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I'm following the proof of the fact that $ \mathrm{dim}(U) + \mathrm{dim} \mathrm{Ann}(U) = \mathrm{dim}(V) $ for $U \subset V$ and $\mathrm{Ann}(U)$ is an annihilator of $U$, in here (Proposition 2.20 (a)).

But I don't understand how $v^{'}(v_{i}) = c_{i}$.

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By definition of dual base $ v_i'(v_j) = \begin{cases} 1 & \text{if } i = j\\ 0 & \text{if } i\neq j \end{cases} $ Now $ v'(v_i) = \sum_j c_j v_j'(v_i) = \sum_{j\neq i} c_j v_j'(v_i) + c_i v_i'(v_i) = c_i v_i'(v_i) = c_i $