A question about long exact sequence of finite dim vector spaces Q) Prove that if $0\rightarrow V_1\rightarrow V_2\rightarrow...\rightarrow V_k\rightarrow 0$ is an exact sequence of finite dimensional vector spaces, then $\sum_{i=1}^{k}(-1)^idim (V_i) = 0$.
For $k=3$, I see that the statement is true. For $k>3$, am I correct in understanding that $V_2 = V_1\oplus V_3 (\implies dim(V_2)=dim(V_1)+dim(V_3))$ and $V_3 = V_2\oplus V_4(\implies dim(V_3)=dim(V_2)+dim(V_4))$ and so on? 
If it is true, then for $k=4$, $\sum_{i=1}^{4}(-1)^idim (V_i) = -m_1+m_1+m_3-m_2-m_4+m_4$ and am not sure why $m_2=m_3$ for the sum to be $0$? where $m_i = dim(V_i)$. Thanks.
 A: If $\alpha : V \to W$ is a linear map, then we have a short exact sequence
$$0 \to \ker\alpha \to V \overset{\alpha}{\longrightarrow} \operatorname{im}\alpha \to 0.$$
If $V$ is finite-dimensional, then by the rank-nullity theorem $\dim V = \dim\ker\alpha + \dim\operatorname{im}\alpha$.
Now note that if $0 \to V_1 \overset{\alpha_1}{\longrightarrow} V_2 \overset{\alpha_2}{\longrightarrow} \dots \overset{\alpha_{k-1}}{\longrightarrow} V_k \to 0$ is an exact sequence of vector spaces (i.e. $\operatorname{im}\alpha_i = \ker\alpha_{i+1}$), we can break it up into a series of short exact sequences:
\begin{align*}
0 \to V_1 \to & V_2 \overset{\alpha_2}{\longrightarrow} \operatorname{im}\alpha_2 \to 0\\
0 \to \ker\alpha_3 \to & V_3 \overset{\alpha_3}{\longrightarrow} \operatorname{im}\alpha_3 \to 0\\
&\ \, \vdots \\
0 \to \ker\alpha_{k-2} \to & V_{k-2} \overset{\alpha_{k-2}}{\longrightarrow} \operatorname{im}\alpha_{k-2} \to 0\\
0 \to \ker\alpha_{k-1} \to & V_{k-1} \overset{\alpha_{k-1}}{\longrightarrow} V_k \to 0.
\end{align*}
Note that $V_1 = \ker\alpha_2$ and $V_k = \operatorname{im}\alpha_{k-1}$, so we have
$$0 \to \ker\alpha_i \to V_i \overset{\alpha_i}{\longrightarrow} \operatorname{im}\alpha_i \to 0$$
for $i = 2, \dots, k - 1$. If $V_i$ is finite-dimensional, then $\dim V_i = \dim\ker\alpha_i + \dim\operatorname{im}\alpha_i$. So we have
\begin{align*}
\sum_{i=2}^{k-1}(-1)^i\dim V_i &= \sum_{i=2}^{k-1}(-1)^i(\dim\ker\alpha_i + \dim\operatorname{im}\alpha_i)\\
&= \sum_{i=2}^{k-1}(-1)^i\dim\ker\alpha_i + \sum_{i=2}^{k-1}(-1)^i\dim\operatorname{im}\alpha_i\\
&= \sum_{i=2}^{k-1}(-1)^i\dim\operatorname{im}\alpha_{i-1} + \sum_{i=2}^{k-1}(-1)^i\dim\operatorname{im}\alpha_i\\
&= \sum_{i=1}^{k-2}(-1)^{i+1}\dim\operatorname{im}\alpha_i + \sum_{i=2}^{k-1}(-1)^i\dim\operatorname{im}\alpha_i\\
&= -\sum_{i=1}^{k-2}(-1)^i\dim\operatorname{im}\alpha_i + \sum_{i=2}^{k-1}(-1)^i\dim\operatorname{im}\alpha_i\\
&= \dim\operatorname{im}\alpha_1 -\sum_{i=2}^{k-2}(-1)^i\dim\operatorname{im}\alpha_i + \sum_{i=2}^{k-2}(-1)^i\dim\operatorname{im}\alpha_i + (-1)^{k-1}\dim\operatorname{im}\alpha_{k-1}\\
&= \dim V_1 + (-1)^{k-1}\dim V_k
\end{align*}
where we have used that $\alpha_1 : V_1 \to V_2$ is injective, so $\operatorname{im}\alpha_1 \cong V_1$, and $\alpha_{k-1} : V_{k-1} \to V_k$ is surjective, so $\operatorname{im}\alpha_{k-1} \cong V_k$.
Rearranging, we see that
$$\sum_{i=1}^k(-1)^i\dim V_i = 0.$$
