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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

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    @Firsttime: No. The way it is done on StackExchange is to ask on one site, and if the responses are not sufficient there, you flag the moderators to migrate the question to another site.2012-11-02

2 Answers 2

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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 ($x=y=0$).

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    Look left, next to an answer to your question. There are (in grey) an up-arrow, a down-arrow and a check-mark, with a black number between the two arrows. The arrows are to upvote/downvote (the number will increase/decrease) and the check-mark is to accept the answer (the one you find best). The check-mark only appears for your own questions, of course, but you can upvote/downvote any answer, even to other people's questions (maybe the latter requires having some points yourself, I forgot)2012-11-02
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$\newcommand{spann}{\operatorname{span}}$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:

  1. $\dim(\spann(S))=\dim(\spann(S'))=1$ (i.e. every pair of vectors is linearly dependent)
  2. $\dim(\spann(S))=\dim(\spann(S'))=|S|=|S'|$ (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 $\{0\}$.

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    @ wj32 : You see my example on $S,S'$. Now, you chosse $A_1,..,A_k$ and $A'_1,..,A'_k$. First, if you chosse $A_1=e_1, e_3, e_1+e_2-e_3$ then you have chosse $A'_1=e_1, e_3, e_1+e_2-5e_3$(1-1 corresponding).2012-11-01