# 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 : Intersection of three vector spaces

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What do you mean by "corresponding subset"? What if $A_1=\emptyset$ and $A_1'=S'$? – wj32 Nov 1 '12 at 3:10
What does " $\,A_i\,,\,A'_i\,$ correspondent in $\,S\,,\,S'\,$ " mean? – DonAntonio Nov 1 '12 at 3:11
Doesn't the question you link to essentially tell you that it's false? – EuYu Nov 1 '12 at 3:12
No, "corresponding subset" means : if $A_1=v_1, v_3$ then $A'_1=v'_1, v'_3$. – Firsttime Nov 1 '12 at 3:12
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. – EuYu Nov 1 '12 at 3:15

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 :$$dimspan(A_1)\cap⋯\cap span(A_k)=dimspan(A′_1)\cap⋯\cap span(A′_k)$$? – Firsttime Nov 2 '12 at 12:49
This is my other question : math.stackexchange.com/questions/227504/… .Please see my link above. – Firsttime Nov 2 '12 at 13:07
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. – Marc van Leeuwen Nov 2 '12 at 15:11
If you don't like my questions then you can ignore them :)) – Firsttime Nov 2 '12 at 15:21
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. – Marc van Leeuwen Nov 2 '12 at 15:38

$\newcommand{spann}{\operatorname{span}}$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 $\dim(\spann(E))=2$.

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 $\{0\}$.
No, if you chosse $A_2={e2}$ then $A'_2={e_2}$ because "corresponding subset" – Firsttime Nov 1 '12 at 3:21
Yes, first we fix an order on $S,S'$ then "corresponding subsets" = they have the same index sets. – Firsttime Nov 1 '12 at 3:27
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. – Firsttime Nov 1 '12 at 5:25