all bases for a vector space have the same cardinality

In this entry, we want to show the following property of bases for a vector spaceMathworldPlanetmath:

Theorem 1.

All bases for a vector space V have the same cardinality.

Let B be a basis for V (B exists, see this link ( If B is infiniteMathworldPlanetmath, then all bases for V have the same cardinality as that of B (proof ( So all we really need to show is where V has a finite basis.

Before proving this important property, we want to prove something that is almost as important:

Lemma 1.

If A and B are subsets of a vector space V such that A is linearly independentMathworldPlanetmath and B spans V, then |A||B|.


If A is finite and B is infinite, then we are done. Suppose now that A is infinite. Since A is linearly independent, there is a supersetMathworldPlanetmath C of A that is a basis for V. Since A is infinite, so is C, and therefore all bases for V are infinite, and have the same cardinality as that of C. Since B spans V, there is a subset D of B that is a basis for V. As a result, we have |A||C|=|D||B|.

Now, we suppose that A and B are both finite. The case where A= is clear. So assume A. As B spans V, B. Let A={a1,,an} and


and assume m<n. So ai0 for all i=1,,n. Since B spans V, a1 can be expressed as a linear combinationMathworldPlanetmath of elements of B. In this expression, at least one of the coefficients (in the field k) can not be 0 (or else a1=0). Rename the elements if possible, so that b1 has a non-zero coefficient in the expression of a1. This means that b1 can be written as a linear combination of a and the remaining b’s. Set


As every element in V is a linear combination of elements of B, it is therefore a linear combination of elements of B1. Thus, B1 spans V. Next, express a2 as a linear combination of elements in B1. In this expression, if the only non-zero coefficient is in front of a1, then a1 and a2 would be linearly dependent, a contradictionMathworldPlanetmathPlanetmath! Therefore, there must be a non-zero coefficient in front of one of the b’s, and after some renaming once more, we have that b2 is the one with a non-zero coefficient. Therefore, b2, likewise, can be expressed as a linear combination of a1,a2 and the remaining b’s. It is easy to see that


spans V as well. Continue this process until all of the b’s have been replaced, which is possible since m<n. We have finally arrived at the set


which is a proper subsetMathworldPlanetmathPlanetmath of A. In addition, Bm spans V. But this would imply that A is linearly dependent, a contradiction. ∎

Now we can completePlanetmathPlanetmathPlanetmath the proof of theorem 1.


Suppose A and B are bases for V. We apply the lemma. Then |A||B| since A is linearly independent and B spans V. Similarly, |B||A| since B is linearly independent and A spans V. An application of Schroeder-Bernstein theorem completes the proof. ∎

Title all bases for a vector space have the same cardinality
Canonical name AllBasesForAVectorSpaceHaveTheSameCardinality
Date of creation 2013-03-22 18:06:51
Last modified on 2013-03-22 18:06:51
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 6
Author CWoo (3771)
Entry type Result
Classification msc 16D40
Classification msc 13C05
Classification msc 15A03