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The Schur Product Theorem basically states that the Hadamard product of two semidefinite matrices is semidefinite. The proof from Wikipedia:

==== Proof of positivity ====

Let $M = \sum \mu_i m_i m_i^T$ and $N = \sum \nu_i n_i n_i^T$. Then $$ M \circ N = \sum_{ij} \mu_i \nu_j (m_i m_i^T) \circ (n_j n_j^T) = \sum_{ij} \mu_i \nu_j (m_i \circ n_j) (m_i \circ n_j)^T $$ Each $(m_i \circ n_j) (m_i \circ n_j)^T$ is positive (but, except in the 1-dimensional case, not positive definite, since they are rank 1 matrices) and $\mu_i \nu_j > 0$, thus the sum giving $M \circ N$ is also positive.

Here, I am not sure how they got the form of $M$ and $N$ with the $\mu$ and $\nu$ in front. Could anyone be kind enough to give me an explanation of what is going on here? thank you!

Mark Yasuda
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user123276
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1 Answers1

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In the Schur product theorem, the positive semi-definite matrices $M$ and $N$ are Hermitian and therefore unitarily diagonalizable. That is, the spectral theorem guarantees that we can write $M = U \Lambda U^*$ for $U$ unitary and $\Lambda$ diagonal, and we can take the columns of $U$ to be a set of orthonormal eigenvectors for $M$ (call them $m_1,\dots,m_n$) and diagonal elements for $\Lambda$ to be the corresponding eigenvalues $\mu_1,\dots,\mu_n.$ From this decomposition, it immediately follows that $M = \sum \mu_i m_i m_i^*.$ Of course, the same observation shows that we can also write $N = \sum \nu_i n_i n_i^*.$

For some reason, the Wikipedia article on the Schur product theorem is writing out this proof using transposes rather than conjugate transposes, apparently making the unnecessary restriction to real positive semi-definite matrices. I may remedy that at some point in the future, if someone else doesn’t make those minor edits before me.

Mark Yasuda
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