Intersection of some vector spaces Let $\newcommand\span{\operatorname{span}}S=\{v_1,\ldots,v_m\}$ and $S'=\{v'_1,\ldots,v'_m\}\,$ be two sets of vectors in $V$ such that any two corresponding subsets  (meaning $\{\,v_i:i\in I\,\}$ and $\{\,v'_i:i\in I\,\}$ for some subset $I\subseteq\{1,2,\ldots,m\}$) of them have same rank. Now, choose corresponding sequences of subsets $A_1,\ldots,A_k$ and $A'_1,\ldots,A'_k$ in $S$ and $S'$, respectively. Is the following true or false ? 
$$
\dim\span(A_1)\cap\cdots\cap \span(A_k)=\dim\span(A'_1)\cap\cdots\cap \span(A'_k).
$$ 
Thanks.
PS: see more : Dimension of Intersection of three vector spaces satisfying specific postulates
 A: $\newcommand{spann}{\operatorname{span}}$Assume that $V$ is finite-dimensional, $S,S'\neq\emptyset$, $0\notin S\cup S'$, and $|S|=|S'|$. Also assume that $A_1,\dots,A_k$ and $A_1',\dots,A_k'$ are disjoint. Your conditions on $S$ and $S'$ imply that exactly one of the following holds:


*

*$\dim(\spann(S))=\dim(\spann(S'))=1$ (i.e. every pair of vectors is linearly dependent)

*$\dim(\spann(S))=\dim(\spann(S'))=|S|=|S'|$ (i.e. both are linearly independent)


To see this, first assume that $|S|,|S'|>1$, for otherwise the result is obvious. If either


*

*$\dim(\spann(S))=1$ but $\dim(\spann(S'))>1$, or

*$\dim(\spann(S))<|S|$ but $\dim(\spann(S'))=|S'|$,


then:


*

*There exists a subset $D \subseteq S$ with $|D|=2$ such that $\dim(\spann(D))=1$, and

*There exists a subset $E \subseteq S$ with $|E|=2$ such that $\dim(\spann(E))=2$.


This is a contradiction according to your conditions.
Now we can prove your result. If (1) above holds, then every intersection is going to be one-dimensional. If (2) above holds, then since $A_1,\dots,A_k$ are disjoint and $A_1',\dots,A_k'$ are disjoint, the intersections will be $\{0\}$.
A: This is false in general, as I have indicated in my answer to the question linked to above. Apparently my recipe was too hard to execute, so I'll do so here.$\newcommand\span{\operatorname{span}}$
We want to define four planes in $K^3$ (where $K$ is the base field), given by equations $x=0$, $y=0$, $z=0$ and $x+y=0$ respectively, each as the span of two out of $8$ vectors $v_1,\ldots,v_8$, where no triple of these vectors are linearly dependent. This can be done (for $K=\mathbf Q$) by taking $v_j$ to be column $j$ of the following matrix
$$
  \begin{pmatrix}
  0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 \\
  1 & 1 & 0 & 0 & 1 & 2 &-1 &-1 \\
  1 & 2 & 1 & 2 & 0 & 0 & 5 & 6 \\
  \end{pmatrix},
$$
for which one can check that all $56$ of its $3\times 3$ minors are nonzero. As a consequence the span of any $d$ distinct vectors $v_j$ is of dimension $\min(d,3)$.
Now taking $S=\{v_1,v_2,v_3,v_4,v_5,v_6\}$ and $S'=\{v_1,v_2,v_3,v_4,v_7,v_8\}$ (so $v'_i=v_i$ for $i\leq 4$ and $v'_5=v_7, v'_6=v_8$), and then
$A_1=A'_1=\{v_1,v_2\}$, $A_2=A'_2=\{v_3,v_4\}$, $A_3=\{v_5,v_6\}$ and $A'_3=\{v'_5,v'_6\}=\{v_7,v_8\}$, one has
$$
  0=\dim\span(A_1)\cap\span(A_2)\cap\span(A_3)\neq\dim\span(A'_1)\cap\span(A'_2)\cap \span(A'_3)=1.
$$
It may be noted that an intersection of at least three subspaces is needed, since
$$
  \dim(A\cap B)=\dim A+\dim B-\dim(A+B).
$$
Note also that although the intersection $A_1\cap A_2$ occurs on both sides, I have avoided choosing any of the $v_i$ on that line ($x=y=0$).
