I suppose I should've posted this first but whatever. Here's what I'm trying to prove:
If $(v_1,v_2,…,v_n)$ and $(w_1,w_2,….,w_m)$ are bases of $V$, where V is a vector space, then $m = n$.
My Proof Attempt:
Let $(v_1,v_2,….,v_n)$ and $(w_1,w_2,….,w_m)$ be bases. Then, by the Exchange Lemma, there exists a $w_j$ for each $v_i$ such that if I replace $v_i$ with $w_j$, the resulting list still forms a basis for $V$. In that case, we need $(w_1,w_2,…,w_m)$ to have at least n vectors. Hence, $m \geq n$.
Similarly, by the Exchange Lemma, there exists a $v_i$ for each $w_j$ such that if I replace $w_j$ with $v_i$, the resulting list still forms a basis for $V$. In that case, we need $(v_1,v_2,…,v_n)$ to have at least m vectors Hence, $n \geq m$.
Since, $m \geq n$ and $n \geq m$, we conclude that $m = n$.
Once again, the proof given in the book was different but I tried proving this on my own first. Can someone have a look and see if it's fine? Thanks.
Best Answer
The shortest method is that in the textbook of A. Maltsev “Foundations of Linear Algebra”. However, it is in Russian. An English adaptation of that proof is given as Theorem 12.24 on page 118 in lecture notes “Linear Algebra, Theory and Algorithms”.