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Let $a_1,a_2,\cdots ,a_{2n}$ be complex numbers. We construct a $2n \times 2n$ matrix, say $A$ which is skew symmetric and entries are from complex numbers. $A=(\alpha_{ij})$, where $\alpha_{ij}=a_ia_j$ for $i<j$. To find the determinant of the matrix $A$.


Since $A$ is a even order skew-symmetric matrix, we have determinant of $A$ a perfect square.

My Intuition: $\det A = a_1^2 \times a_2^2 \times \cdots \times a_{2n}^2$.

I was trying to see what happens when $n=2$, i.e. we have $4 \times 4$ matrix $A$.

Thus we have complex numbers $a_1,a_2,a_3 \ \text{and} \ a_4$ and \ $A= \begin{bmatrix} 0 & a_1a_2 & a_1a_3 & a_1a_4 \\ -a_1a_2 & 0 & a_2a_3 & a_2a_4 \\ -a_1a_3 & -a_2a_3 & 0 & a_3a_4\\ -a_1a_4 & -a_2a_4 & -a_3a_4 & 0 \end{bmatrix} $

We can see $ A= \left[ \begin{array}{c|c} D_1 & B \\ \hline -B^T & D_2 \end{array} \right] $, where

$D_1 = \begin{bmatrix} 0 & a_1a_2\\ -a_1a_2 & 0 \end{bmatrix} $,

$D_2 = \begin{bmatrix} 0 & a_3a_4\\ -a_3a_4 & 0 \end{bmatrix} $ and

$ B = \begin{bmatrix} a_1a_3 & a_1a_4\\ a_2a_3 & a_2a_4 \end{bmatrix} $

Also I have noted that $\det B =0$.


Can we use the result of determinant of block matrices? Can someone shed some light how to do the problem?


Let $ D = \text{diag}(a_1,a_2,\cdots,a_{2n})$ , and $C = (c_{ij})$, where $C$ is a skew symmetric matrix with $c_{ij} = 1$ when $i<j$. Then one can easily see that $A=DCD$. So we are let to prove that $\det C =1$.

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  • $\begingroup$ Have you tried to find the eigenvalues of $C$? $C$ is real skew-symmetric, so all eigenvalues are pure imaginary. They're the roots of real polynomial, so they come in conjugate pairs. Perhaps if you look at the first 2 or 3 examples, you may find a pattern. $\endgroup$ – John Hughes Sep 9 '18 at 18:26
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Yes, the determinant is $a_1^2 \cdots a_{2n}^2$. To see this, notice that if you divide the $i$'th row by $a_i$ for all $i$, and then divide the $i$'th column by $a_i$ for all $i$, then you get a matrix with entries in $\{0,1,-1\}$ whose determinant is easily seen (do some row-reduction!) to be 1.

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  • $\begingroup$ When you say is easily seen to be equal to $1$, can you elaborate? I tried and I don’t find it easy. The easy part is the one you cover factoring $a_1^2 \dots a_{2n}^2$. $\endgroup$ – mathcounterexamples.net Sep 10 '18 at 19:07
  • $\begingroup$ Subtract row 2 from row 1. Then subtract row 3 from row 2, etc. Then the first row becomes (1 1 0 ... 0) and the second row becomes (0 1 1 0 ... 0), the second last is (0 .. 0 1 1) etc. $\endgroup$ – Mark Sep 11 '18 at 0:07
  • $\begingroup$ Then at the end, add rows 1, 3, 5,... to the last row so that it becomes (0 ... 0 1). Then you have an upper triangular matrix with 1's on the diagonal. $\endgroup$ – Mark Sep 11 '18 at 11:18

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