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Let $\hat\rho$ denote the density operator. Then, its matrix representation $\rho^\beta$ in basis $\vert\beta_n\rangle$ is given by $$\hat\rho = \sum \rho^\beta_{mm'}\vert\beta_m\rangle\langle\beta_{m'}\vert\ .$$ By substituting $\vert\beta_m\rangle = \sum S_{mn}\vert\alpha_n\rangle$, we obtain $$\hat\rho = \sum ... 1 You may find the following paper useful: A Symbolic Solution of the Hubbard Model for Small Clusters, by J. Yepez. You may also want to review group theory for condensed matter physics, because your questions essentially span the basics of group and representation theory. Many texts give good overviews of the fundamentals of group theory as applied to ... 2 Take the expansion of an arbitrary vector in an orthonormal basis like so$$ |\alpha\rangle = \sum |i\rangle\langle i |\alpha\rangle=\left(\sum |i\rangle\langle i |\right)|\alpha\rangle$$But, this is true for any |\alpha\rangle, so the sum over projections must be the identity operator. 3 Act with your sum of projection operators on an arbitrary state psi. Use completeness to expand psi into a sum of basis vectors. Use orthonormality to simplify the sum (with \langle n |m\rangle=\delta_{ij} ). Simplify. The sum you're left with is the original vector psi. By the arbitrariness of the initial vector, the operator must be the identity matrix. 2 Is this what you want?$$U = \sum_{i,j} U_{ij} |{e_i}\rangle \langle e_j| = \sum_{i,j} |e_i\rangle U_{ij} \langle e_j| \\ = \sum_{i,j} |e_i\rangle \langle e_i|U|e_j\rangle \langle e_j| = \sum_{i,j} |e_i\rangle \langle e_i|\bar{e}_j\rangle \langle e_j| = \sum_j |\bar{e}_j \rangle \langle e_j|