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I would like to show that from a generating family of a finite dimensional vector space $V$ we can always extract a basis.

Here is my attempt

Let $(u_i)_{i=1,...,p}$ be a generating family of $V$. This means that for any vector $v$ of $V$

$v=\sum_{i=1}^{p}\lambda_i u_i,\quad (\lambda_i)_{i=1,...,p}\in\mathbb{K}^{p}$

If this family is not linearly independent, then there exists $\lambda_p \neq 0$ such that

$u_p=-\frac{1}{\lambda_p}\sum_{i=1}^{p-1}\lambda_i u_i$

(the choice of the index p can always be done by a change of the order of the vectors of this family which will have no incidence thereafter) which means that $u_p$ is not indispensable in this family to be generative, we end up with the following generative family $(u_i)_{i=1,...,p-1}$ and we reproduce the same procedure until we arrive at an independent family : a generating and independent family and thus a basis of $V$.

Is this seems correct to you ?

Thank you a lot

EDIT : Thank's to Jean-Claude Arbault a big mistake has been removed.

G2MWF
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    Basically yes. Note that the generating family may already be a base, i.e. "Since this family is not linearly independent" is not correct: if it's not linearly independent, then you can remove a vector and continue the process, otherwise you stop before removing any vector. – Jean-Claude Arbaut Nov 26 '22 at 22:01
  • You are totally right . Thank you a lot ! – G2MWF Nov 26 '22 at 22:26

1 Answers1

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Yes, this works as long as the generating set you start with is finite (which is necessary for your labeling of them as $u_1, \dots, u_p$). Keep in mind that even in a finite dimensional vector space, there are infinite generating sets (as long as the base field is infinite): for example the whole vector space is a generating set.

It is still true that any generating set contains a basis, see eg An infinite generating set of a finite dimensional vector space contains a basis.

ronno
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