I cam accross the following definition in Wikipedia https://en.m.wikipedia.org/wiki/Skew-symmetric_matrix , which is attached below , for a reference:
However, I dont get how the two definitions i.e one focussing upon the property $A^t=A$ and another giving a condition upon the elements $a_{ij}=-a_{ji}$ are equivalent. I don't get how those two conditions given , are equivalent to one another ?
My understanding behind this fact goes as follows:
If $A=(a_{ij})_{1\leq i\leq m,1\leq j\leq n}$, then $A^t=(b_{ij})_{1\leq i\leq m,1\leq j\leq n}$, where $b_{ij}=a_{ji}$, so, $A^t=(b_{ij})_{1\leq i\leq m,1\leq j\leq n}=(a_{ji})_{1\leq i\leq m,1\leq j\leq n}$. Now,if $A=(a_{ij})_{1\leq i\leq m,1\leq j\leq n}=(a_{ij})$, then $A^t=(a_{ji})_{1\leq i\leq m,1\leq j\leq n}=(a_{ji})$ and hence if $-A^t=A$, then $(-A^t=)(-a_{ji})=(A=)(a{ij})$ or $a_{ij}=-a_{ji}$.So we can say, if , $-A^t=A$ then $a_{ij}=-a_{ji}$. Now, conversely if $a_{ij}=-a_{ji}$ then it follows trivially that $-A^t=A$. So, we can say $-A^t=A$ iff $a_{ij}=-a_{ji}$.
Is my understanding of the proof correct?
Just write the matrices in terms of their entries. Recall that if
$$A = (a_{i,j})_{1 \le i \le m, 1 \le j \le n}$$
then the transpose is given by
$$A^{\mathsf{T}} = (a_{j,i})_{1 \le j \le n, 1 \le i \le m}$$
Hence, if $A^{\mathsf{T}} = -A$ then
$$a_{j,i} = - a_{i,j} \text{ for all } i \in \{1,2,\cdots,m\}, j \in \{1,2,\cdots,n\}$$
(that is, corresponding entries in each position are equal).
In particular, owing to the ranges in which $i,j$ are defined, we automatically know that $A$ is square if this condition is satisfied (it needs to be the "same shape" before and after) so we can write (in a less cumbersome manner)
$$A = (a_{i,j})_{1\le i,j \le m} \qquad A^{\mathsf{T}} = (a_{j,i})_{1\le i,j \le m}$$
and hence $A^{\mathsf{T}} = -A$ if and only if
$$a_{j,i} = - a_{i,j} \text{ for all } i,j \in \{1,2,\cdots,m\}$$