what does it mean for the transpose of a matrix to be the negative of the matrix?

Question

Say I have matrix A,if $A^{T} = A$ then A is symmetrical. But what does it mean if $A^{T} = -A$ and can I know something about such a matrix inverse?

Answer 1

$A^T+A=0$ implies that $-a_{i,j}=a_{j,i}$ for all $i,j$, and in particular, $a_{i,i}=0$ for all $i$.

$\det(A)=\det(A^T)=\det(-A)=(-1)^n\det(A)$, so if $n$ is odd the determinant is 0, i.e. $A$ is non-invertible.

Answer 2

Assuming that the matrix is over the reals the following item springs to mind. If $A^T=-A$, then consider the set of matrices $$ G=\{\exp(tA)\mid t\in\Bbb{R}\}. $$ Here $\exp$ is the usual exponential function defined as a (everywhere convergent) power series $$ \exp(X)=\sum_{k=0}^\infty\frac{X^k}{k!}. $$ Whenever two matrices $X,Y$ commute, i.e. $XY=YX$, then we have $$ \exp(X+Y)=\exp(X)\exp(Y). $$ Compare with the usual result $e^{x+y}=e^x\cdot e^y$.

In particular this means that $\exp(tA)\exp(-tA)=\exp(0)=I$, so $\exp(tA)^{-1}=\exp(-tA)$. Here (finally!) we get to use the fact $-A=A^T$. Together with the preceding point we see that $$ \exp(tA)^{-1}=\exp(t(-A))=\exp(tA^T)=\exp(tA)^T. $$ In other words all the matrices in the set $G$ are orthogonal - their inverse is equal to the transpose.

TL;DR; The set $G$ is a group of orthogonal matrices.