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Basis and DimensionIn the preceding section, we observed that
e1 = (1,0,0,...,0)
e2 = (0,1,0,...,0)
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·
·
en = (0,0,0,...,1)
is a set of generators for the vector space Fn. Thus, every element v of Fn can be written in the form (1)
v = iei, i F.
Actually, the set S of generators has a very special property: The elements i in (1) are uniquely determined. Indeed,
iei = (1, 2, ..., n),
so that if (1) holds, and Definition 1: Let V be a vector space over F. A basis of V is a subset {ei} of V (finite or infinite) with the property that every element v of V can be uniquely written in the form
v = iei.
(Here we always assume that if {ei} is infinite, then Example 1: {ei, ..., en} is a basis for Fn. Example 2: {e1 - e2,e1 + e2} is a basis for F2 since
v = (v1 - v2)/2(e1 - e2) + (v1 + v2)/2(e1 + e2),
and if The main goal of this section is to show that any two bases of a vector space V have the same number of elements, in the following sense: If one basis of V has a finite number of elements, then every basis has a finite number of elements, and every basis has the same number. If one basis of V is infinite, then every basis is infinite. Assuming this result for the moment, let us make the following definition.
Definition 2: Let V be a vector space. If V has a finite set of generators, then we say that V is finite-dimensional and we define its dimension, denoted For example, let Assume no that
v = v1 + ... + nvn (i F).
We assert that this representation is unique. For if
v1 = 2v2 + ... + nvn,
where
w = 1v1 + ... + nvn, (i F)
= (2 + 12)v2 + ... + (n + 1n)vn,
so that w can be written as a linear combination of v2, ..., vn. Then, since
Definition 3: Let
1s1 + 2s2 + ...+ nsn = 0.
If S is not linearly dependent, then we say that S is linearly independent. Example 3: Let
(-2) · e1 + 1 · (2e1) = 0.
Example 4: Let V be arbitrary and let Example 5: Let V be arbitrary and let S be a basis of V. Then S is linearly independent, since if
1s1 + 2s2 + ... + nsn = 0, i F, si S,
then since
0 = 0 · s1 + 0 · s2 + ... + 0 · sn,
the fact that 0 is uniquely representable in terms of the basis implies that 1 = 2 = ... = n = 0.
Proposition 4: Let V be a vector space and let Proof: If S is a basis, then S is clearly a set of generators for V. Moreover, S is linearly independent by Example 5 above.
Let
v = ss,
where all but a finite number of the s are 0. Let us show that the representation (2) is unique. Indeed if
v = ss, s F,
then
0 = v - v = (s - s)s.
But since S is linearly independent, we see that Theorem 5: Let V be a vector space. Then any two bases of V contain the same number of elements. Proof: Let us consider the special case where V has a finite basis {e1, ..., en}. We will prove that if {fi}iI is any other basis of V, then it is finite and contains n elements. Consideration of the special case suffices, for then if V has one infinite basis, then the special case implies that all bases must be infinite. Since {fi}iI is linearly independent, all the fi are nonzero. Let f1 be one of the fi. Since {e1, ..., en} is a basis for V, there exist
f1 = e1 + ... + nen.
Since f1 0, not all j are zero. Let us renumber the ej so that
e1 = 1-1f1 - 1-12e2 - ... - 1-1nen.
Let us show that {f1,e2,...,en} is a basis of V. Let
v = 1e1 + ... + nen.
Therefore, by (3),
v = 1'f1 + 2'e2 + ... + n'fn.
Thus, {f1,e2,...,en} is a set of generators for V. If {f1,e2,...,en} is linearly dependent, then there exist
1f1 + 2e2 + ... + nen = 0,
where 1,...,n are not all 0. If
1e1 + (2 + 2)e2 + ... + (n + n)en = 0, 1 0,
which contradicts the fact that {e1,...,en} are linearly independent. The contradiction proves that {f1,e2...,en} is linearly independent, so that by Proposition 4, {f1,e2...,en} is a basis of V. If
f2 = 1f1 + 2e2 + ... + nen,
1,...,n not all 0. If Remarks: 1. It is possible to strengthen Theorem 5 in case V has infinite dimension. Namely, it is possible to show that not only are any two bases of V is infinite, but given any two bases of V, there exists a bijection of one of these into the other. 2. It is not clear from what we have said that any vector space has a basis. In order to prove this assertion, it is necessary to appeal to Zorn's lemma. If V is finite-dimensional, then the existence of a basis of V was proved. In most of what follows, we will be concerned only with finite-dimensional vector spaces, so we will not give proof of the existence of a basis in the most general case. |
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