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 $\det C =1$.
Best Answer
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.