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What is the relationship between the determinants of the square matrices of equal dimensions $\mathbf{A}$ and $\mathbf{B}$ where each element of $\mathbf{B}$ is equal to the corresponding element of $\mathbf{A}$, times some constant ($\alpha$) raised to the absolute value of the difference of the row and column numbers? E.g. the $3 \times 3$ case is:

$$ \mathbf{A} = \begin{vmatrix} a_{1,1} & a_{1,2} & a_{1,3} \\ a_{2,1} & a_{2,2} & a_{2,3} \\ a_{3,1} & a_{3,2} & a_{3,3} \end{vmatrix} \qquad\&\qquad\mathbf{B} = \begin{vmatrix} \alpha^0 a_{1,1} & \alpha^1 a_{1,2} & \alpha^2 a_{1,3} \\ \alpha^1 a_{2,1} & \alpha^0 a_{2,2} & \alpha^1 a_{2,3} \\ \alpha^2 a_{3,1} & \alpha^1 a_{3,2} & \alpha^0 a_{3,3} \end{vmatrix} $$

I haven't been able to get beyond trivial answers that are directly obvious from the problem statement.

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    $\begingroup$ So $\mathbf B$ is a Hadamard product... the relationship between the determinant of a Hadamard product and the determinants of the individual matrices does not look to be simple... $\endgroup$ Oct 10, 2011 at 17:29
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    $\begingroup$ If you expand it out it isn't really anything dependent on $\det A$ (unless $\alpha=1$ of course), though I guess you could get a "nontrivial" answer by augmenting the Leibniz formula with $\sum |\sigma(k)-k|$ as a factor. If you considered the non-absolute differences between row and column index the determinant of $A$ would be preserved. $\endgroup$
    – anon
    Oct 10, 2011 at 17:49
  • $\begingroup$ Yes, that's what I meant by preserved. I'll type something up. $\endgroup$
    – anon
    Oct 10, 2011 at 18:45

1 Answer 1

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Expanding gives $$\begin{vmatrix} \alpha^0 a_{11} & \alpha^1 a_{12} & \alpha^2 a_{13} \\ \alpha^1 a_{21} & \alpha^0 a_{22} & \alpha^1 a_{23} \\ \alpha^2 a_{31} & \alpha^1 a_{32} & \alpha^0 a_{33} \end{vmatrix}$$ $$=a_{11}a_{22}a_{33}\alpha^0-(a_{11}a_{32}a_{23}+a_{21}a_{12}a_{33})\alpha^2+(a_{21}a_{32}a_{13}+a_{31}a_{12}a_{23}-a_{31}a_{22}a_{13})\alpha^4.$$ This doesn't really have any relationship with $\det A$ except for trivial equality when $\alpha=\pm1$. One "nontrivial" thing that can be done is define $|\sigma-\rho|=\sum_{k=1}^n |\sigma(k)-\rho(k)|$ for permutations $\sigma,\rho$, in which case the Leibniz formula can be augmented to read (for the $n\times n$ case) $$\det B=\sum_{\sigma}(-1)^{\sigma}\alpha^{|\sigma-\mathrm{Id}|}\prod_{j=1}^n a_{j,\sigma(j)}.$$

If instead we use the signed differences between column and row index, the matrix $B$ is the result of multiplying the second and third rows by $\alpha$ and $\alpha^2$ respectively and the second and third columns by $\alpha^{-1}$ and $\alpha^{-2}$ respectively of the matrix $A$, which preserves the determinant. (The multi-dimensional generalization obtains in the same straightforward manner.)

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