# A interesting question on Skew-symmetric matrix...finding the determinant.

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 $\det C =1$.

• 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. Sep 9, 2018 at 18:26

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.
• 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 $a_1^2 \dots a_{2n}^2$. Sep 10, 2018 at 19:07