3.2 Subspaces of Fn

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DEFINITION 3.2.1 A subset S of Fn is called a subspace of Fn if

1. The zero vector belongs to S; (that is, 0 2 S);

2. If u 2 S and v 2 S, then u + v 2 S; (S is said to be closed under

vector addition);

3. If u 2 S and t 2 F, then tu 2 S; (S is said to be closed under scalar

multiplication).

EXAMPLE 3.2.1 Let A 2 Mm£n(F). Then the set of vectors X 2 Fn

satisfying AX = 0 is a subspace of Fn called the null space of A and is

denoted here by N(A). (It is sometimes called the solution space of A.)

55

56 CHAPTER 3. SUBSPACES

Proof. (1) A0 = 0, so 0 2 N(A); (2) If X; Y 2 N(A), then AX = 0 and

AY = 0, so A(X + Y ) = AX + AY = 0 + 0 = 0 and so X + Y 2 N(A); (3)

If X 2 N(A) and t 2 F, then A(tX) = t(AX) = t0 = 0, so tX 2 N(A).

For example, if A =

·

1 0

0 1

¸

, then N(A) = f0g, the set consisting of

just the zero vector. If A =

·

1 2

2 4

¸

, then N(A) is the set of all scalar

multiples of [¡2; 1]t.

EXAMPLE 3.2.2 Let X1; : : : ;Xm 2 Fn. Then the set consisting of all

linear combinations x1X1 + ¢ ¢ ¢ + xmXm, where x1; : : : ; xm 2 F, is a sub-

space of Fn. This subspace is called the subspace spanned or generated by

X1; : : : ;Xm and is denoted here by hX1; : : : ;Xmi. We also call X1; : : : ;Xm

a spanning family for S = hX1; : : : ;Xmi.

Proof. (1) 0 = 0X1 + ¢ ¢ ¢ + 0Xm, so 0 2 hX1; : : : ;Xmi; (2) If X; Y 2

hX1; : : : ;Xmi, then X = x1X1 + ¢ ¢ ¢ + xmXm and Y = y1X1 + ¢ ¢ ¢ + ymXm,

so

X + Y = (x1X1 + ¢ ¢ ¢ + xmXm) + (y1X1 + ¢ ¢ ¢ + ymXm)

= (x1 + y1)X1 + ¢ ¢ ¢ + (xm + ym)Xm 2 hX1; : : : ;Xmi:

(3) If X 2 hX1; : : : ;Xmi and t 2 F, then

X = x1X1 + ¢ ¢ ¢ + xmXm

tX = t(x1X1 + ¢ ¢ ¢ + xmXm)

= (tx1)X1 + ¢ ¢ ¢ + (txm)Xm 2 hX1; : : : ;Xmi:

For example, if A 2 Mm£n(F), the subspace generated by the columns of A

is an important subspace of Fm and is called the column space of A. The

column space of A is denoted here by C(A). Also the subspace generated

by the rows of A is a subspace of Fn and is called the row space of A and is

denoted by R(A).

EXAMPLE 3.2.3 For example Fn = hE1; : : : ;Eni, where E1; : : : ;En are

the n{dimensional unit vectors. For if X = [x1; : : : ; xn]t 2 Fn, then X =

x1E1 + ¢ ¢ ¢ + xnEn.

EXAMPLE 3.2.4 Find a spanning family for the subspace S of R3 de¯ned

by the equation 2x ¡ 3y + 5z = 0.

3.2. SUBSPACES OF FN 57

Solution. (S is in fact the null space of [2; ¡3; 5], so S is indeed a subspace

of R3.)

If [x; y; z]t 2 S, then x = 3

2y ¡ 5

2z. Then

2

4

x

y

z

3

5 =

2

4

3

2y ¡ 5

2z

y

z

3

5 = y

2

4

3

21

0

3

5 + z

2

4 ¡5

2

0

1

3

5

and conversely. Hence [ 3

2 ; 1; 0]t and [¡5

2 ; 0; 1]t form a spanning family for

S.

The following result is easy to prove:

LEMMA 3.2.1 Suppose each of X1; : : : ;Xr is a linear combination of

Y1; : : : ; Ys. Then any linear combination of X1; : : : ;Xr is a linear combi-

nation of Y1; : : : ; Ys.

As a corollary we have

THEOREM 3.2.1 Subspaces hX1; : : : ;Xri and hY1; : : : ; Ysi are equal if

each of X1; : : : ;Xr is a linear combination of Y1; : : : ; Ys and each of Y1; : : : ; Ys

is a linear combination of X1; : : : ;Xr.

COROLLARY 3.2.1 Subspaces hX1; : : : ;Xr; Z1; : : : ;Zti and hX1; : : : ;Xri

are equal if each of Z1; : : : ;Zt is a linear combination of X1; : : : ;Xr.

EXAMPLE 3.2.5 If X and Y are vectors in Rn, then

hX; Y i = hX + Y; X ¡ Y i:

Solution. Each of X + Y and X ¡ Y is a linear combination of X and Y .

Also

X =

1

2

(X + Y ) +

1

2

(X ¡ Y ) and Y =

1

2

(X + Y ) ¡

1

2

(X ¡ Y );

so each of X and Y is a linear combination of X + Y and X ¡ Y .

There is an important application of Theorem 3.2.1 to row equivalent

matrices (see De¯nition 1.2.4):

THEOREM 3.2.2 If A is row equivalent to B, then R(A) = R(B).

Proof. Suppose that B is obtained from A by a sequence of elementary row

operations. Then it is easy to see that each row of B is a linear combination

of the rows of A. But A can be obtained from B by a sequence of elementary

operations, so each row of A is a linear combination of the rows of B. Hence

by Theorem 3.2.1, R(A) = R(B).

58 CHAPTER 3. SUBSPACES

REMARK 3.2.1 If A is row equivalent to B, it is not always true that

C(A) = C(B).

For example, if A =

·

1 1

1 1

¸

and B =

·

1 1

0 0

¸

, then B is in fact the

reduced row{echelon form of A. However we see that

C(A) =

¿·

1

1

¸

;

·

1

1

¸À

=

¿·

1

1

¸À

and similarly C(B) =

¿·

1

0

¸À

.

Consequently C(A) 6= C(B), as

·

1

1

¸

2 C(A) but

·

1

1

¸

62 C(B).

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