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The component-wise product (Hadamard product) of two positive definite matrices is a positive definite matrix (Schur product theorem). I encountered the following proof of it:

$A=(a_{ij})$ and $B=(b_{ij})$ are positive definite. Let $a_{ij} = \displaystyle \sum_{k=1}^N \lambda_k t_{ik} t_{jk}$ where $T=(t_{ij})$ is an orthogonal matrix and $\lambda_k$ are the eigenvalues of $A$. $$\sum_{i,j=1}^N a_{ij} b_{ij} x_i x_j = \sum_{i,j=1}^N b_{ij} x_i x_j \left( \sum_{k=1}^N \lambda_k t_{ik} t_{jk} \right) = \sum_{k=1}^N \lambda_k \left( \sum_{i,j=1}^N b_{ij} x_i t_{ik} x_j t_{jk} \right)$$ and so then showing $$\sum_{i,j=1}^N b_{ij} x_i t_{ik} x_j t_{jk} >0$$ is enough.

Is there a simple geometrical relationship between the eigenvalues and eigenvectors of $C=(a_{ij} b_{ij})$ and the eigenvalues and eigenvectors of $A$ and $B$ that explains the geometrical motivation behind this proof?

mle
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user782220
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1 Answers1

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Generally speaking, no, there is no relationship among the eigenvalues/vectors of $A$, $B$, and their Hadamard product $A\circ B$. See, for example, the upvoted comment here.

I think the Wackypedia article you cite on the Schur product theorem has a nice section on how to use the eigenvalues/vectors of $A$ and $B$ to show $A\circ B$ is positive definite. It starts with knowing $A = \sum \alpha_i x_i x_i^T$ and $B = \sum \beta_i y_i y_i^T$. Then $$A \circ B = \sum_{ij} \alpha_i \beta_j (x_i x_i^T) \circ (y_j y_j^T) = \sum_{ij} \alpha_i \beta_j (x_i \circ y_j) (x_i \circ y_j)^T$$ Each term in the sum is positive semidefinite.

In your notation, $T$ is a matrix whose columns are the (normalized) eigenvectors of $A$, the $x_i$'s.

Community
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Armadillo Jim
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  • Why $A = \sum \alpha_i x_i x_i^T$? What is $x_i$ and $a_i$? Are you making summation on $i$? – Logan Apr 22 '17 at 15:12
  • @Logan If $A$ is positive definite, it is symmetric and has a complete set of eigenvalues $\alpha_i$ and eigenvectors $x_i$. The sum $A = \sum \alpha_i x_i x_i^T$ is the eigendecomposition of $A$. See https://en.wikipedia.org/wiki/Eigendecomposition_of_a_matrix along with the link to the Schur product theorem (in the answer above). – Armadillo Jim May 21 '17 at 15:59