Dimensions of vector spaces in an exact sequence

I've read the following formula in wikipedia:

Given finite dimensional vector spaces $V_i$ and an exact sequence $\cdots\rightarrow V_i\rightarrow V_{i+1}\rightarrow\cdots$, we have

$$\sum_{n\in 2\mathbb{Z}}\dim V_n = \sum_{n\in 2\mathbb{Z}+1}\dim V_n$$

Is there a name for this theorem? Could anyone please tell me where to find a proof of this in the literature?

Thank you!

-
the Euler characteristic of the complex is $0$ –  user8268 Dec 10 '12 at 11:41
@user8268: maybe I'm wrong but isn't the Euler characteristic of a complex the alternated sum of lengths of the homologies of the complex? here the complex is just acyclic so the homologies are all trivial. –  Simone Dec 10 '12 at 12:28

Let us prove the version of the statement in the Wikipedia entry:

If $0 \to V_1 \to V_2 \cdots \to V_r\to 0$ is an exact sequence of finite dimensional vector spaces then $$\sum_{i=1}^{r} (-1)^i \dim V_i=0.$$

Let $f_0$ be the first map in the sequence, $f_i$ be the map from $V_i$ to $V_{i+1}$, etc. By the Rank Nullity theorem, we have $\dim V_i = \dim\ker f_i + \dim \operatorname{im} f_i.$ Thus the left hand side is

$$\sum_{i=1}^{r} (-1)^i \dim\ker V_i+\sum_{i=1}^{r} (-1)^i \dim\operatorname{im} V_i.$$

Now by the defining property of an exact sequence, $\operatorname{im} f_i = \ker f_{i+1}.$ Place that information into one of the sums, and the two sums then cancel out.

Note that in order for the series in question to converge, the sequence must be of the form in this answer, perhaps with extra $0$ s.

-
the same argument works for exact sequences that are not bounded, as the ones in the question. In particular, this statement does not work only for the dimesion of vector spaces but for any function $L:\mathfrak A\to \mathbb R_{\geq 0}\cup\{\infty\}$, where $\mathfrak A$ is an abelian category, such that $L(A)=L(B)+L(C)$ whenever $0\to B\to A\to C\to 0$ is a s.e.s. in $\mathfrak A$. (Examples: composition length, torsion-free rank for modules over left Ore domains, log of the cardinality in Abelian groups, ...) –  Simone Dec 10 '12 at 12:38
Thank you all - that was more easy than I thought... –  Sh4pe Dec 10 '12 at 13:21