Suppose $\sum_{i=1}^n \dim(\textsf{W}_i) > (n-1)\dim(\textsf{V})$. Then prove that $\bigcap_{i=1}^n \textsf{W}_i$ is non-empty Suppose that a vector space $\textsf V$ has some subspaces $\textsf{W}_1,\textsf{W}_2,\dots,\textsf{W}_n$ such that
$$\sum_{i=1}^n \dim(\textsf{W}_i) > (n-1)\dim(\textsf{V})$$
then $\displaystyle \bigcap_{i=1}^n \textsf{W}_i$ is non-empty.
I have no idea how to start this question, any hint on this? Thank you.
 A: Let $A = \{\alpha_j\}$ be a basis for $V$. Suppose that $\displaystyle\bigcap_{i=1}^{n} W_i = \emptyset$. Then we know that for all $\alpha \in A, \alpha \notin W_i $ for some $i$. This means that $\sum \dim(W^c) \geq \dim(V)$ since every basis vector will appear the complement of at least one subspace. Thus, $$\sum \operatorname{dim}(W_i) = \sum\left(\dim(V) - \dim(W_i^c)\right) \leq n\dim(V) - \dim(V) = (n-1)\dim(V)$$
Since the contrapositive statement is true, the original statement is true as well.
EDIT: $W^c_i$
 is not a vector space, but I abuse notation and let $\dim(W_i^c) = |W_i^c \cap A|$.  
A: Let $f:\textsf{V}\to \bigoplus_{i=1}^n \textsf{V}/\textsf{W}_i$ be the linear map defined by
$$f(v)=(v+\textsf{W}_1,v+\textsf{W}_2,\ldots,v+\textsf{W}_n).$$
Since $$\dim\left(\bigoplus_{i=1}^n \textsf{V}/\textsf{W}_i\right)=\sum_{i=1}^n\Big(\dim \textsf{V}-\dim\textsf{W}_i\Big)=n\dim\textsf{V}-\sum_{i=1}^n\dim\textsf{W}_i<\dim \textsf{V},$$
$f$ cannot be injective.  So $\ker f\neq 0$.  Hence we can find $w\in V$ such that $w\ne 0$ and $f(w)=0$.  But this means $w\in \textsf{W}_i$ for all $i$, so $\bigcap_{i=1}^n\textsf{W}_i$ is non-trivial (I guess you meant to say "non-trivial" rather than "non-empty" as $0$ belongs to $\bigcap_{i=1}^n\textsf{W}_i$).
