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[parent] all bases for a vector space have the same cardinality (Result)

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). If $ B$ is infinite, 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 independent and $ B$ spans $ V$, then $ \vert A\vert\le \vert B\vert$.
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 $ \vert A\vert\le \vert C\vert=\vert D\vert\le \vert B\vert$.

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

$\displaystyle B=\lbrace b_1,\ldots, b_m \rbrace$
and assume $ m<n$. So $ a_i\ne 0$ for all $ i=1,\ldots, 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
$\displaystyle B_1=\lbrace a_1, b_2,\ldots, b_m\rbrace.$
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
$\displaystyle B_2=\lbrace a_1, a_2, b_3, \ldots, b_m\rbrace$
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
$\displaystyle B_m = \lbrace a_1, \ldots, a_m\rbrace$
which is a proper subset of $ A$. In addition, $ B_m$ spans $ V$. But this would imply that $ A$ is linearly dependent, a contradiction. $ \qedsymbol$

Now we can complete the proof of theorem 1.

Proof. Suppose $ A$ and $ B$ are bases for $ V$. We apply the lemma. Then $ \vert A\vert\le \vert B\vert$ since $ A$ is linearly independent and $ B$ spans $ V$. Similarly, $ \vert B\vert\le \vert A\vert$ since $ B$ is linearly independent and $ A$ spans $ V$. An application of Schroeder-Bernstein theorem completes the proof. $ \qedsymbol$



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Cross-references: Schroeder-Bernstein theorem, application, proof, complete, imply, addition, proper subset, contradiction, linearly dependent, field, coefficients, expression, linear combination, clear, superset, spans, linearly independent, subsets, property, finite, infinite, basis, cardinality, bases, vector space, property of bases
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This is version 3 of all bases for a vector space have the same cardinality, born on 2008-06-05, modified 2008-06-05.
Object id is 10660, canonical name is AllBasesForAVectorSpaceHaveTheSameCardinality.
Accessed 189 times total.

Classification:
AMS MSC: 15A03 (Linear and multilinear algebra; matrix theory :: Vector spaces, linear dependence, rank)
 13C05 (Commutative rings and algebras :: Theory of modules and ideals :: Structure, classification theorems)
 16D40 (Associative rings and algebras :: Modules, bimodules and ideals :: Free, projective, and flat modules and ideals)
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