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I need to compute the determinant of a matrix that is calculated as $B \circ A$, with $B$ and $A$ being square matrices and $\circ$ representing their Hadamard product.

One way of doing this is through the eigenvalues of $B \circ A$, since a determinant is equal to the product of eigenvalues. This calculation is part of a simulation, where $A$ remains constant and $B$ changes during the simulation. Since $A$ does not change, its eigenvalues will not change, but $B$'s eigenvalues will.

Is there a way to calculate this determinant by only updating the eigenvalues of $B$? Or, similarly, how can I efficiently determine the eigenvalues of the Hadamard product of two matrices (perferably in R)?

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    $\begingroup$ $B$ and $A$ are the same size. Why is it easier to compute eigenvalues of $B$ than eigenvalues of $A \circ B$? $\endgroup$ – dranxo Jun 19 '12 at 1:29
  • $\begingroup$ it isn't. But it may be that the eigenvalues of B can be determined slightly faster, due to a special structure that B may end up having (depending on the simulation). Also, B is updated in another part of the simulation, which would be a natural place to calculate B's eigenvalues, making it more convenient to pass B's eigenvalues to the function that calculates the determinant of B*A, rather than to pass the entire matrix and then calculate the eigenvalues of the matrices' Hadamard product. $\endgroup$ – Peter Verbeet Jun 19 '12 at 4:51
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    $\begingroup$ In general, the eigenvalues of $B\circ A$ are unrelated to those of $A$ and $B$. So nothing can be gained from knowing the eigenvalues of $B$. $\endgroup$ – Arnold Neumaier Jun 19 '12 at 8:59
  • $\begingroup$ That's a pity, thanks for the answer. This is helpful (albeit unfortunate). $\endgroup$ – Peter Verbeet Jun 19 '12 at 15:40
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In general, the eigenvalues of $B\circ A$ are unrelated to those of $A$ and $B$ individually. So, in a general case, not a lot can be gained from the knowledge of the eigenvalues of $A$ and $B$.

The following info might be useful:

and confirms that not a lot can be gained for obtaining the eigenvalues of $B\circ A$, and even estimating the spectrum is limited to a very particular cases now.

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