# 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 $i\neq j$

$$A$$ 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 $$\det(S)$$.

• 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. – R zu Jul 29 at 14:33
• Look en.wikipedia.org/wiki/… and math.stackexchange.com/questions/319321/… I think $\det(A)=b_1c_1M^{(1)}_{11}+b_2c_1M^{(1)}_{21}-b_3c_1M^{(1)}_{31}+\cdots+(-1)^{n+1}(-b_nc_1)M^{(1)}_{n1}$ – Ahmed Hossam Jul 29 at 15:30

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

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

• What is the general formula for $f(n)$? – R zu Jul 29 at 14:49
• 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. – vadim123 Jul 29 at 15:01

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