# Proof for Steinitz exchange lemma

Hi there I was wondering if any of you could give me a "easier" proof for Steinitz exchange lemma. I found something on wikipedia here but I can't understand. It is too formal.

Lemma (wikipedia version 8.10.2015) If $$\{v_1, \cdots, v_m\}$$ is a set of m linearly independent vectors in a vector space $$V$$, and $$\{w_1, \cdots , w_n\}$$ span $$V$$ then $$m\le n$$ and, possibly after reordering the $$w_i$$, the set $$\{v_1, \cdots, v_m, w_{m + 1},\cdots, w_n\}$$ spans $$V$$.

Let $A=\{v_1,\ldots,v_m\}$ be a set of linearly independant verctors, $B=\{w_1,\ldots,w_m\}$ a set of vectors that generate our vector space $V$. Among all sets $C$ such that $A\subseteq C\subseteq A\cup B$ and $C$ generates $V$, pick one that is minimal. This $C$ is of the form $C=\{v_1,\ldots, v_m,w_{i_1},\ldots, w_{i_r}\}$. Assume $$a_1v_1+\ldots+a_mv_m+b_1w_{i_1}+\ldots+b_rw_{i_r}=0$$ If any of the $b_k$ is nonzero, then $w_{i_k}$ is in the span of the remaining elements of $C$, hence $C\setminus\{w_{i_k}\}$ is a smaller generating set between $A$ and $A\cup B$, contradicting minimality of $C$. Hence $b_1=\ldots =b_r=0$. But then from the linear independence of the $v_i$ also $a_1=\ldots =a_m=0$. We conclude that $C$ is linearly indepenedant. And it is generating. Hence it is a basis.
• Where have you proved that $m\leqslant n$? Jul 6, 2023 at 14:11