Simplifying the determinant of the matrix whose $(i,j)$-th entry is $b_i c_j$ for $i=j$ and $-b_i c_j$ for $i\neq j$

Question

$A$ 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 $\det(A)$?

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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 $\det(S)$.

Answer 1

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 $$ \det(S) = 2^{n-1}(2-n). $$

Answer 2

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: $f(n)=2^n(n-1)$.

Answer 3

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 $$\prod_{j=0}^{n-1}f(\omega_j)=f(\omega_0)\prod_{j=1}^{n-1}f(\omega_j)=(2-n)\prod_{j=1}^{n-1} 2=(2-n)2^{n-1}$$