# Basis Extension Theorem

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.

• Are you necessarily working in finite dimensions here? Apr 16 '21 at 7:11
• Isn't Basis Extension Theorem is just for finite dimension? Apr 16 '21 at 7:15
• 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). Apr 16 '21 at 7:17
• 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. Apr 16 '21 at 7:24
• @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. Apr 16 '21 at 7:43

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 $$\mathcal{B}$$.
• So there is no other way than checking the whole $V \setminus \operatorname{span} \mathcal{B}$ after modifying $B$ at every successful finding? Apr 16 '21 at 7:44
• 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 $\operatorname{span}\{w_1, \ldots, w_k\} \cap \operatorname{span}\mathcal{B} = \{0\}$. Apr 16 '21 at 7:54