@robjohn when I try to run the series, Mathematica gives me an error and a window appears with a message like that "Mathematica Kernel for Windows has encountered a problem and needs to close."
@PeterTamaroff That is where I was at. I denoted $R_1,\cdots,R_i,\cdots, R_j, \cdots, R_n$ the rows of A Then I considered a linear combination transformation $R_j$ into $R_j+\lambda R_i$ such that $R_1,\cdots,R_i,\cdots, R_j+\lambda R_i, \cdots, R_n$ are the rows of A'
@Jean-FrancoisRossignol You can do this, if you like. First, let's agree to use $\operatorname{span}(v_1,\ldots,v_n)$. Let $V=\{v_1,\ldots,v_n\}$ and $W=\{w_1,\ldots, w_r\}$. Try to show this $$\operatorname{span}V \subseteq \operatorname{span}W $$ if, and only if $v_i\in \operatorname{span}(w_1,\ldots,w_r)$ for each $i=1,\ldots,n$.
is cubical homology the name referring to the homology theory, or to taking the homology of the chain complex that arises from the above commutative cubes?
