Let $V$ be a finite dimensional vector space with basis $\mathcal{B} = \{\mathbf{b}_1, \ldots, \mathbf{b}_n\}$ with dual basis $\mathcal{B}^* = \{\mathbf{b}_1^*, \ldots, \mathbf{b}_n^* \}.$ Let $\mathcal{C} = \{ \mathbf{c}_1, \ldots, \mathbf{c}_n \}$ be another basis with corresponding $\mathcal{C}^* = \{\mathbf{c}_1^*, \ldots, \mathbf{c}_n^*\}. $
Prove that for every linear map $T$, $$\sum_{i=1}^{n}\left\langle b_{i}^{*}, T b_{i}\right\rangle=\sum_{i=1}^{n}\left\langle c_{i}^{*}, T c_{i}\right\rangle.$$
I have tried using the identities $v=\sum_{i=1}^{n}\left\langle b_{i}^{*}, v\right\rangle b_{I}$ and $ \ell=\sum_{i=1}^{n}\left\langle\ell, b_{i}\right\rangle b_{i}^{*}, $ but I am not able to simplify the summations when expanding $\mathbf{b}_i$ and $\mathbf{b}_i^*$ in terms of $\mathbf{c}_i$ and $\mathbf{c}_i^*.$
I have also tried using the transpose operator, $\langle T^t\mathbf{b}_i^*, \mathbf{b} \rangle = \langle \mathbf{b}_i^*, T\mathbf{b} \rangle.$
A little late, but here is: \begin{align} \sum_{i=1}^n \langle \mathbf b_i^*,T (\mathbf b_i) \rangle &= \sum_{i=1}^n \Big\langle \sum_{j=1}^n \langle \mathbf b_i^*, \mathbf c_j \rangle \mathbf c_j^*, T (\mathbf b_i) \Big\rangle & \mathbf b_i^* = \sum_{j=1}^n \langle \mathbf b_i^*, \mathbf c_j \rangle \mathbf c_j^* \\ &= \sum_{i=1}^n \sum_{j=1}^n \langle \mathbf b_i^*, \mathbf c_j \rangle \langle \mathbf c_j^*, T(\mathbf b_i) \rangle & \textrm{using bilinearity of $\langle \cdot, \cdot \rangle$} \\ &= \sum_{i=1}^n \sum_{j=1}^n \langle \mathbf b_i^*, \mathbf c_j \rangle \langle T^t(\mathbf c_j^*), \mathbf b_i \rangle & \textrm{definition of transpose} \\ &= \sum_{j=1}^n \Big\langle T^t(\mathbf c_j^*), \sum_{i=1}^n \langle \mathbf b_i^*, \mathbf c_j \rangle \mathbf b_i \Big\rangle & \textrm{again, bilinearity} \\ &= \sum_{j=1}^n \langle T^t(\mathbf c_j^*),\mathbf c_j \rangle & \mathbf c_j = \sum_{i=1}^n \langle \mathbf b_i^*, \mathbf c_j \rangle \mathbf b_i\\ &= \sum_{j=1}^n \langle \mathbf c_j^*, T(\mathbf c_j) \rangle. \end{align}