Dimension theorem for vector spaces

From Wikipedia, the free encyclopedia

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.

[edit] 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.

[edit] 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.

[edit] 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.

[edit] Alternative Proof

Warning: The proof above is based on results which are less fundamental than the Theorem itself. A proof based only on the definitions of Linear Independence and Basis is also possible (here is a proof to the final dimentsion case). Otherwise, a circular argument is created (many results in the theory of Matrices are proved from this theorem.) You can prove the following fact: if $\{a_1,\dots,a_n\}$ is a basis of a vector space $V$ , and $\{b_1,...,b_r\}\subset V$ is a subset such that $Span\{b_1,\dots,b_r\}=V$ , then $n\leq r$ .^[1] The argument is as follows: suppose, by means of contradiction, that $n > r$ . Then,

(1) clearly the set $\{a_1,b_1,\dots,b_r\}$ is linearly dependent; since $a_1\neq 0$ , then there is an element $t_1\in\{1,\dots,r\}$ such that $b_{t_1}$ can be written as a linear combination of $\{a_1,b_1,\dots,b_{t_1-1}\}$ ; thus, $Span(\{a_1,b_1,\dots,b_r\}\setminus\{b_{t_1}\})=V$ ;

(2) suppose that $s$ elements from $\{b_1,\dots,b_r\}$ have been replaced by elements of $\{a_1,\dots,a_n\}$ and the resulting set still spans $V$ , i.e., there are $\{t_1,\dots,t_s\}\subseteq\{1,\dots,r\}$ such that $Span(\{a_1,a_2,\dots,a_s,b_1,\dots,b_r\}\setminus\{b_{t_1},\dots,b_{t_s}\})=V$ ; clearly the set $\{a_1,\dots,a_s,a_{s+1},b_1,\dots,b_r\}\setminus\{b_{t_1},\dots,b_{t_s}\}$ is linearly dependent; since $\{a_1,\dots,a_n\}$ is a basis, then there is $t_{s+1}\in\{1,\dots,r\}\setminus\{t_1,\dots,t_s\}$ such that $b_{t_{s+1}}$ can be written as a linear combination of $\{a_1,\dots,a_{s+1},b_1,\dots,b_r\}\setminus\{b_{t_1},\dots,b_{t_{s+1}}\}$ ; thus, $Span(\{a_1,\dots,a_{s+1},b_1,\dots,b_r\}\setminus\{b_{t_1},\dots,b_{t_{s+1}}\})=V$

(3) The procedure above stops after $r$ steps and shows that $Span(\{a_1,\dots,a_{r}\})=V$ ; this is a contradiction because $n > r$ and $\{a_1,\dots,a_n\}$ is a basis of $V$ .

[edit] 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.

[edit] See also

basis (linear algebra)

[edit] References

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

[0] S. Axler, "Linear Algebra Done Right," Springer, 2000.

[1]

Dimension theorem for vector spaces

From Wikipedia, the free encyclopedia

Contents

[edit] Proof

[edit] Case 1

[edit] Case 2

[edit] Alternative Proof

[edit] Kernel extension theorem for vector spaces

[edit] See also

[edit] References

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