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Let $w_1, w_2,....w_M$ be a set of vectors. There exists no pair of parallel vectors and $D$ vectors are linearly independent. Let $C_k$ be the number of $k$-tuples of linearly independent vectors. I mean that each $k$-tuple lies on a different vectorial subspace. Then I want to find an upper limit for $C_k$, given $C_{k+1}$.

In particular, I would be interested to the case that $D$ is smaller than $M/2$.

edited Dec 12 2010 at 22:56

Andrey Rekalo
14k●1●34●69

asked Dec 12 2010 at 22:31

alberto
1

1

EDIT: removed algebraic geometry tag. – J.C. Ottem Dec 12 2010 at 22:34

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I see the possibility of C_k being M choose k, and C_(k+1) being any number between 0 and M choose (k+1). Similarly, C_(k+1) being nonzero means C_k is at least k+1, but it is hard to see what would stop all k-subsets from being linearly independent, knowing nothing else about the set. Gerhard "Ask Me About System Design" Paseman, 2010.12.12 – Gerhard Paseman Dec 12 2010 at 22:56

Also, it is not clear what "different vectorial subspace" entails. Perhaps you can clarify with some examples. Gerhard "Ask Me About System Design" Paseman, 2010.12.12 – Gerhard Paseman Dec 12 2010 at 22:57

Since there are D linearly independent vectors, C_k is at least equal to D!/(D-k)!/k! – alberto Dec 12 2010 at 23:01

I second Gerhard: the way I interpret "each *k*-tuple lies on a different vectorial subspace" contradicts the definition of $C_k$. E.g. with vectors (0,1), (1,0) and (1,1), then $C_2=3$ (or I guess $C_2=6$ since we're considering ordered pairs), but all pairs of vectors span the same space $R^2$. – Thierry Zell Dec 13 2010 at 0:35

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number of sets of linearly independent vectors

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