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$A$ is a $n \times n$ real matrix.

$A_{ij} = \begin{cases} \phantom{-}b_{i}c_{j} & \text{if } i = j \\ -b_{i}c_{j} & \text{if } i \ne j \end{cases}$

How to simplify $\det(A)$?


Update:

Can I simplify the determinant with elementary row and column operations as described at http://www.maths.nuigalway.ie/~rquinlan/MA203/section2-5.pdf?

  1. Divide each row by $b_{i}$ (elementary row operation)
  2. Divide each column by $c_{j}$ (elementary column operation)

$\det(A) = \left(\prod_{i=1}^{n} b_{i} \right) \left(\prod_{j=1}^{n} c_{j} \right) \det(S)$

where $S_{ij} = \begin{cases} + 1 & \text{if } i = j \\ -1 & \text{if } i \ne j \end{cases}$

So, the problem reduces to finding $\det(S)$.

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  • $\begingroup$ I guess that the sign pattern doesn't matter if the column vectors of signs are linearly independent from each other. But I am not sure. $\endgroup$
    – R zu
    Commented Jul 29, 2019 at 14:33
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    $\begingroup$ Look en.wikipedia.org/wiki/… and math.stackexchange.com/questions/319321/… I think $\det(A)=b_1c_1M^{(1)}_{11}+b_2c_1M^{(1)}_{21}-b_3c_1M^{(1)}_{31}+\cdots+(-1)^{n+1}(-b_nc_1)M^{(1)}_{n1}$ $\endgroup$ Commented Jul 29, 2019 at 15:30

3 Answers 3

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For $n\ge 3$ there is a clear pattern. For $n=3$ we have $$ \det(A)=- 4b_1b_2b_3c_1c_2c_3. $$ For $n=4$ we have $$ \det(A)=- 16b_1b_2b_3b_4c_1c_2c_3c_4. $$ So we should have $\det(A)=-f(n)\prod_{i=1}^n b_i\prod_{i=1}^ n c_i$ for all $n\ge 3$ with a positive integer $f(n)$. We have $$ f(3)=4,\; f(4)=16,\; f(5)=48,\; f(6)=128,\; f(7)=320. $$ Conjecture: $f(n)=2^n(n-1)$.

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  • $\begingroup$ What is the general formula for $f(n)$? $\endgroup$
    – R zu
    Commented Jul 29, 2019 at 14:49
  • $\begingroup$ Note that, by OEIS, your conjectured values satisfy: a(n) = -det(M(n+1)) where M(n) is the n X n matrix with m(i,i)=1, m(i,j)=-i/j for i != j. $\endgroup$
    – vadim123
    Commented Jul 29, 2019 at 15:01
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We can write $A = BSC$, where $S$ is the matrix you describe and $$ B = \pmatrix{b_1\\ & b_2 \\ && \ddots\\ &&& b_n}, \quad C = \pmatrix{c_1\\ & c_2 \\ && \ddots\\ &&& c_n}, $$ so that as you noted, we have $\det(A) = \det(B)\det(C) \det(S)$.

In order to compute $\det(S)$, it suffices to note that $$ S = 2I - \pmatrix{1 & \cdots & 1\\ \vdots & \ddots & \vdots \\ 1 & \cdots & 1}, $$ which is to say that $S$ is a rank-one update of a scalar matrix. One method to compute the determinant of such a matrix is by considering its eigenvalues: because $S$ has eigenvalues $2$ with multiplicity $n-1$ and $2-n$ with multiplicity $1$, we compute $$ \det(S) = 2^{n-1}(2-n). $$

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Diedrich Burde's conjecture (with $n$ shifted by one) is confirmed.

After pulling out the $b_i$'s and $c_i$'s, what's left is a circulant matrix whose first row is $[1, -1, -1, \ldots, -1]$, and associated polynomial $f(x)=1-x-x^2-\cdots-x^{n-1}$. The determinant of a circulant matrix is known to be $\prod_{j=0}^{n-1} f(\omega_j)$, where $\omega_j$ are the $n$ different complex $n$-th roots of unity, i.e. $\omega_j=\exp(\frac{2j\pi i}{n})$.

Now, $f(\omega_0)=f(1)=1-1-1-\cdots-1=2-n$. For all other $j$, $\omega_j\neq 1$. We have $$f(x)=2-(1+x+x^2+\cdots + x^{n-1})=2-\frac{x^n-1}{x-1}$$ and hence, for $j>0$, $f(\omega_j)=2-\frac{0}{\omega_j-1}=2$.

Substituting into our determinant formula, we get $$\prod_{j=0}^{n-1}f(\omega_j)=f(\omega_0)\prod_{j=1}^{n-1}f(\omega_j)=(2-n)\prod_{j=1}^{n-1} 2=(2-n)2^{n-1}$$

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