Relationship on a complex of vector spaces 
Fact: If $U,V,W$ are $k-$vector spaces and there is a short exact sequence 
  $$0 \rightarrow U \rightarrow V \rightarrow W \rightarrow 0$$
  then the following relation holds:
  $$\textrm{dim}~V =\textrm{dim}~U + \textrm{dim}~W.$$

Show that if 
$$C:0 \rightarrow V_{n} \rightarrow V_{n+1} \rightarrow \dots \rightarrow V_1 \rightarrow V_{0} \rightarrow 0$$
is a complex of $k-$vecotr spaces, then
$$\sum_{i=0}^{n} (-1)^i \textrm{dim}~V_i = \sum_{i=0}^{n} (-1)^{i} \textrm{dim}~H_i(C).$$
While studying for my preliminaries I came across this problem. Any help is much appreciated. 
 A: For a compltex $C$ of the form $0 \to V_1 \to V_0 \to 0$, we have $\textrm{dim } V_0 - \textrm{dim } V_1 = \textrm{dim } H_0 (C) - \textrm{dim } H_1 (C)$. 
By induction, we assume that this result holds for $0 \to V_{n} \to \cdots \to V_2 \to V_1 \to V_0 \to 0$. Then for $0 \to V_{n+1} \to V_{n} \to \cdots \to V_2 \overset{f}{\to} V_1 \to V_0 \to 0$, we decompose it two short complexes $$0 \to V_{n+1} \to V_{n} \to \cdots \to V_2 \to \text{im f} \to 0$$ and $$0  \to \text{im f} \to V_1 \to V_0 \to 0.$$ Then we can get the result.
A: Note that by definition of homology of a complex
$$\cdots \to V_{i+1} \xrightarrow{d_{i+1}} V_i \xrightarrow{d_i} V_i \to \cdots$$
you have short exact sequences
$$0 \to \operatorname{im} d_{n+i} \to \ker d_i \to H_i (V_\bullet) \to 0$$
And so (by your Fact, which is also known as the rank-nullity theorem)
$$\tag{*} \dim H_i (V_\bullet) = \dim \ker d_i - \dim \operatorname{im} d_{i+1}.$$
On the other hand, you have short exact sequences
$$0 \to \ker d_i \to V_i \to \operatorname{im} d_i \to 0$$
and hence relations
$$\tag{**} \dim V_i = \dim \ker d_i + \dim \operatorname{im} d_i.$$
Now just put together (*) and (**).

By the way, the number $\chi (V_\bullet) = \sum_i (-1)^i \, \dim V_i = \sum_i (-1)^i \, \dim H_i (V_\bullet)$ is called the Euler characteristic of $V_\bullet$.
