Dimensions: $\bigcap^{k}_{i=1}V_i \neq \{0\}$ 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. 
 A: 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 $\bigcap\limits_{i=1}^k V_i \neq \{0\}$.
A: 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 $V_i$. 
A: 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
$$\sum_{i=1}^k W_i > (k-1)n \implies \dim\left(\bigcap_{i=1}^k W_i\right) \ge \sum_{i=1}^k\dim(W_i) - (k-1)n > 0$$
A: 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
A: 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 $\dim( V_1 + V_2) + \dim V_1 \cap V_2 = \dim V_1 + \dim V_2$.
