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!
 A: 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
$$\sum_{i=1}^{r} (-1)^i \dim\ker f_i+\sum_{i=1}^{r} (-1)^i \dim\operatorname{im} f_i.$$
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.
A: For a finite exact sequence of finite dimensional vector spaces, the claimed
equality is also provable by induction. For the induction anchor ($n = 1$),
observe that if $0 \rightarrow V_1 \rightarrow 0$ exact, then the second arrow
is injective, and thus $\dim(V_1) = 0$, such that the claimed equality holds.
For the induction step, the exact sequence in question is (up to adjustments
for $n = 2$ or $n = 3$):
$$
    0 \rightarrow V_1 \rightarrow V_2 \rightarrow V_3
    \rightarrow \ldots \rightarrow V_n \rightarrow 0
$$
from which the following, shorter sequence can be naturally constructed
(by considering the domain of the the third arrow, from $V_2$ to $V_3$, modulo
its kernel, which is, by exactness of the sequence in question, isomorphic to
$V_1$, and thus identified with $V_1$):
$$
    0 \rightarrow V_2 \mathbin{/} V_1 \rightarrow V_3
    \rightarrow \ldots \rightarrow V_n \rightarrow 0
$$
which is still exact, because the second arrow is now an injective map, while
it has, by construction, the same image in $V_3$ as the corresponding arrow of
the original exact sequence. Combining
$\dim(V_2 \mathbin{/} V_1) = \dim(V_2) - \dim(V_1)$ and the induction
hypothesis yields the claimed equality.
