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Let $V$ be a vector space of dimension $n$ and let $V_1,V_2,\ldots,V_k \subset V$ be subspaces of $V$. Assume that

\begin{eqnarray} \sum^{k}_{i=1} \dim(V_i) > n(k-1). \end{eqnarray}

To show that $\bigcap^{k}_{i=1}V_i \neq \{0\}$, what must be done? Also, could there be an accompanying schematic/diagram to show the architecture of the spaces' form; that is, something like what's shown here.

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I think there is a general outline as to what's going on here in Hoffman's "Linear Algebra" on page 209. I'm just having difficulty bringing everything together. – Trancot Apr 18 '13 at 6:52
It certainly would be interesting to see a diagram of what exactly the subspace/vector space containment arragement looks like. Something like this. – Trancot Apr 18 '13 at 6:57
Just curious: the Seirios's answer is perfect; but is it possible to show this proposition by the PIE? – Orat Apr 18 '13 at 11:42
@Seirios Unfortunately, I'm most likely not allowed to use material too much outside my professor's closed-form lecture notes. – Trancot Apr 18 '13 at 20:00
@Seirios If it is not too taxing, perhaps you could wind out what you've propounded in terms more native to the simple, elementary content of an honors linear algebra course. – Trancot Apr 18 '13 at 20:05
up vote 4 down vote accepted

Lemma: Let $W_1,\ \cdots,\ W_k\subset V$. Then $$\dim\left(\bigcap_{i=1}^{k}W_i\right)\ge \sum_{i=1}^k\dim(W_i) - (k-1)n$$ Proof: First, recall the following equation relating the dimension of the intersection to the dimension of the sum. $$\dim(W_1 + W_2) + \dim(W_1\cap W_2) = \dim(W_1) + \dim(W_2)$$ Since $W_1,\ W_2\subset V$ it follows that $W_1 + W_2\subseteq V$ and therefore $\dim(W_1+W_2)\le n$. We then have $$\dim(W_1 \cap W_2) \ge \dim(W_1) + \dim(W_2) - n$$ This is our base case. Let us proceed to prove via induction. Suppose the inequality holds for $k\ge 2$ and consider $k+1$. Then $$\dim\left(\bigcap_{i=1}^{k+1}W_i\right) = \dim\left(W_{k+1}\cap\bigcap_{i=1}^kW_i\right)$$ By the base case, the above satisfies $$\dim\left(W_{k+1}\cap\bigcap_{i=1}^kW_i\right) \ge \dim(W_{k+1}) + \dim\left(\bigcap_{i=1}^kW_i\right)-n$$ By the inductive hypothesis, we then have $$\dim(W_{k+1}) + \dim\left(\bigcap_{i=1}^kW_i\right)-n \ge \dim(W_{k+1}) + \left(\sum_{i=1}^k\dim(W_i) - (k-1)n\right) - n$$ Combining, we of course have $$\dim\left(\bigcap_{i=1}^{k+1}W_i\right) \ge \sum_{i=1}^{k+1}\dim(W_i) - kn$$ The proposition follows by mathematical induction. $\square$

Now your result immediately follows since $$\sum_{i=1}^k W_i > (k-1)n \implies \dim\left(\bigcap_{i=1}^k W_i\right) \ge \sum_{i=1}^k\dim(W_i) - (k-1)n > 0$$

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+1 for elementary solution. BTW how do you come up with that lemma? Comparing the hypothesis and the conclusion? (Since I cannot feel the lemma is true intuitively.) – Orat Apr 19 '13 at 1:36
@Taro I immediately had the idea of using the identity $$\dim(W) + \dim(V) = \dim(W\cap V) + \dim(W +V)$$ since it's one of the main results which relate vectors spaces to their intersection. The natural way to get rid of the $W+V$ term in the above is to bound it above by the dimension of the ambient space. That gives the case $k=2$. The generalization to $k>2$ is also quite natural I think. – EuYu Apr 19 '13 at 1:56
Thank you for explanation. I also thought in that direction. For $k = 3$, it becomes $$\dim(V_1 \cap V_2 \cap V_3) = \dim(V) - \dim(V_1) - \dim(V_2) - \dim(V_3) + \dim(V_1 \cap V_2) + \dim(V_2 \cap V_3) + \dim(V_1 \cap V_3).$$ That's why I mentioned to the PIE in the comment. – Orat Apr 19 '13 at 13:10

For $1 \leq i \leq k$, let $p_i : V \to V/V_i$ be the canonical projection. Now, consider the morphism $\varphi = p_1 \times \dots \times p_k : V \to \prod\limits_{i=1}^k V/V_i$; notice that $\text{ker}(\varphi)= \bigcap\limits_{i=1}^k V_i$.

But $\dim \big( \prod\limits_{i=1}^k V/V_i \big) = \sum\limits_{i=1}^k \left( \dim(V)-\dim(V_i) \right)<n$. Therefore, $\varphi$ cannot be injective, hence $\bigcap\limits_{i=1}^k V_i \neq \{0\}$.

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Perhaps you could be more explicit. – Trancot Apr 18 '13 at 6:50
@Trancot : what do you mean ? How could it be made “more explicit” than this ? I do not think it can. – Ewan Delanoy Apr 18 '13 at 7:27
@Trancot: Is there something you don't understand? – Seirios Apr 18 '13 at 7:45
@Seirios Yes, well, no. I wouldn't say "not understand," but maybe "not familiar yet." I have to do a bit more combing through textbooks and the like to see what you're really saying. – Trancot Apr 18 '13 at 8:10
@Trancot: If you want, you can take $p_i : V \to W_i$, where $V= V_i \oplus W_i$ and then $\varphi : V \to \prod\limits_{i=1}^k W_i$. It is equivalent. – Seirios Apr 18 '13 at 8:15

First write the equation in the form $$ n>\sum_{i=1}^k\bigl(n-\dim V_i\bigr) $$ so that you can get some idea of what it really says: the codimensions of the subspaces $V_i$ add up to less than the dimension of the whole space. Now the codimension of a subspace gives the number of linear equations needed to define the subspace. If we choose such a set of equations for each subspace $V_i$, and take together all equations thus obtained, we have fewer equations than $n=\dim V$. This means (by the rank-nullity theorem if you like) that there is some nonzero vector that satisfies all equations at once. This vector lies in $\bigcap_{i=1}^kV_i$.

Added, to address comments by OP. To obtain from the inequality given in the question the one I gave above, first write the right hand side of the former as $nk-n$, then the term $nk$ as $\sum_{i=1}n$, now move the summation on the left to the right hand side, integrating its terms $\dim V_i$ as $-\dim V_i$ into the summation just created, and finally move the term $-n$ to the left, becoming $n$. The inequality remains a strict "$>$" throughout these manipulations. As for the fact that the codimension of a subspace equals the number of equations needed to define it, that is a basic fact from linear algebra: if you got a list of homogeneous linear equations (right hand side is $0$), then each new equation decreases the dimension of the solution space by $1$, unless it is linearly dependent of the previous equations, in which case one can drop the new equation. If you want a more formal argument for this, choose a basis of $d=\dim V_i$ vectors for the subspace $V_i$, extend it by $n-d$ more vectors to a basis of the whole space (incomplete basis theorem); then consider the $n$ coordinate functions for this basis, and the $n-d$ equations setting the last $n-d$ coordinates to $0$, which gives a system whose solution is precisely $V_i$.

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Don't you mean $\leq$ the dimension of the whole space? – Trancot Apr 18 '13 at 21:22
or anyone else reading this comment, could you explain how the summation equation reduces to this? – Trancot Apr 18 '13 at 21:29
Could you expound on your comment "Now the codimension of a subspace gives the number of linear equations needed to define the subspace," please? – Trancot Apr 18 '13 at 21:53

Hint: take complements. That is, pick vector spaces $W_i$ with $\dim(W_i) + \dim(V_i) = n$ and $W_i \cap V_i = \{0\}$.

EDIT: thanks, Ted

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Should be $W_i \cap V_i = \{0\}$. – Ted Apr 12 '13 at 3:11
@Ted, Oops, and thanks! – user66345 Apr 12 '13 at 3:20
I really don't see what you could do with subspaces $W_i$. Unless you consider the projections onto the $W_i$ parallel to the $V_i$ and combine them to a linear map to the external direct sum of the $W_i$ (a rather non-obvious thing to do, it is really not clear that you hinted at this), in which case the argument reduces to that of Serios. – Marc van Leeuwen Apr 18 '13 at 6:56

Hint: Can you do the case where $k = 2$? Then try induction on $k$, that involves replacing $V_1,V_2$ by $V_1 \cap V_2$, using the identity $\dim( V_1 + V_2) + \dim V_1 \cap V_2 = \dim V_1 + \dim V_2$.

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