Linear algebra MOC

Dimension of a vector space

The dimension¹ of a vector space is the cardinality of any basis for that space. linalg Thus all possible bases for a vector space have the same cardinality.²

Proof

Let be a vector space. First we show that if are linearly independent in and , then .

Let and We iterate the following steps, starting with and incrementing until exhaustion:

1. Move a vector out of into .
2. Since , is a linear combination of other elements of , so one of the can be removed from and still . Without loss of generality by reïndexing we remove from . remains linearly independent.

If , we eventually exhaust all and and will partition . But then , i.e. some are linear combinations of other , which is a contradiction. Therefore, .

It follows immediately that if a vector space has any finite spanning set, then any two bases of have the same size.

Now consider with no finite spanning set. Let and be two distinct bases for . Then any can be written as a finite linear combination of with nonzero coëfficients, say

but because is a basis, it follows

for if the vectors in can be expressed as finite linear combinations of vectors in a proper subset , then , which is a contradiction. Since for all , it follows from Upper bound on the cardinality of an arbitrary union that

By the same token thus by the Schröder-Bernstein theorem .

The vector spaces of a given dimension form an Isomorphism class.³

Vector spaces with multiple dimensionalities

Unless a vector space is over a prime field, there are typically multiple dimensionalities assignable to a vector space, depending on which ground field is being considered. This is distinguished by a subscript, for example but . If no ground field is specified, assume the topical field .

---

tidy | sembr | en

Footnotes

1. sometimes dimensionality, especially in plural since dimensions is confusing ↩

2. 2008. Advanced Linear Algebra, pp. 48ff. ↩

3. 2008. Advanced Linear Algebra, pp. 63–64 ↩