3.4 Basis of a subspace

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We now come to the important concept of basis of a vector subspace.

DEFINITION 3.4.1 Vectors X1; : : : ;Xm belonging to a subspace S are

said to form a basis of S if

(a) Every vector in S is a linear combination of X1; : : : ;Xm;

(b) X1; : : : ;Xm are linearly independent.

Note that (a) is equivalent to the statement that S = hX1; : : : ;Xmi as we

automatically have hX1; : : : ;Xmi µ S. Also, in view of Remark 3.3.1 above,

(a) and (b) are equivalent to the statement that every vector in S is uniquely

expressible as a linear combination of X1; : : : ;Xm.

EXAMPLE 3.4.1 The unit vectors E1; : : : ;En form a basis for Fn.

62 CHAPTER 3. SUBSPACES

REMARK 3.4.1 The subspace f0g, consisting of the zero vector alone,

does not have a basis. For every vector in a linearly independent family

must necessarily be non{zero. (For example, if X1 = 0, then we have the

non{trivial linear relation

1X1 + 0X2 + ¢ ¢ ¢ + 0Xm = 0

and X1; : : : ;Xm would be linearly dependent.)

However if we exclude this case, every other subspace of F n has a basis:

THEOREM 3.4.1 A subspace of the form hX1; : : : ;Xmi, where at least

one of X1; : : : ;Xm is non{zero, has a basis Xc1 ; : : : ;Xcr , where 1 · c1 <

¢ ¢ ¢ < cr · m.

Proof. (The left{to{right algorithm). Let c1 be the least index k for which

Xk is non{zero. If c1 = m or if all the vectors Xk with k > c1 are linear

combinations of Xc1 , terminate the algorithm and let r = 1. Otherwise let

c2 be the least integer k > c1 such that Xk is not a linear combination of

Xc1 .

If c2 = m or if all the vectors Xk with k > c2 are linear combinations

of Xc1 and Xc2 , terminate the algorithm and let r = 2. Eventually the

algorithm will terminate at the r{th stage, either because cr = m, or because

all vectors Xk with k > cr are linear combinations of Xc1 ; : : : ;Xcr .

Then it is clear by the construction of Xc1 ; : : : ;Xcr , using Corollary 3.2.1

that

(a) hXc1 ; : : : ;Xcr i = hX1; : : : ;Xmi;

(b) the vectors Xc1 ; : : : ;Xcr are linearly independent by the left{to{right

test.

Consequently Xc1 ; : : : ;Xcr form a basis (called the left{to{right basis) for

the subspace hX1; : : : ;Xmi.

EXAMPLE 3.4.2 Let X and Y be linearly independent vectors in Rn.

Then the subspace h0; 2X; X; ¡Y; X +Y i has left{to{right basis consisting

of 2X; ¡Y .

A subspace S will in general have more than one basis. For example, any

permutation of the vectors in a basis will yield another basis. Given one

particular basis, one can determine all bases for S using a simple formula.

This is left as one of the problems at the end of this chapter.

We settle for the following important fact about bases:

3.4. BASIS OF A SUBSPACE 63

THEOREM 3.4.2 Any two bases for a subspace S must contain the same

number of elements.

Proof. For if X1; : : : ;Xr and Y1; : : : ; Ys are bases for S, then Y1; : : : ; Ys

form a linearly independent family in S = hX1; : : : ;Xri and hence s · r by

Theorem 3.3.2. Similarly r · s and hence r = s.

DEFINITION 3.4.2 This number is called the dimension of S and is

written dim S. Naturally we de¯ne dim f0g = 0.

It follows from Theorem 3.3.1 that for any subspace S of F n, we must have

dim S · n.

EXAMPLE 3.4.3 If E1; : : : ;En denote the n{dimensional unit vectors in

Fn, then dim hE1; : : : ;Eii = i for 1 · i · n.

The following result gives a useful way of exhibiting a basis.

THEOREM 3.4.3 A linearly independent family of m vectors in a sub-

space S, with dim S = m, must be a basis for S.

Proof. Let X1; : : : ;Xm be a linearly independent family of vectors in a

subspace S, where dim S = m. We have to show that every vector X 2 S is

expressible as a linear combination of X1; : : : ;Xm. We consider the following

family of vectors in S: X1; : : : ;Xm; X. This family contains m+1 elements

and is consequently linearly dependent by Theorem 3.3.2. Hence we have

x1X1 + ¢ ¢ ¢ + xmXm + xm+1X = 0; (3.1)

where not all of x1; : : : ; xm+1 are zero. Now if xm+1 = 0, we would have

x1X1 + ¢ ¢ ¢ + xmXm = 0;

with not all of x1; : : : ; xm zero, contradictiong the assumption that X1 : : : ;Xm

are linearly independent. Hence xm+1 6= 0 and we can use equation 3.1 to

express X as a linear combination of X1; : : : ;Xm:

X = ¡x1

xm+1

X1 + ¢ ¢ ¢ + ¡xm

xm+1

Xm:

64 CHAPTER 3. SUBSPACES

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