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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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    What do you mean by "corresponding subset"? What if $A_1=\emptyset$ and $A_1'=S'$?2012-11-01
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    What does " $\,A_i\,,\,A'_i\,$ *correspondent* in $\,S\,,\,S'\,$ " mean?2012-11-01
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    Doesn't the question you link to essentially tell you that it's false?2012-11-01
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    No, "corresponding subset" means : if $A_1=v_1, v_3$ then $A'_1=v'_1, v'_3$.2012-11-01
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    From what I can tell, this question is just a generalization of the question you linked to which was already proven false. So what exactly are you asking for? As you've defined the subsets, all the properties in the other question still hold here.2012-11-01
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    No, this question is difference in my link, you need to separate them2012-11-01
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    Also posted to MO, http://mathoverflow.net/questions/111112/intersection-of-some-vector-spaces without notice to either site. Very rude.2012-11-01
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    @ Gerry Myerson : If you have a problem equivalent to you have a disease. You need to go to all hospitals :))2012-11-01
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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.

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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    Thank Marc van Leeuwen for your specific example. Your answer is very helpful. It is sufficient to check only 40 of its 3×3 minors are nonzero instead of 56. Here my next question : what condition of $A_i's$ so that the conclusion is true :$$dimspan(A_1)\cap⋯\cap span(A_k)=dimspan(A′_1)\cap⋯\cap span(A′_k)$$?2012-11-02
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    This is my other question : http://math.stackexchange.com/questions/227504/intersection-of-direct-sums .Please see my link above.2012-11-02
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    How much times do you need "no" for an anwer to stop keeping asking these questions? The upshot is: just knowing dimensions of subspaces and of all possible sums you can from from them does _not_ allow you to do the same for intersections. Nothing given "upwards" allows you to deduce dimensions downwards, it just don't work that way.2012-11-02
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    If you don't like my questions then you can ignore them :))2012-11-02
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    And if you like my answers, you can just accept (or at least upvote) them ;-) That's what the check-mark and up-arrow next to an answer are for.2012-11-02
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    How to accept, upvote ? I'm newbie here therefore i don't know. If this website has "like symbol" as facebook then i will click 1000 for you :))2012-11-02
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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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    No, if you chosse $A_2={e2}$ then $A'_2={e_2}$ because "corresponding subset"2012-11-01
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    @Firsttime: As "corresponding subset" is not standard terminology, I have used my best guess. If you want a more accurate answer, please *define* what you mean by this term before using it.2012-11-01
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    Yes, first we fix an order on $S,S'$ then "corresponding subsets" = they have the same index sets.2012-11-01
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    @Firsttime: I've updated my answer.2012-11-01
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    Thank wj32 : my main question not assume that $A_1,…,A_k$ and $A′_1,…,A′_k$ are disjoint. Their intersects of several sets are non empty.2012-11-01
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    You can see my sets $S,S'$ as follows: $V=\mathbb{R}^3$, $S=e_1,e_2,e_3,e_1+e_2,e_1+e_2-e_3$;$S'=e_1,e_2,e_3,e_1+e_2,e_1+e_2-5e_3$. Therefore, every two corresponding sets are same rank.2012-11-01
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    @Firsttime: That doesn't make any sense. If you don't want to assume that they're disjoint, then my previous counterexample works. How about you edit your question and use some proper English sentences along with clear statements of what conditions your sets satisfy? Sorry, but I can't read your mind.2012-11-01
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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