I am currently reviewing linear algebra from Steven Roman's Advanced Linear Algebra, and I have gotten stuck on problem 1.23 for quite some time. The question asks given bases $B = \{b_1, ..., b_n\}, C = \{c_1,...,c_n\}$ for a vector space $V$ and some $1 \leq m \leq n-1$, prove there exist some permutation $\pi$, such that $$\{b_1, ..., b_m, c_{\pi(m+1)}, ..., c_{\pi(n)}\} \text{ and } \{c_{\pi(1)}, ..., c_{\pi(m)}, b_{m+1}, ..., b_n\}$$ are both bases for $V$.

I tried this problem for a few hours, then found a solution here https://staff.math.su.se/mleites/books/prasolov-1994-problems.pdf (Theorem 7.2). However, in this solution, they rewrite the form of the determinant

$$M(b_1, ..., b_n) = \sum_{A \subset V} \pm M(b_1,...,b_m, A)M(V\setminus A, b_{m+1}, ..., b_n)$$

Where $M(x_1,...,x_n)$ is the determinant of a matrix with rows $x_i$ written in coordinates with respect to the basis $C$. I am wondering what identity/theorem is applied here to rewrite the determinant in this way?

  • 1
    $\begingroup$ It's the generalized Laplace expansion in the first $k$ rows, using an auxiliary identity for minor determinants. But you don't need to use something so fancy. $\endgroup$
    – blargoner
    May 14 at 5:15


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