**Section 2.1 **

**Independence and Dimension**

`> `
**with(linalg):**

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

`> `

**Definition of Independence**

**************************************************************

A set of vectors
,
,
,.....,
of a vector space V is
** linearly dependent**
if there

exists scalars , , ,...., not all zero such that

+ + + ...... + = 0

that is, at least one of the vectors is a linear combination of the remaining vectors.

The set of vectors is
** linearly independent**
if any zero combination

+ + + ...... + = 0

implies that = = = ... = = 0.

***************************************************************

**Examples of Linear Dependent and Independent Sets**

**Example 1**
: Consider the three vectors

`> `
**v1:=matrix(3,1,[1,2,3]):
v2:=matrix(3,1,[4,3,1]):
v3:=matrix(3,1,[5,5,4]):**

`> `
**evalm(v1),evalm(v2),evalm(v3);**

`> `

Set up the equations.

`> `
**c1*evalm(v1)+c2*evalm(v2)+c3*evalm(v3) = matrix(3,1,[0,0,0]);**

`> `

Notice that this forms a homogenous system of equations.

`> `
**augment(v1,v2,v3)*matrix(3,1,[c1,c2,c3])=matrix(3,1,[0,0,0]);**

`> `

Solve the linear system.

`> `
**rref(augment(v1,v2,v3,matrix(3,1,[0,0,0])));**

`> `

Notice that we can write one of the vectors as a linear combination of the other vectors.

**Example 2**
: Consider the three vectors

`> `
**v1 := matrix(4,1,[1,2,3,1]): v2:= matrix(4,1,[2,2,1,3]): v3:=matrix(4,1,[-1,2,7,3]):**

`> `
**evalm(v1),evalm(v2),evalm(v3);**

Set up the equations.

`> `
**c1*evalm(v1)+c2*evalm(v2)+c3*evalm(v3)= matrix(4,1,[0,0,0,0]);**

`> `

Notice that this forms a homogenous system of equations.

`> `
**augment(v1,v2,v3)*matrix(3,1,[c1,c2,c3])=matrix(4,1,[0,0,0,0]);**

`> `

Solve the linear system.

`> `
**rref(augment(v1,v2,v3,matrix(4,1,[0,0,0,0])));**

`> `

Notice that we can
**NOT **
write one of the vectors
as a linear combination of the other vectors.

**Procedure**

Determining the linear dependence or independence of a set of vectors reduces to solving

a homogeneous system of linear equations. If the homogeneous system has only the trivial

solution, then the set of vectors is linearly independent; otherwise, if the homogeneous

system has infinitely many solutions, the set of vectors is linearly dependent.

To determine the linear dependence or independence of a set of vectors:

1. Write the linear combination of the vectors and set it equal to zero.

2. Write the associated homogeneous system of linear equations.

3. Solve the homogeneous system.

**Definition of Bases**

*************************************************

Let W be a subspace of . A subset { , . . ., }of W is

a
** basis**
for W if and only if the following two conditions are met:

1.) The vectors , . . ., span W

2.) The vectors , . . ., are linearly independent

*************************************************

** Fact 1:**
A basis for a subspace W is not unique.

** Fact 2:**
All basis for W contain the same number of vectors.

**Examples of**
** **
**Bases**

**Example 3**
: Consider the subspace S:

`> `
**v1:=matrix(3,1,[1,2,3]):
v2:=matrix(3,1,[-1,3,1]):
v3:=matrix(3,1,[5,1,4]): v4 := matrix(3,1,[1,2,0]):**

`> `

`> `
**S = span(evalm(v1),evalm(v2),evalm(v3),evalm(v4));**

`> `

Do the vectors form a basis for S?

`> `
**rref(augment(v1,v2,v3,v4,matrix(3,1,[0,0,0])));**

`> `

************************************************************

Finding a basis for W = span{ , . . ., }.

1.) Form a matrix A whose jth column vector is .

2.) Row reduce R to row echelon form H.

3.) The set of all such that the jth column of H contains a pivot is a basis for W.

*************************************************************

`> `
**rref(augment(v1,v2,v3));**

`> `

**Example 4: **
Consider the subspace S:

`> `
**v1:=matrix(5,1,[1,-1,0,2,1]):
v2:=matrix(5,1,[0,-3,2,4,2]):
v3:=matrix(5,1,[2,1,-2,0,0]): v4 := matrix(5,1,[3,3,-4,-2,-1]):**

`> `

`> `
**S = span(evalm(v1),evalm(v2),evalm(v3),evalm(v4));**

`> `

`> `
**rref(augment(v1,v2,v3,v4));**

`> `

**Dimension of a Subspace**

********************************************

__FACTS:__

1.) Let W be a subspace of . Let , . . ., be vectors

in W that span W, and let , . . ., be vectors in W

that are linear independent. Then .

2.) Any two bases of a subspace W of contain the same

number of vectors.

3.) Let W be a subspace of . The number of elements in a basis

for W is the
** dimension**
of W and denoted by dim(W).

********************************************

The vector space has dimension n, i.e. dim( ) = n.

dim({0}) = 0.

**Example 3**
(revisited): Consider the subspace S:

`> `
**v1:=matrix(3,1,[1,2,3]):
v2:=matrix(3,1,[-1,3,1]):
v3:=matrix(3,1,[5,1,4]): v4 := matrix(3,1,[1,2,0]):**

`> `

`> `
**S = span(evalm(v1),evalm(v2),evalm(v3),evalm(v4));**

`> `

What is the dimension of S?

`> `
**rref(augment(v1,v2,v3,v4));**

`> `

**********************************************

__FACTS:__

1.) Every subspace W of has a basis and .

2.) Every independent set of vectors in can be enlarged, if

necessary to become a basis for .

3.) If W is a subspace of and , then

a.) every independent set of k vectors in W is a basis for W and

b.) every set of k vectors in W that spans W is a basis for W.

**************************************************

**Example 5:**
Enlarge the independent set of the two vectors to form a basis for

`> `
**v1 := matrix(3,1,[1,1,-1]); v2:=matrix(3,1,[1,2,-2]);**

`> `
**v3 := matrix(3,1,[?,?,?]);**

`> `

`> `
**rref(augment(v1,v2,v3));**

`> `

**Exercises**

7, 9, 11, 12, 15, 19, 21, 22, 25, 27, 28.