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I know that using basis extension theorem we can extend a set of LI vector to the basis of vector space, but how we actually do it?

I have only theoretically used this theorem in proof, how to actually extend a given set of LI vectors of a vector space to its basis?

Thanks.

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  • $\begingroup$ Are you necessarily working in finite dimensions here? $\endgroup$
    – Theo Bendit
    Commented Apr 16, 2021 at 7:11
  • $\begingroup$ Isn't Basis Extension Theorem is just for finite dimension? $\endgroup$
    – cheems
    Commented Apr 16, 2021 at 7:15
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    $\begingroup$ Many of the basis theorems work in infinite dimensions too, but some special care needs to be taken with talking about infinite bases, and what it means to be "spanning", etc. Your response tells me that you're just interested in finite dimensions (at least, for now). $\endgroup$
    – Theo Bendit
    Commented Apr 16, 2021 at 7:17
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    $\begingroup$ Let $n=\dim(V)$ and suppose you have $1\leq k\leq n$ lin. ind. vectors $(\vec{v}_i:1\leq i\leq k)$. For each standard basis element $\vec{e}_j$, $1\leq j\leq n$, check if $(\vec{v}_i:1\leq i\leq k)\cup(\vec{e}_j)$ is linearly independent. If it is, then add it to the family of linearly independent vectors. Do this until you have a basis of $n$ linearly independent vectors. $\endgroup$
    – C Squared
    Commented Apr 16, 2021 at 7:24
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    $\begingroup$ @OldMan if you are working in an arbitrary, finitely generated vector space, you know that there is a finite basis of $V$. just keep adding basis elements to your family of lin. ind. vectors and checking if it’s still linearly independent. $\endgroup$
    – C Squared
    Commented Apr 16, 2021 at 7:43

1 Answer 1

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The pertinent fact here is the following:

If $\mathcal{B}$ is a (finite) linearly independent set in vector space $V$, and $w \in V \setminus \operatorname{span} \mathcal{B}$, then $\mathcal{B} \cup \{w\}$ is also linearly independent.

To prove this, suppose $\mathcal{B} = \{v_1, \ldots, v_n\}$, and suppose $a_1, \ldots, a_n, a$ are scalars such that $$a_1 v_1 + \ldots + a_n v_n + aw = 0.$$ Note that, if $a = 0$, then linear independence of $\mathcal{B}$ implies $a_1 = \ldots = a_n = 0$ as well, and we are done. Otherwise, if $a \neq 0$, then $$w = -\frac{a_1}{a} v_1 - \frac{a_2}{a} v_2 - \ldots - \frac{a_n}{a} \in \operatorname{span} \mathcal{B},$$ against assumption, proving the result.

What this tells us is that we can extend a linearly independent set $\mathcal{B}$ to a strictly larger set whenever $\operatorname{span} \mathcal{B} \neq V$, simply by choosing some $w \in V \setminus \operatorname{span} \mathcal{B}$, and adjoining it to $\mathcal{B}$.

The procedure for extending a linearly independent set to a basis is really this simple: keep adding vectors that are not in the span (which will maintain linear independence) until you run out of vectors to add. At that point, the span of your linearly independent set is the entire space, i.e. your set is a basis.

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  • $\begingroup$ So there is no other way than checking the whole $V \setminus \operatorname{span} \mathcal{B}$ after modifying $B$ at every successful finding? $\endgroup$
    – cheems
    Commented Apr 16, 2021 at 7:44
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    $\begingroup$ It's hard to find concrete methods for doing anything when $V$ is an arbitrary vector space over an arbitrary field. You could add a batch of linearly independent vectors $\{w_1, \ldots, w_k\}$ to $\mathcal{B}$ if you can show that $\operatorname{span}\{w_1, \ldots, w_k\} \cap \operatorname{span}\mathcal{B} = \{0\}$. $\endgroup$
    – Theo Bendit
    Commented Apr 16, 2021 at 7:54

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