3.3 Linear dependence

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We now recall the de¯nition of linear dependence and independence of a

family of vectors in Fn given in Chapter 2.

DEFINITION 3.3.1 Vectors X1; : : : ;Xm in Fn are said to be linearly

dependent if there exist scalars x1; : : : ; xm, not all zero, such that

x1X1 + ¢ ¢ ¢ + xmXm = 0:

In other words, X1; : : : ;Xm are linearly dependent if some Xi is expressible

as a linear combination of the remaining vectors.

X1; : : : ;Xm are called linearly independent if they are not linearly depen-

dent. Hence X1; : : : ;Xm are linearly independent if and only if the equation

x1X1 + ¢ ¢ ¢ + xmXm = 0

has only the trivial solution x1 = 0; : : : ; xm = 0.

EXAMPLE 3.3.1 The following three vectors in R3

X1 =

2

4

1

2

3

3

5 ; X2 =

2

4 ¡1

1

2

3

5 ; X3 =

2

4 ¡1

7

12

3

5

are linearly dependent, as 2X1 + 3X2 + (¡1)X3 = 0.

3.3. LINEAR DEPENDENCE 59

REMARK 3.3.1 If X1; : : : ;Xm are linearly independent and

x1X1 + ¢ ¢ ¢ + xmXm = y1X1 + ¢ ¢ ¢ + ymXm;

then x1 = y1; : : : ; xm = ym. For the equation can be rewritten as

(x1 ¡ y1)X1 + ¢ ¢ ¢ + (xm ¡ ym)Xm = 0

and so x1 ¡ y1 = 0; : : : ; xm ¡ ym = 0.

THEOREM 3.3.1 A family of m vectors in Fn will be linearly dependent

if m > n. Equivalently, any linearly independent family of m vectors in F n

must satisfy m · n.

Proof. The equation

x1X1 + ¢ ¢ ¢ + xmXm = 0

is equivalent to n homogeneous equations in m unknowns. By Theorem 1.5.1,

such a system has a non{trivial solution if m > n.

The following theorem is an important generalization of the last result

and is left as an exercise for the interested student:

THEOREM 3.3.2 A family of s vectors in hX1; : : : ;Xri will be linearly

dependent if s > r. Equivalently, a linearly independent family of s vectors

in hX1; : : : ;Xri must have s · r.

Here is a useful criterion for linear independence which is sometimes

called the left{to{right test:

THEOREM 3.3.3 Vectors X1; : : : ;Xm in Fn are linearly independent if

(a) X1 6= 0;

(b) For each k with 1 < k · m; Xk is not a linear combination of

X1; : : : ;Xk¡1.

One application of this criterion is the following result:

THEOREM 3.3.4 Every subspace S of Fn can be represented in the form

S = hX1; : : : ;Xmi, where m · n.

60 CHAPTER 3. SUBSPACES

Proof. If S = f0g, there is nothing to prove { we take X1 = 0 and m = 1.

So we assume S contains a non{zero vector X1; then hX1i µ S as S is a

subspace. If S = hX1i, we are ¯nished. If not, S will contain a vector X2,

not a linear combination of X1; then hX1; X2i µ S as S is a subspace. If

S = hX1; X2i, we are ¯nished. If not, S will contain a vector X3 which is

not a linear combination of X1 and X2. This process must eventually stop,

for at stage k we have constructed a family of k linearly independent vectors

X1; : : : ;Xk, all lying in Fn and hence k · n.

There is an important relationship between the columns of A and B, if

A is row{equivalent to B.

THEOREM 3.3.5 Suppose that A is row equivalent to B and let c1; : : : ; cr

be distinct integers satisfying 1 · ci · n. Then

(a) Columns A¤c1 ; : : : ;A¤cr of A are linearly dependent if and only if the

corresponding columns of B are linearly dependent; indeed more is

true:

x1A¤c1 + ¢ ¢ ¢ + xrA¤cr = 0 , x1B¤c1 + ¢ ¢ ¢ + xrB¤cr = 0:

(b) Columns A¤c1 ; : : : ;A¤cr of A are linearly independent if and only if the

corresponding columns of B are linearly independent.

(c) If 1 · cr+1 · n and cr+1 is distinct from c1; : : : ; cr, then

A¤cr+1 = z1A¤c1 + ¢ ¢ ¢ + zrA¤cr , B¤cr+1 = z1B¤c1 + ¢ ¢ ¢ + zrB¤cr :

Proof. First observe that if Y = [y1; : : : ; yn]t is an n{dimensional column

vector and A is m £ n, then

AY = y1A¤1 + ¢ ¢ ¢ + ynA¤n:

Also AY = 0 , BY = 0, if B is row equivalent to A. Then (a) follows by

taking yi = xcj if i = cj and yi = 0 otherwise.

(b) is logically equivalent to (a), while (c) follows from (a) as

A¤cr+1 = z1A¤c1 + ¢ ¢ ¢ + zrA¤cr , z1A¤c1 + ¢ ¢ ¢ + zrA¤cr + (¡1)A¤cr+1 = 0

, z1B¤c1 + ¢ ¢ ¢ + zrB¤cr + (¡1)B¤cr+1 = 0

, B¤cr+1 = z1B¤c1 + ¢ ¢ ¢ + zrB¤cr :

3.4. BASIS OF A SUBSPACE 61

EXAMPLE 3.3.2 The matrix

A =

2

4

1 1 5 1 4

2 ¡1 1 2 2

3 0 6 0 ¡3

3

5

has reduced row{echelon form equal to

B =

2

4

1 0 2 0 ¡1

0 1 3 0 2

0 0 0 1 3

3

5 :

We notice that B¤1; B¤2 and B¤4 are linearly independent and hence so are

A¤1; A¤2 and A¤4. Also

B¤3 = 2B¤1 + 3B¤2

B¤5 = (¡1)B¤1 + 2B¤2 + 3B¤4;

so consequently

A¤3 = 2A¤1 + 3A¤2

A¤5 = (¡1)A¤1 + 2A¤2 + 3A¤4:

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