For a given $k,n$ such that $0<k \leq n/2$, is there a number $N\_{n,k}$, such that if one has $N$ different $k$-dimensional subspaces $V\_1, V\_2,...,V\_N$ in $\mathbb{R}^n$ satisfying:


1) $\bigcap\_{i = 1}^N V\_i = \{0\}$


2) $dim (\bigoplus\_{i = 1}^N V\_i) \geq 2k$.


Then there are at least two subspaces $V\_i, V\_j$ such that $V\_i \cap V\_j = \{0\}$.


For example if $k =2$ and $n = 4$, then $N\_{n,k}$ is $\frac{1}{2} {4 \choose 2} + 1$, my guess was that for $n = 2k$, the maximal number would be $\frac{1}{2} {2k \choose k} + 1$ just by considering canonical subspaces for the canonical base of $\mathbb{R}^{2k}$.


I'm not very interested in the actual number $N\_{n,k}$ but in its existence actually.


Related results are a theorem of Erdos-Ko-Rado that states that in a set of $n$ elements there are at most $n-1 \choose k-1$ subsets of $k$ elements such that all of them have non-trivial intersection. Related results where proved for vector spaces over finite fields by Hsieh, Katona and others but I'm not able to find whether this statement or something similar holds.



Maximal number of intersecting subspaces of a finite dimensional vector space
The number $N\_{6,3}$ does not exist:
let $V\_1=\mathrm{span}(e\_1,e\_2,e\_3)$, $V\_2=\mathrm{span}(e\_3,e\_4,e\_5)$, $V\_3=\mathrm{span}(e\_1,e\_6,e\_5)$, and $V\_j=\mathrm{span}(e\_1,e\_3, e\_4+j\cdot e\_5)$ for $j\ge 4$.


