Let {\( T_i\)}, \(i=1,2,…,n\) be the linear transformations, \( T_i: V_{i-1} \to V_i\), \(T_0: 0 \to V_1\), \(T_{n+1}: V_n \to 0\).
Since \(\textrm{Im} T_i = \textrm{Ker} T_{i+1}, i=0,1,…,n\), we have \(\textrm{Rank} T_i =\textrm{Nullity} T_{i+1}\)
Since \(\textrm{Ker} T_0 = 0, \textrm{Im} T_{n+1} = 0\), we obtain \(\textrm{Nullity} T_0 = 0,\textrm{Rank} T_{n+1} = 0,\textrm{Rank} T_0 = \textrm{dim} 0 – \textrm{Nullity} T_0 = 0\)
Thus, \(\sum\limits_{k=0}^{n}{(-1)^k\textrm{dim} V_k}\)
\(=\sum\limits_{k=0}^{n}{ (-1)^k(\textrm{Rank} T_{k+1} + \textrm{Nullity} T_{k+1}) }\)
\(=\sum\limits_{k=0}^{n}{ (-1)^k(\textrm{Rank} T_{k+1} + \textrm{Rank} T_k )}\)
\(=\sum\limits_{k=1}^{n+1}{-(-1)^k(\textrm{Rank} T_k } + \sum\limits_{k=0}^{n}{(-1)^k\textrm{Rank} T_k }\)
\(=(-1)^0\textrm{Rank} T_0 – (-1)^{n+1}(\textrm{Rank} T_{n+1} )\)
\(=0\)