Given bases $b_1$, $b_2$ of $\mathbb{R}^n$ and a subset of $b_1$, show there is a subset of $b_2$ with which it can be exchanged to generate new bases Consider two bases in $\mathbb{R}^{n}$ and suppose $k$ is smaller than $n$. Take $k$ vectors from the first basis. Prove that we can exchange them with $k$ vectors from the second base such that in the end we have two new bases.
I found this answer, but I don't get it:
https://www.artofproblemsolving.com/community/c7h101634p573784
 A: This is more an example or an intuition pump than a complete answer, but I thought it might be helpful. (On the other hand it might be beside the point, depending on what aspect of the proof you don't understand.)
The key step in the Art of Problem Solving proof is the final claim that the determinant can be expanded as a linear combination of products of minors of the form $\Delta_k\Delta_{n-k}$, weighted with coefficients that are either $1$ or $-1$. This fact is a generalization of the familiar Laplace expansion, which is the special case of the AoPS claim when $k=1$. (This generalization is mentioned, but not proved, at the end of the Wikipedia article under the heading "General Statement.")
I like to think about these facts in terms of wedge products. Let $A$ be an $n\times n$ matrix, let $e_i$ denote the standard basis, and let $b_i$ denote the $i$-th column of $A$, $b_i:=Ae_i$. Now recall that the determinant of $A$ is the (unique, well-defined) scalar $\det A$ such that
$$b_1\wedge \cdots\wedge b_n=(\det A)\,e_1\wedge\cdots\wedge e_n$$
The point is that we get different representations of $\det A$ as linear combinations of minors depending on how we carve up the wedge product on the lefthand side.

For concreteness, I'll illustrate how this works for Laplace expansion along the first row and then suggest how it works for expansion along the first two rows. 
If we expand $b_1=Ae_1=a_{11}e_1+\cdots+a_{1n}e_n$, linearity of the wedge product says the lefthand side is
$$a_{11}\,e_1\wedge b_2\wedge\cdots\wedge b_n+a_{12}\,e_2\wedge b_2\wedge\cdots\wedge b_n+\cdots+a_{1n}\,e_n\wedge b_2\wedge\cdots\wedge b_n\tag{$\star$}$$
But $e_i\wedge b_2\wedge\cdots\wedge b_n$ is just a scalar multiple times the wedge product of $e_i$ with the other $e_j$'s with indices in ascending order, and that scalar multiple is precisely the minor $M_{1i}$ (the determinant of the submatrix formed by deleting row 1 and column $i$). That is, 
$$e_i\wedge b_2\wedge\cdots\wedge b_n=M_{1i}\,\color{red}{e_i\wedge(e_1\wedge\cdots\wedge\tilde{e_i}\wedge\cdots\wedge e_n)}$$
where the tilde indicates that $e_i$ should be left out. By permuting the red expression so the indices are ascending, i.e. by permuting to get $e_1\wedge e_2\wedge\cdots\wedge e_n$, we simply flip the sign of the coefficient. Putting all this together, $(\star)$ exhibits $\det A$ as a linear combination of products of the form $a_{1i}M_{1i}$, where the coefficients of these products are either $1$ or $-1$, depending on the sign of the permutation required in the last step. That's the Laplace expansion of $\det A$ along the first row.
To get a different representation of $\det A$, we can proceed further with $(\star)$ by expanding another column, say $b_2=a_{21}e_1+\cdots+a_{2n}e_n$. I leave it to you to see that we will get a sum of the form
$$\sum_{i<j}c_{ij}(e_i\wedge e_j)\wedge(b_3\wedge\cdots\wedge b_n)\tag{$\square$}$$
where $c_{ij}$ is the $2\times 2$ determinant of the submatrix consisting of the entries in the first two rows and columns $i$ and $j$. Moreover, we will have
$$(e_i\wedge e_j)\wedge(b_3\wedge\cdots\wedge b_n)=d_{ij}\color{blue}{(e_i\wedge e_j)\wedge(e_1\wedge\cdots\wedge\tilde{e_i}\wedge\cdots\wedge\tilde{e_j}\wedge\cdots\wedge e_n)}$$
and it can be shown the $d_{ij}$ are the determinants of the complement matrices of the $2\times 2$ matrices that give the $c_{ij}$. Thus, by permuting again, $(\square)$ exhibits $\det A$ as a linear combination of products of the form $c_{ij}d_{ij}$, again with weights $\pm1$. In the notation of the AoPS proof, this is the representation of $\det A$ in terms of products of the form $\Delta_2\Delta_{n-2}$.
