Dimension theorem for vector spaces

From Wikipedia, the free encyclopedia
Jump to: navigation, search

In mathematics, the dimension theorem for vector spaces states that all bases of a vector space have equally many elements. This number of elements may be finite, or given by an infinite cardinal number, and defines the dimension of the space.

Formally, the dimension theorem for vector spaces states that

Given a vector space V, any two linearly independent generating sets (in other words, any two bases) have the same cardinality.

If V is finitely generated, then it has a finite basis, and the result says that any two bases have the same number of elements.

While the proof of the existence of a basis for any vector space in the general case requires Zorn's lemma and is in fact equivalent to the axiom of choice, the uniqueness of the cardinality of the basis requires only the ultrafilter lemma,[1] which is strictly weaker (the proof given below, however, assumes trichotomy, i.e., that all cardinal numbers are comparable, a statement which is also equivalent to the axiom of choice). The theorem can be generalized to arbitrary R-modules for rings R having invariant basis number.

The theorem for finitely generated case can be proved with elementary arguments of linear algebra, and requires no forms of the axiom of choice.


Assume that { ai: iI } and { bj: jJ } are both bases, with the cardinality of I bigger than the cardinality of J. From this assumption we will derive a contradiction.

Case 1[edit]

Assume that I is infinite.

Every bj can be written as a finite sum

b_j = \sum_{i\in E_j} \lambda_{i,j} a_i , where E_j is a finite subset of I.

Since the cardinality of I is greater than that of J and the Ej's are finite subsets of I, the cardinality of I is also bigger than the cardinality of \bigcup_{j\in J} E_j. (Note that this argument works only for infinite I.) So there is some i_0\in I which does not appear in any E_j. The corresponding a_{i_0} can be expressed as a finite linear combination of b_j's, which in turn can be expressed as finite linear combination of  a_i's, not involving a_{i_0}. Hence  a_{i_0} is linearly dependent on the other a_i's.

Case 2[edit]

Now assume that I is finite and of cardinality bigger than the cardinality of J. Write m and n for the cardinalities of I and J, respectively. Every ai can be written as a sum

a_i = \sum_{j\in J} \mu_{i,j} b_j

The matrix  (\mu_{i,j}: i\in I, j\in J) has n columns (the j-th column is the m-tuple  (\mu_{i,j}: i\in I)), so it has rank at most n. This means that its m rows cannot be linearly independent. Write r_i = (\mu_{i,j}: j\in J) for the i-th row, then there is a nontrivial linear combination

 \sum_{i\in I}  \nu_i r_i = 0

But then also \sum_{i\in I} \nu_i a_i = \sum_{i\in I} \nu_i \sum_{j\in J} \mu_{i,j} b_j = \sum_{j\in J} \biggl(\sum_{i\in I} \nu_i\mu_{i,j} \biggr) b_j = 0, so the  a_i are linearly dependent.

Alternative Proof[edit]

The proof above uses several non-trivial results. If these results are not carefully established in advance, the proof may give rise to circular reasoning. Here is a proof of the finite case which requires less prior development.

Theorem 1: If A = (a_1,\dots,a_n) \subseteq V is a linearly independent tuple in a vector space V, and B_0 = (b_1,...,b_r) is a tuple that spans V, then n\leq r.[2] The argument is as follows:

Since B_0 spans V, the tuple (a_1,b_1,\dots,b_r) also spans. Since a_1\neq 0 (because A is linearly independent), there is at least one t \in \{1,\ldots,r\} such that b_{t} can be written as a linear combination of B_1 = (a_1,b_1,\dots,b_{t-1}, b_{t+1}, ... b_r). Thus, B_1 is a spanning tuple, and its length is the same as B_0's.

Repeat this process. Because A is linearly independent, we can always remove an element from the list B_i which is not one of the a_j's that we prepended to the list in a prior step (because A is linearly independent, and so there must be some nonzero coefficient in front of one of the b_i's). Thus, after n iterations, the result will be a tuple B_n = (a_1, \ldots, a_n, b_{m_1}, \ldots, b_{m_k}) (possibly with k=0) of length r. In particular, A \subseteq B_n, so |A| \leq |B_n|, i.e., n \leq r.

To prove the finite case of the dimension theorem from this, suppose that V is a vector space and S = \{v_1, \ldots, v_n\} and T = \{w_1, \ldots, w_m\} are both bases of V. Since S is linearly independent and T spans, we can apply Theorem 1 to get m \geq n. And since T is linearly independent and S spans, we get n \geq m. From these, we get m=n.

Kernel extension theorem for vector spaces[edit]

This application of the dimension theorem is sometimes itself called the dimension theorem. Let


be a linear transformation. Then

dim(range(T)) + dim(kernel(T)) = dim(U),

that is, the dimension of U is equal to the dimension of the transformation's range plus the dimension of the kernel. See rank-nullity theorem for a fuller discussion.


  1. ^ Howard, P., Rubin, J.: "Consequences of the axiom of choice" - Mathematical Surveys and Monographs, vol 59 (1998) ISSN 0076-5376.
  2. ^ S. Axler, "Linear Algebra Done Right," Springer, 2000.