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 space:

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 (http://planetmath.org/ZornsLemmaAndBasesForVectorSpaces)). If $B$ is infinite, then all bases for $V$ have the same cardinality as that of $B$ (proof (http://planetmath.org/CardinalitiesOfBasesForModules)). 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 independent and $B$ spans $V$, then $\mathrm{|}A\mathrm{|}\mathrm{\le }\mathrm{|}B\mathrm{|}$.

Proof.

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 superset $C$ 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 $|A|\le |C|=|D|\le |B|$.

Now, we suppose that $A$ and $B$ are both finite. The case where $A=\mathrm{\varnothing }$ is clear. So assume $A\ne \mathrm{\varnothing }$. As $B$ spans $V$, $B\ne \mathrm{\varnothing }$. Let $A=\{a_1,\mathrm{\dots },a_n\}$ and

$$B=\{b_1,\mathrm{\dots },b_m\}$$

and assume . So $a_i\ne 0$ for all $i=1,\mathrm{\dots },n$. Since $B$ spans $V$, $a_1$ can be expressed as a linear combination of elements of $B$. In this expression, at least one of the coefficients (in the field $k$) can not be $0$ (or else $a_1=0$). Rename the elements if possible, so that $b_1$ has a non-zero coefficient in the expression of $a_1$. This means that $b_1$ can be written as a linear combination of $a$ and the remaining $b$’s. Set

$$B_1=\{a_1,b_2,\mathrm{\dots },b_m\}.$$

As every element in $V$ is a linear combination of elements of $B$, it is therefore a linear combination of elements of $B_1$. Thus, $B_1$ spans $V$. Next, express $a_2$ as a linear combination of elements in $B_1$. In this expression, if the only non-zero coefficient is in front of $a_1$, then $a_1$ and $a_2$ would be linearly dependent, a contradiction! 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 $b_2$ is the one with a non-zero coefficient. Therefore, $b_2$, likewise, can be expressed as a linear combination of $a_1,a_2$ and the remaining $b$’s. It is easy to see that

$$B_2=\{a_1,a_2,b_3,\mathrm{\dots },b_m\}$$

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

$$B_m=\{a_1,\mathrm{\dots },a_m\}$$

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

Now we can complete the proof of theorem 1.

Proof.

Suppose $A$ and $B$ are bases for $V$. We apply the lemma. Then $|A|\le |B|$ since $A$ is linearly independent and $B$ spans $V$. Similarly, $|B|\le |A|$ 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