Section 3.2

Basic Concepts of Vector Spaces

>

> with(linalg):

```Warning, the protected names norm and trace have been redefined and unprotected
```

>

Linear Combinations, Spans, and Subspaces

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Definition 3.2

Given vectors ,..., in a vector space V and scalars , ...., in R

the vector

+ . . . +

is a linear combination of the vectors ,..., with scalar coefficients

, ...., .

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Example 1: Given the set S = { x , , , 5 x -6}. Linear combination of

the vectors in S is given as .

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Definition 3.3

Let X be a subset of a vector space V. The span of X is the set of all linear combinations

of vectors in X, and is denoted by sp(X). If X is a finite set, so that X={ ,..., } then

we also write sp(X) as sp( ,..., ). If W = sp(X), the vectors in X span or generate W.

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Recall spans in Section 1.1

Example 2: Let

>

> W = `span`(matrix(2,2,[1,0,0,0]),matrix(2,2,[0,1,0,0]));

>

Example 3: Let S = { ,..., }. Then the sp(S) = , i.e. the set of all polynomials of

degree <= n ( p(x) = + . . . + ).

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Definition 3.4

A subset W of a vector space V is a subspace of V if W itself fulfills the requirements of a vector

space, where addition and scalar multiplication of vectors in W produce vector the same vectors

as these operations did in V.

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Recall subspaces in Section 1.6

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Theorem 3.2 (Test for a Subspace)

A subset W of a vector space V is a subspace of V if and only if W

1.) W is non-empty

2.) If v and w are in W, then v+w is in W. (Closure under vector addition.)

3.) If r is a scalar in R and v is in W, then rv is in W. (Closure under scalar multiplication.)

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Example 4: Let S be the set of all polynomials of degree <=n such that p(0) = 0. Then S is

a subspace of . Why? (Usual addition and scalar multiplication.)

Example 5: Let S be the set of all continuous functions on [a,b] such that f(a)=f(b). Then S

is a subspace of C[a,b]. Why? (Usual addition and scalar multiplication.)

Example 6: Let S be the set of all polynomials of degree <= 3 with integer coefficients. Then

S is not a subspace of . Why? (Usual addition and scalar multiplication.)

Linear Independence

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Definition 3.5

Let X be a set of vectors in a vector space V. A dependence relation in

this set X is an equation of the form

+ . . . + some

where in V for i=1,2,...,k. If such a dependence relation exists, then X

is a linearly dependent set of vectors. Otherwise, the set X of vectors is

linearly independent .

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Example 7: Let

>

> W = `span`(matrix(2,2,[1,1,0,2]),matrix(2,2,[-1,1,0,2]),matrix(2,2,[1,1,1,1]));

Are the vectors in W linearly independent or linearly dependent?

We must solve the following to determine if the only solution for is = .

>

> r[1]*matrix(2,2,[1,1,0,2])+r[2]*matrix(2,2,[-1,1,0,2])+r[3]*matrix(2,2,[1,1,1,1]) = matrix(2,2,[0,0,0,0]);

>

You have the following system of equations,

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> r[1] -r[2] + r[3] = 0; r[1]+r[2]+r[3] = 0; r[3]=0; 2*r[1]+2*r[2]+r[3]=0;

>

or,

>

> matrix(4,3,[1,-1,1,1,1,1,0,0,1,2,2,1])*matrix(3,1,[r[1],r[2],r[3]]) = matrix(4,1,[0,0,0,0]);

>

Augmented matrix,

>

> Aug:= matrix(4,4,[1,-1,1,0,1,1,1,0,0,0,1,0,2,2,1,0]);

>

Which has the following solution,

>

> rref(Aug);

>

Example 8: Let S = { }. Are the vectors in S linearly independent

or linearly dependent?

We must solve the following to determine if the only solution for is = .

( NOTE : This equation must hold FOR ALL x!!!!!)

>

> r1*(1+2*x^2)+r2*(4+x+5*x^2)+r3*(3+2*x) = 0;

> p:= r1*(1+2*x^2)+r2*(4+x+5*x^2)+r3*(3+2*x):

>

Equate coefficients on both sides of the equation to get,

>

> coeff(p,x,0)=0; coeff(p,x,1)=0; coeff(p,x,2)=0;

>

This produces the following linear system,

>

> matrix(3,3,[1,4,3,0,1,2,2,5,0])*matrix(3,1,[r1,r2,r3]) = matrix(3,1,[0,0,0]);

>

Augmented matrix,

>

> Aug := augment(matrix(3,3,[1,4,3,0,1,2,2,5,0]),matrix(3,1,[0,0,0]));

>

Which has the following solution,

>

> rref(Aug);

>

Example 9: Let S = {cos(x), sin(x)}. Are the vectors in S linearly independent

or linearly dependent?

The equation,

>

> r*cos(x) + s*sin(x) = 0;

>

must be true for all values of x. Therefore, if I pick two values of x, I will be able to determine

if there exist constant r and s not equal to zero that solve the equation. If the vectors are linealy

dependent then nonzero constants must exist for all choices of x.

Let x = 0, then

>

> r*cos(0) + s*sin(0) = 0;

>

Let x = , then

>

> r*cos(Pi/2) + s*sin(Pi/2) = 0;

>

The only solution is r=s=0. Which implies sin(x) and cos(x) are linearly independent.

Basis

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Definition 3.6

Let V be a vector space. A set of vectors in V is a basis for V if the following conditions

are met:

1.) The set of vectors spans V.

2.) The set is linearly independent.

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Example 10: Determine a basis for all 2 x 2 matrices.

>

> S = `span`(matrix(2,2,[1,0,0,0]),matrix(2,2,[0,1,0,0]),matrix(2,2,[0,0,1,0]),matrix(2,2,[0,0,0,1]));

>

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Definition 3.7

Let V be a finitely generated vector space. The number of elements in a basis for V is the

dimension of V, and is denoted by dim(V).

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Recall dimension in Section 2.1

Example 10 (revisited): What is the dimension of S the set of all 2 x 2 matrices?

Example 11: What is the dim( ) ?

Exercises

1, 3, 5, 8, 11, 13, 25.