Brian Bi
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Exercise 3.6.1 Let \(\{w_1, \ldots, w_m\}\) be a basis of \(W\). Let \(\{x_1, \ldots, x_n\}\) be a basis of \(V/W\). It is easy to construct a basis of \(V\) by combining the two bases. Explicitly, a basis of \(V\) is given by \(\{y_1, \ldots, y_{m+n}\}\) where \(y_1 = w_1, \ldots, y_m = w_m\), and for each \(i = 1, 2, \ldots, n\), \(y_{m+i}\) is arbitrarily chosen from \(\pi^{-1}(x_i)\) (where \(\pi\) denotes the projection map from \(V\) to \(V/W\)).

Let \(a \in A\) be given. For each \(i = 1, 2, \ldots, m+n\), write \begin{equation} ay_i = \sum_{j=1}^{m+n} c_{ij} y_j \label{eqn:components} \end{equation} Then, \begin{equation} \chi_V(a) = \sum_{i=1}^{m+n} c_{ii} \label{eqn:chiV} \end{equation} Observe that \(aw_i = \sum_{j=1}^m c_{ij} w_j\) since \(w_i = y_i\). There are no contributions from higher \(j\) since \(W\) is a subrepresentation. Thus, \begin{equation} \chi_W(a) = \sum_{i=1}^m c_{ii} \label{eqn:chiW} \end{equation} Fix \(k \in \{1, 2, \ldots, n\}\) and apply the projection map to both sides of \((\ref{eqn:components})\), putting \(i = m + k\). We obtain \begin{equation*} ax_k = a\pi(y_i) = \pi(ay_i) = \sum_{j=m+1}^{m+n} c_{ij} x_{j-m} \end{equation*} where there are no contributions from lower \(j\) since \(\pi(y_j) = 0\) for all \(j \leq m\). The coefficient of \(x_k\) on the RHS is obtained by setting \(i = m + k\) and \(j-m = k\) in \(c_{ij}\), therefore: \begin{equation*} \chi_{V/W}(a) = \sum_{k=1}^n c_{m+k,m+k} = \sum_{i=m+1}^{m+n} c_{ii} \end{equation*} Together with \((\ref{eqn:chiV})\) and \((\ref{eqn:chiW})\) this implies that \(\chi_V = \chi_W + \chi_{V/W}\).