Basic "rank" question for $\binom{m}{k}$

Question

In my differential manifolds text, I've seen $\binom{m}{k}$ or a similar matrix form for rank of a vector space or topological space, and I haven't been able to get what this means straight in my head. Any help?

Here is an example for context. Let $\dim V=n$ and consider direct sum $\oplus \Lambda^p V$ as one linear space. Then $$\dim (\oplus_{p=0}^n \Lambda^p V)= \sum_{p=0}^n \binom{n}{p}= 2^n.$$ Therefore, if the matrix form of the rank is "pre"-numerical rank, how would I get $2^n$ from $\binom{n}{p}$?

Or, maybe the rank has two degrees under matrix form such as the (m-r)(n-r) question: https://math.stackexchange.com/questions/518202/what-is-the-codimension-of-matrices-of-rank-r-as-a-manifold. — Michael T Chase, Apr 13 '21 at 02:05
A concise notation for $\binom{m}{k}$ is \binom{m}{k} or {m\choose k}. I've edited the question using the former one. — metamorphy, Apr 13 '21 at 04:53

score 1 · Accepted Answer · answered Apr 13 '21 at 03:41

1

I believe that these terms are simply the binomial coefficients

$$ {n \choose p} \;\; =\;\; \frac{n!}{p!(n-p)!}. $$

It can be shown that the dimension of the vector space of alternating $p$-forms $\Lambda^pV$ is in fact ${n\choose p}$ and the resulting sum is simply just an application of the binomial theorem:

$$ \sum_{p=0}^n {n\choose p} \;\; =\;\; \sum_{p=0}^n {n \choose p} 1^p 1^{n-p} \;\; =\;\; (1+1)^n \;\; =\;\; 2^n. $$

answered Apr 13 '21 at 03:41

Mnifldz

13,228

Awesome! Much appreciated! – Michael T Chase Apr 13 '21 at 04:17

Basic "rank" question for $\binom{m}{k}$

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