Let $V$ be the vector space of finitely nonzero sequences or real numbers. Find a basis $B$ of $V$.

My attempt I have to find a set $B$ such that: (i) $\text{span}(B)=V$ and (ii) if $x\in B$, then $x\not\in \text{span}(B-\{x\})$.

This set is very abstract and I have no idea where to start from. Any help or hint?


1 Answer 1


Let me give you some hints. The vector space $V$ has generic element $x = (a_1,a_2,a_3,\ldots)$, where $a_i\in F$ and $a_i=0$ for finitely many $i$. For instance, an element of $V$ is $y = (0,0,1,2,0,0,\ldots)$, with only two non-zero elements.

Now let $B = \{e_0,e_1,e_2,\ldots\}$, where $e_0=(0,0,0,\ldots)$, $e_1=(1,0,0,0,\ldots)$, $e_2 = (0,1,0,0,\ldots)$ and so on. Consider $x\in V$, with $x = (a_1, a_2,a_3,\ldots)$ a sequence with $n$ non-zero terms and let the nonzero position be $i_1,\ldots,i_n$.

For example, if $x = (0,0,1,2,0,1.2,0\ldots)$, then the non-zero terms are at the positions 2, 3 and 5, thus $i_1=1, i_2=3,i_3=5$.

Now check that properties (i) and (ii) hold.

  • 2
    $\begingroup$ so in order to express $x$ as linear combination of the basis do I have to do something like $x = a_{i_1}e_{i_1}+a_{i_2}e_{i_2}+\cdots+a_{i_n}e_{i_n}$ ? $\endgroup$
    – yahiro
    Dec 7, 2022 at 8:44
  • 1
    $\begingroup$ that's right! that shows the basis is generating... $\endgroup$
    – utobi
    Dec 7, 2022 at 8:45
  • 2
    $\begingroup$ thank you very very much!! Bad I don't have enough reps to upvote your anser... $\endgroup$
    – yahiro
    Dec 7, 2022 at 8:47

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