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I know that by Zorn's Lemma we can prove that every vector space has a Hamel Basis. Where Hamel Basis means a maximal Linearly independent set.

My question is, if this is the Finite Dimension case, then we will be able to write any element of the vector space as a linear combination of the Basis vectors.
But is it also true for the infinite Dimension case?

I came across this doubt when reading the last line of this attached picture (Funtional Analysis By J.B. Conway):
So for me it looks like they have written any vector of $X$ as a sum of the Hamel Basis Elements..
Funtional Analysis By J.B. Conway

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2 Answers 2

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Yes, by the definition of Hamel basis. But note that, in this context, each expression of a vector $v$ as a linear combination of elements of the basis only involves finite sums: $v$ can be written as $a_1v_1+\cdots+a_nv_n$, where each $v_1,\ldots,v_n$ is an element of the basis.

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Yes, that is immediate from the definition of a Hamel basis. Every element is a finite linear combination of elments in the Hamel basis.

If a vector $x$ is not a finite linear combination of elements of the Hamel basis $H$ then we can see that $H \cup \{x\}$ would be linearly independent, contradicting the fact that $H$ is a maximal linearly independent set.

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  • $\begingroup$ So is that the definition of Hamel Basis? I thought it was the maximal Iinearly independent vectors. $\endgroup$
    – Charith
    Oct 24, 2019 at 23:49
  • $\begingroup$ @gune I have added an explanation. $\endgroup$
    – Kavi Rama Murthy
    Oct 24, 2019 at 23:50
  • $\begingroup$ Can you please have a look at the image I just attached.. $\endgroup$
    – Charith
    Oct 24, 2019 at 23:57
  • $\begingroup$ @gune They are using the fact that any vector is a linear combination of elements of the Hamel basis and that is what I have proved in my answer. I don't understand what further confusion you have now. $\endgroup$
    – Kavi Rama Murthy
    Oct 25, 2019 at 0:00
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    $\begingroup$ @gune They are using $\sum_i$ and $\sum_j$ denote finite sums. They could have used a more explicit notation but what they have written is also OK. $\endgroup$
    – Kavi Rama Murthy
    Oct 25, 2019 at 0:05

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