# MATH 323 Lecture 12

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Substitute lecture

## Contents

## Chapter 3.4: Basis and Dimension

### Definitions

We say that vectors are **linearly independendent** if

We say that span if any can be written as a linear combination of .

Let be a vector space.

The set is called a **basis** in if

- are linearly independent
- They span

The number of vectors in the basis is called the **dimension** of

### Theorem

All bases in have the same number of elements called the dimension of .

**Note:**it is possible for a "finitely defined" Vector space to have infinite dimension.

### Theorem 3.4.1

If is a spanning set of , , then any vectors are linearly independent.

*Proof.* (substitution/elimination process).

If , is a 0 vector, then there is nothing to prove:

If , then . At least one of the 's is nonzero. Without loss of generality, we assume that it is which is nonzero.

Therefore , so is a spanning set for .

Repeat the process with the above spanning set and . When we can no longer continue, we end up with as a spanning set for , and

#### Example

has bases , and , .

Let , , and

**Note:**is a spanning set of .

…

#### Corollary 3.4.2

If and are two bases in , then .

*Proof.* Assume 's are spanning set and , then from Thm. 3.4.1, we know that are linearly dependent. Not a basis (contradiction)

Flip and by same logic → contradiction. Therefore .

### Infinite Dimension

If the vector space has no basis of finitely many vectors, we say that has infinite dimension.

If , we say that

### Example

Find the dimension of the space of solutions to the system

Three variables, two constraints:

Note that the solutions form a vector space, and the set of solutions will be the nullspace of the coefficient matrix .

The solution is of the form , so the basis has only one vector. Therefore

### Example

Find the dimension of the nullspace for the operation over

In other words, does . For a polynomial of degree ≤ 1, . If , then , and is anything. Therefore, , the basis is , and .

### Example

A basis in would be:

### Theorem 3.4.3

Let be a vector space, and . Then

- any linearly independent vectors span
- any vectors that span are linearly independent.

*Proof.*

- assume they do not span : there is a vector which is not a linear combination of 's, then would be linearly independent. this contradicts #Theorem 3.4.1 since there are vectors with only in a basis.
- assume they are linearly
*dependent*: then , where are not all 0. This means that one vector could be written as a linear combination of other vectors, so vectors would span . This contradicts the definition of a basis, stating that vectors must be linearly independent.

### Theorem 3.4.4

- No set of fewer than vectors spans
- Any linearly independent vectors could be extended (by more vectors) to form a basis.
- If , , span , we can pare them down to vectors, which would be a basis.