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 π are linearly independent in { π£ π } π π = 1 and π , then π = s p a n β‘ { π π } π π = 1 . π β€ π Let
and π΄ = { π£ π } π π = 1 We iterate the following steps, starting with π΅ = { π π } π π = 1 and incrementing until exhaustion: π = 1
- Move a vector
out of π£ π into π΄ . π΅ - Since
, s p a n β‘ ( π΅ ) = π is a linear combination of other elements of π£ π , so one of the π΅ can be removed from π π and still π΅ . Without loss of generality by reΓ―ndexing we remove s p a n β‘ ( π΅ ) = π from π π . π΅ remains linearly independent. π΄ If
, we eventually exhaust all π < π and π π and π΄ will partition π΅ . But then { π£ π } π π = 1 , i.e. some π΄ β s p a n β‘ ( π΅ ) are linear combinations of other π£ π , which is a contradiction. Therefore, π£ π . π β€ π It follows immediately that if a vector space
has any finite spanning set, then any two bases of π have the same size. π Now consider
with no finite spanning set. Let π and B = { π π } π β πΌ be two distinct bases for C = { π π } π β π½ . Then any π can be written as a finite linear combination of π π with nonzero coΓ«fficients, say π π π π = β π β π π π π π π but because
is a basis, it follows C β π β π½ π π = πΌ for if the vectors in
can be expressed as finite linear combinations of vectors in a proper subset C , then B β² β B , which is a contradiction. Since π = s p a n β‘ B β² for all | π π | < β΅ 0 , it follows from Upper bound on the cardinality of an arbitrary union that π β π½ | B | = | πΌ | β€ β΅ 0 | C | = | C | By the same token
thus by the SchrΓΆder-Bernstein theorem | C | β€ | B | . | B | = | C |
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 d i m β β‘ β = 1 . If no ground field is specified, assume the topical field d i m β β‘ β = 2 . π
Footnotes
-
sometimes dimensionality, especially in plural since dimensions is confusing β©
-
2008. Advanced Linear Algebra, pp. 48ff. β©
-
2008. Advanced Linear Algebra, pp. 63β64 β©