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Let $x_0, x_1, \ldots x_{n-1}$ be arbitrary vectors in a complex Hilbert space. Define the $n \times n$ symmetric real matrix $M$ by $M_{ij} = \lvert \langle x_i, x_j \rangle \rvert^2$. Must $M$ be positive semidefinite?

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Yes. The matrix $A$ with $a_{i,j} = \langle x_i,x_j \rangle$ is a Gram matrix and thus positive semidefinite, so $A^T = \overline{A}$ is positive semidefinite too. It then follows from the Schur product theorem that your matrix $M = A \circ \overline{A}$ (where $\circ$ denotes the entrywise/Hadamard product) is positive semidefinite too.

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