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Basic Analysis – Gently Done Topological Vector Spaces

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5. <strong>Vector</strong> <strong>Spaces</strong><br />

We shall collect together here some of the basic algebraic results we will need<br />

concerning vector spaces. Wewish toconsider vector spaces either over R, thefield<br />

of real numbers, or over C, the complex numbers, so for notational convenience,<br />

we use the symbol K to stand for either R or C. It is often immaterial which of<br />

these fields of scalars is used, but when it is important we will indicate explicitly<br />

which is meant.<br />

Definition 5.1 A finite set of elements x 1 ,...,x n in a vector space X over K is<br />

said to be linearly independent if and only if<br />

α 1 x 1 +···+α n x n = 0<br />

with α 1 ,...,α n ∈ K implies that α 1 = ··· = α n = 0. A subset A in a vector space<br />

is said to be linearly independent if and only if each finite subset of A is.<br />

Definition 5.2 A linearly independent subset A in a vector space X is called a<br />

Hamel basis of X if and only if any non-zero element x ∈ X can be written as<br />

x = α 1 u 1 +···+α m u m<br />

for some m ∈ N, non-zero α 1 ,...,α m ∈ K and distinct elements u 1 ,...,u m ∈ A.<br />

In other words, A is a Hamel basis of X if it is linearly independent and if any<br />

element of X can be written as a finite linear combination of elements of A.<br />

Note that if A is a linearly independent subset of X and if x ∈ X can be<br />

written as x = α 1 u 1 +···+α m u m , as above, then this representation is unique.<br />

To see this, suppose that we also have that x = β 1 v 1 + ···+β k v k , for non-zero<br />

β 1 ,...,β k ∈ C and distinct elements v 1 ,...,v k ∈ A. Taking the difference, we get<br />

0 = α 1 u 1 +···+α m u m −β 1 v 1 −···−β k v k .<br />

Suppose that m ≤ k. Now v 1 is not equal to any of the other v j ’s and so,<br />

by independence, cannot also be different from all the u i ’s. In other words, v 1<br />

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