Every Linearly Independent List Extends to a Basis

Every linearly independent list extends to a basis

Every linearly independent list of vectors in a finite-dimensional vector space can be extended to a basis of the vector space.

Proof. Suppose $u_1,\dots ,u_m$ is linearly independent in a finite-dimensional vector space $V$. Let $w_1,\dots ,w_n$ be a list of vectors in $V$ that spans $V$. Thus the list

$$u_1,\dots ,u_m,w_1,\dots ,w_n$$

spans $V$. Applying the procedure from the above result (every spanning list contains a basis) to reduce this list to a basis of $V$ produces a basis consisting of the vectors $u_1,\dots ,u_m$ and some $w$‘s. (None of the $u$‘s get deleted in this procedure since $u_1,\dots ,u_m$ is linearly independent).

As an example in ${\mathbb{F}}^3$, suppose we start with the linearly independent list $(2,3,4),(9,6,8)$. If we take $w_1,w_2,w_3$ to be the standard basis of ${\mathbb{F}}^3$, then applying the procedure in the proof above produces the list

$$(2,3,4),(9,6,8),(0,1,0)$$

which is a basis of ${\mathbb{F}}^3$.