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Given a real symmetric positive semi-definite matrix $\mathbf{A}$, is $\mathbf{A}^{\cdot k},k\in\mathbb{N}$ also positive semi-definite?

Matrix $\mathbf{A}\in\mathbb{R}^{n\times n}$ is positive semi-definite if $\mathbf{x}^T\mathbf{A}\mathbf{x}\geq0,\forall\mathbf{x}\in\mathbb{R}^n$.

The elements of $\mathbf{A}^{\cdot k}$ are $\{\mathbf{A}^{\cdot k}\}_{ij}=A_{ij}^k,1\leq i,j \leq n$.

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  • $\begingroup$ Oop, misread the exponentiation used :) $\endgroup$
    – rschwieb
    Feb 7, 2014 at 15:23
  • $\begingroup$ Yeah, I know that it's true for normal exponentiation. $\endgroup$
    – sk1ll3r
    Feb 7, 2014 at 15:26

1 Answer 1

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We prove that if $A$ and $B$ are positive semi-definite, then $A\circ B$ (the component-wise multiplication) is also positive semi-definite.

Note that the eigenvalues of then Kronecker product $A\otimes B$ are product of eigenvalues of $A$ and $B$, so $A\otimes B$ is positive semi-definite. Now note that $A\circ B$ is a principle sub-matrix of $A\otimes B$.

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  • $\begingroup$ How do the eigenvalues of a matrix relate to those of its principle sub-matrix? $\endgroup$
    – sk1ll3r
    Feb 7, 2014 at 19:05
  • $\begingroup$ The eigenvalue of a principle sub-matrix must lie between the minimal and maximal eigenvalues of the original matrix. $\endgroup$
    – Chen Wang
    Feb 7, 2014 at 19:39

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