LetA = 2 2

�

. Find a basis for Col(A).

2 4 2 3 6 4 3

Solution: By row reducing A, we get

2 1 2 0 0

1 2 2 1 0 0 1 0 2 4 2 3 0 0 0 1 3 6 4 3 0 0 0 0

The first, third, and fourth columns of the reduced row echelon form of A are linearly independent. Therefore, by Theorem 2, the first, third, and fourth columns of matrix A

form a basis for Col(A). Thus, a basis for Col(A) is

{ �

^{1 ,}

�

^{1 ,}

� .

¹

}

3 4 3

Notice in Example 13 that every vector in the columnspace of the reduced row echelon form of A has a last coordinate 0, which is not true in the columnspace of A, so the two columnspaces are not equal. Thus, the first, third, and fourth columns of the reduced row echelon form of A do not form a basis for Col(A).

Let A=

[ �

_-1 ^{-1 0} ₂ ²

^�

^-

^� ]

·Find a basis for Col(A).

-3 -2

There is an alternative procedure for finding a basis for the columnspace of matrix A, which uses the fact that the columnspace of a matrix A is equal to the rowspace of AT. However, the basis obtained in this manner is sometimes not as useful, as it may

not consist of the columns of A.

Find a basis for the columnspace of the matrix A from Example 13 by finding a basis for the rows pace of AT.

Basis of the Nullspace of a Matrix

In Section 2.2 we saw that if the rank of A was r, then the general solution of the homogeneous system Ax =

0

_was

automatically expressed as a spanning set of n -r vectors. We can now show that these spanning vectors are linearly independent, so that the dimension of the nullspace of A is n -r. Since this quantity is imp01tant, we make the following definition.

Let A be an m x n matrix. We call the dimension of the nullspace of A the nullity of A and denote it by nullity(A).

EXAMPLE 14

Theorem 7

Consider the homogeneous system Ax =

0,

where the coefficient matrix

[1 2 0

³

4]

A=

0 0 1

^{5 6}

is already in reduced row echelon form. By finding a basis for Null(A), determine the nullity of A and relate it to rank(A).

Solution: The general solution is

-4

^-3

-2

0 0 1

x =ti -6 + t2 -5 ⁺ t3

0

0 0

0 0

-4

^-3

-2 0 0 1

Thus, -6 -5

0

is a spanning set for the nullspace of A. We now check for

0 1 0 0 0

linear independence.

Let us look closely at the coordinates of the general solution x corresponding to the free variables (x2, ^X4, and xs in this example):

* * * *

0 0 1

^t3

x =ti ^* + t2 ^{* +} t3 ^* = ^*

0 1 0

^t2

1 0 0

^t1

Clearly, this linear combination is the zero vector only if t1 = t2 = t3 =

0,

^{so the}

spanning vectors are linearly independent and hence form a basis for the nullspace of A. It follows that nullity(A) = 3 = (#of columns) - rank(A). In particular, it is the number of free variables in the system.

Following the method in Example

14,

we prove the following theorem.

Let A be an m x n matrix with rank(A) = r. Then the spanning set for the general solution of the homogeneous system Ax =

0

obtained by the method in Chapter

2

is a basis for Null(A) and the nullity of A is n -r.

Proof: Let {vi, ... , v11_,} be a spanning set for the general solution of Ax =

0

^obtained

in the usual manner and consider

t1V1 + · · · + tn-rVn-r =

0

^(3.5)

Theorem 8

EXAMPLE 15

Section 3.4 Special Subspaces for Systems and Mappings: Rank Theorem 161

Then, the coefficients ^ti are just the parameters to the free variables. Thus, the coordi­

nate associated with the i-th parameter is non-zero only in the i-th vector. Hence, the only possible solution of (3.5) _{is t1 = · · · = t11_, =} 0. Therefore, the set is a linearly independent spanning set for Null(A) and thus forms a basis for Null(A). Thus, the

nullity of A is n

-

^r, as required. •

Putting Theorem 6 and Theorem 7 together gives the following important result.

[Rank Theorem]

If A is any m x n matrix, then

rank(A) + nullity(A) = n

Find a basis for the rowspace, columnspace, and nullspace of A ^�

[-i ^; �]

^and

verify the Rank Theorem in this case.

Solution: Row reducing A gives

2 3

1 [1

⁰

11

3 2

1 1

_� 0 ^{0 0 0}

1

¹

Thus, a basis for the rowspace of A is

{[� l

^·

[�]}-

Also, the first and second columns of the reduced row echelon form of A have leading l s, so the corresponding columns from A form a basis for the columnspace of A. That is, a basis for the columnspace of A

is

{ [-i l

^·

[�I}.

Thus, since the rank of A is equal to the dimension of the columnspace (or rowspace), the rank of A is 2.

By back-substitution, we find that the general solution is 1 _{= t}

[ =; l

^{· t E R}

Hence, a basis for the nullspace of A is

{[ ⁼ ;]}

·Thus, we have nullity( A) ⁼

1

^and

rank(A) + nullity(A) = 2 ₊ 1 ₌ 3 as predicted by the Rank Theorem.

EXERCISE9

Find a basis for the rowspace, columnspace, and nullspace of A=

[� ^: =� �]

^and

verify the Rank Theorem in this case.

In document Introduction to Linear Algebra for Science and Engineering 1st Ed (Page 173-177)