MHB Is the Alternating Sum of Dimensions in an Exact Sequence Zero?

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Here is this week's problem!

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Let

$$\cdots \rightarrow V_{i-1} \rightarrow V_i \rightarrow V_{i+1} \rightarrow \cdots$$

be an exact sequence of finite dimensional vector spaces. Show that that alternating sum of their dimensions is zero, i.e., show that $\sum\limits_i (-1)^{i-1}\operatorname{dim}(V_i) = 0$.

Note: It is assumed $V_i = 0$ for all but finitely many $i$, so that the series $\sum\limits_i (-1)^{i-1}\operatorname{dim}(V_i)$ converges.

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No one answered this week's problem. For clarification, I made a note on the problem that $V_i = 0$ for all but finitely many $i$. You can find my solution below.

Spoiler

For each $i$, let $f_i$ be the map from $V_i$ to $V_{i+1}$. Since $V_i = 0$ for all but finitely many $i$, there are indices $m$ and $n$ with $n > m$ such that the long exact sequence of $V_i$ reduces to the short exact sequence

$$0 \rightarrow V_m \xrightarrow{f_m} V_{m+1}\xrightarrow{f_{m+1}}\cdots \xrightarrow{f_{n-1}} V_{n}\rightarrow 0.$$

By the rank-nullity theorem, given $i\in \Bbb Z$, $\operatorname{dim}(V_i) = \operatorname{rank}(f_i) + \operatorname{nullity}(f_i)$. Since $\operatorname{ker}(f_i) = \operatorname{Im}(f_{i-1})$ we have $\operatorname{nullity}(f_i) = \operatorname{rank}(f_{i-1})$. Hence $\operatorname{dim}(V_i) = \operatorname{rank}(f_i) + \operatorname{rank}(f_{i-1})$ for all $i$, and

$$\sum_i (-1)^{i-1}\operatorname{dim}(V_i) = \sum_{i = m}^{n+1} [(-1)^{i-1}\operatorname{rank}(f_{i}) - (-1)^{i-2}\operatorname{rank}(f_{i-1})] = (-1)^{n}\operatorname{rank}(f_{n+1}) - (-1)^{m}\operatorname{rank}(f_{m-1}) =0 - 0 = 0.$$