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 $\det C =1$.
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.