PROPUESTAS Y EXPERIENCIAS EN LA CONCRECIÓN DE LAS COMPETENCIAS CLAVE

Spanning Set

EXAMPLE3

Spanning Sets and Linear Independence

One of the main ways that subspaces arise is as the set of all linear combinations of some spanning set. We next present an easy theorem and a bit more vocabulary.

If {v 1, ^•••, vk} is a set of vectors in JRn and S is the set of all possible linear combi­

nations of these vectors,

then S is a subspace of ]Rn.

Proof: By properties (1) and (6) of Theorem 1, t1v1 + ^{· · ·} + tkvk E JR.11, so S is a subset of JRn. Taking t; = 0 for 1 ^� i � k, we get

0

= Ov 1 + · · · + Ov k E S, so S is non-empty.

Let x,y ES. Then, for some real numbers s; and t;, 1 � i � k, x = s1v1 +· · ·+skvk andy = t1v1 + ^· · · + tkvk. It follows that

so, x + y E S since (s; + t;) E JR. Hence, S is closed under addition.

Similarly, for all t E JR,

So, S is closed under scalar multiplication. Therefore, S is a subspace of JRn. •

If S is the subspace of JR.11 consisting of all linear combinations of the vectors v 1, ^..•, v k E JR.11, then S is called the subspace spanned by the set of vectors 13 = {v 1, ... , vk}, and we say that the set 13 spans S. The set 13 is called a spanning set for the subspace S.

We denote S by

S = Span{i11, ... , vk} =Span 13

Let v E JR.2 with v ^*

0

and consider the line L with vector equation x = tV, t E JR. Then L is the subspace spanned by {V}, and {V} is a spanning set for L. We write L = Span{V}.

Similarly, for v1, v2 E JR.2, the set M with vector equation x = ti\11 + t2v2 is a subspace oflR2 with spanning set {v1, v2}. That is, M = Span{i11, v2}.

If v E JR.2 with v *

0,

then we can guarantee that Span{v} represents a line in JR.2 that passes through the origin. However, we see that the geometrical interpretation of Span{v1, v2} depends on the choices of v1 and v2. We demonstrate this with some examples.

EXAMPLE4

EXAMPLES

The set

S

1 = Span

{[ �]

^,

[ �]}

has vector equation

Section 1.2 Vectors in JR.11 19

Hence,

S

1 is a line in

JR.2

that passes through the origin.

The set

S 2

= Span

{ [ �]

^,

[-�]}

has vector equation

where

t

=

t1 - 2t2

E JR.. Hence,

S 2

represents the same line as

SI·

That is,

The set

S 3

= Span

{ [ �]

^,

[ �]}

has vector equation

Hence,

S 3

=

JR.2.

That is,

S 3

spans the entire two-dimensional plane.

From these examples, we observe that

{V1, v2}

is a spanning set for

JR.2

if and only if neither v ¹ nor

v2

is a scalar multiple of the other. This also means that neither vector can be the zero vector. We now look at this in

JR.3.

The set

S 1

"Span

{[ =�l

^·

m, [�] }

has vector equation Hence,

EXAMPLES

(continued)

Theorem 3

The set S 2 = Spm

{[ j l

^·

[ _; l

^·

[!]}

has vwor equation

1 =

{�]

⁺

{i]

^{+ t3}

[!]

^• 1,,12,!3 ER which can be written as

1 = t,

[ j l

^{+ !2}

[ _ � l

^{+ !2}

[!]

^{+ !3}

l!l

^{= (t I + 12)}

[

^_

� l

^{+ ( 12 + 13)}

[!]

So, S2 =Span

{[ 3]

^·

[!]}

We extend this to the general case in IR.11•

Let v 1, ... , v k be vectors in IR.11• If v k can be written as a linear combination of V1, ... ,Vk-1, then

Proof: We are assuming that there exists t1, ... , tk-l E IR. such that

Let 1 E Span{v1, ... , vk}. Then, there exists S1, ... , Sk E IR. such that

1 = S1V1 + ... + Sk-lvk-1 + SkVk

= s1v1 + · · · + Sk-1Vk-l + sk(t1v1 + · · · + tk-1Vk-1)

= (s1 + Skt1W1 + · · · + (sk-1 + sktk-1Wk-I

Thus, 1 E Span{\11, ... , vk-d· Hence, Span{vi. ... , vd s;; Span{\11, ... , vk-d· Clearly, we have Span{\11, ... , vk-d s;; Span{v1, ... , vk} and so

as required. •

In fact, any vector which can be written as a linear combination of the other vectors in the set can be removed without changing the spanned set. It is important in linear algebra to identify when a spanning set can be simplified by removing a vector that can be written as a linear combination of the other vectors. We will call such sets linearly dependent. If a spanning set is as simple as possible, then we will call the set linearly independent. To identify whether a set is linearly dependent or linearly independent, we require a mathematical definition.

Definition

Linearly Dependent Linearly Independent

Theorem 4

EXAMPLE6

Section 1.2 Vectors in IR" 21

Assume that the set {V 1, • • • , v k} is linearly dependent. Then one of the vectors, say v;, is equal to a linear combination of some (or all) of the other vectors. Hence, we can find scalars t1, ... tk E IR such that

where t; f. 0. Thus, a set is linearly dependent if the equation

has a solution where at least one of the coefficients is non-zero. On the other hand, if the set is linearly independent, then the only solution to this equation must be when all the coefficients are 0. For example, if any coefficient is non-zero, say t; f. 0, then we can write

Thus, v; E Span{v1, ... , v;_1, V;+1, . • . , v,,}, and so the set can be simplified by using Theorem 3.

We make this our mathematical definition.

A set of vectors { v 1, ... , v k} is said to be linearly dependent if there exist coefficients t1, • • . , tk not all zero such that

A set of vectors {V 1, • • • , v d is said to be linearly independent if the only solution to

is t1 = t2 = · · · = tk = 0. This is called the trivial solution.

If a set of vectors {V 1, ... , v k} contains the zero vector, then it is linearly dependent.

Proof: Assume V; =

0.

Then we have

Hence, the equation

0

^{= t1v1 + · · · +} tk v k has a solution with one coefficient, t;, that is non-zero. So, by definition, the set is linearly dependent. •

show that the set

{[-1 :J . [ 1:Ix4]

^•

[-�]}

is linead y dependent Solution: We consider

EXAMPLE6

(continued)

EXERCISE 3

Definition

Basis

EXAMPLE 7

Using operations on vectors, we get

Since vectors are equal only if their corresponding entries are equal, this gives us three equations in three unknowns

7t1 - lOt2 - t3

⁼ 0

-14t,

₊

15t2

= 0 1

5

6t,

+ 1

4 t2

+

3t3

= 0

Solving using substitution and elimination, we find that there are in fact infinitely many possible solutions. One is

t1

=

�, ^t2

⁼

�, ^t3

⁼ -1. Hence, the set is linearly dependent.

Determine whether

ml

^·

m ^, [ m

is linearly dependent or Ji near I y independent.

Remark

Observe that determining whether a set

{\11, ... 'vk}

in JR.11 is linearly dependent or linearly independent requires determining solutions of the equation

t1 v1

+ · · · +

tkvk

=

0.

However, this equation actually represents n equations (one for each entry of the vectors) ink unknowns

t1,

• • • ,

tk.

In the next chapter, we will look at how to efficiently solve such systems of equations.

What we have derived above is that the simplest spanning set for a subspace S is one that is linearly independent. Hence, we make the following definition.

If

{v,, ... ,vk}

is a spanning set for a subspace S of JR.11 and

{V1, ... ,vk}

is linearly independent, then

{V1,

• • • ,

vk}

is called a basis for S.

Let

"· =

[-H

^{"' =}

r \l· ^v,

⁼

[-H

and let s be the subspace of

JR3

given by S =Span

{V1, i12, v3}.

Find a basis for S.

Solution: Observe that

{V 1, i12, i13}

is linearly dependent, since

EXAMPLE 7

(continued)

Definition

Standard Basis for'.?."

EXAMPLE 8

Section 1.2 Vectors in !fll.11 23

In particular, we can write v3 as a linear combination of v 1 and i12. Hence, by Theorem 3,

Moreover, observe that the only solution to

is t1 = t2 = 0 since neither v 1 nor i12 is a scalar multiple of the other. Hence, {V 1, i12} is linearly independent.

Therefore, {v 1, v2} is linearly independent and spans S and so it is a basis for S.

Bases (the plural of basis) will be extremely important throughout the remainder of the book. At this point, however, we just define the following very important basis.

In !fll.11, let e; represent the vector whose i-th component is 1 and all other components are 0. The set { e1' ... ' e,,} is called the standard basis for !fli.11•

Observe that this definition matches that of the standard basis for !fli.2 _and !fll.3 _given in Section 1.1.

The standard basis for R3 is re,, e,, e, J =

{ [ �]

^•

[!]

^.

[m

It is linearly independent since the only solution to

is 11 = 12 = 13 = 0. Moreover, it is a spanning set for R3 since every vector X =

[�:l

^E

!fll.3 can be written as a linear combination of the basis vectors. In particular,

[�:l

^{= X1}

[�]

^{+ X2}

[!]

^{+ X3}

m

Remark

Compare the result of Example 8 with the meaning of point notation P(a, b, c). When we say P(a, b, c) we mean the point P _having a amount in the x-direction, b _amount in they-direction, and c amount in the z-direction. So, observe that the standard basis vectors represent our usual coordinate axes.

EXERCISE4

Definition

Linein3i."

Definition

Plane in J."

Definition

Hyperplane in R"

EXAMPLE9

EXAMPLE 10

State the standard basis for

JR5.

Prove that it is linearly independent and show that it is a spanning set for

JR5.

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