Dimension theorem for vector spaces: Information from Answers.com

In mathematics, the dimension theorem for vector spaces states that a vector space has a definite, well-defined number of dimensions. This may be finite, or an infinite cardinal number.

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, the result says that any two bases have the same number of elements.

The cardinality of a basis is called the dimension of the vector space.

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, which is strictly weaker. The theorem can be generalized to arbitrary R-modules for rings R having invariant basis number.

For the finitely generated case it can be done with elementary arguments of linear algebra, requiring no forms of choice.

1 Proof
2 Kernel extension theorem for vector spaces
3 See also
4 References

Proof

Assume that { a_i: i ∈ I } and { b_j: j ∈ J } are both bases, with the cardinality of I bigger than the cardinality of J. From this assumption we will derive a contradiction.

Case 1

Assume that I is infinite.

Every b_j 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 E_j'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

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 a_i 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

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 $A = (a_1,\dots,a_n)$ (considered as an ordered $n$ -tuple) is a basis of a vector space $V$ , and $B_0 = (b_1,...,b_r)\subset V$ is an $r$ -tuple of vectors such that $\operatorname{span}\{b_1,\dots,b_r\}=V$ (a spanning list), then $n\leq r$ .^[1] The argument is as follows:

Since $B 0$ spans $V$ , the list $(a_1,b_1,\dots,b_r)$ is linearly dependent. Since $a_1\neq 0$ (because $A$ is a basis), there is a $t \in \{1,\ldots,r\}$ such that $b t$ can be written as a linear combination of $\{a_1,b_1,\dots,b_{t-1}\}$ . Thus, $\operatorname{span}(\{a_1,b_1,\dots,b_r\}\setminus\{b_{t}\})=V$ . Define $B_1 = (a_1,b_1,\dots,b_{t-1},b_{t+1},\ldots,b_r)$ . Then $B 1$ is a spanning $r$ -tuple (i.e., its length is unchanged from this process).

Repeat this process. Because $A$ is linearly independent, the element that we remove from the list $B i$ as we create $B i + 1$ will never be one of the $a j$ 's that we prepended to the list in a prior step (because no $a j$ is in the span of the other $a$ 's). Thus, after $n$ iterations, the result will be an $r$ -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 a basis and $T$ is spanning, we can apply Theorem 1 to get $m \geq n$ . Since $T$ is a basis and $S$ is spanning, we get $n \geq m$ . From these, we get $m = n$ , which is what we wanted to prove.

Kernel extension theorem for vector spaces

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

T: U → V

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.

References

^ S. Axler, "Linear Algebra Done Right," Springer, 2000.

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Dimension theorem for vector spaces

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