|
|
If $A$ is a complex $n \times n$ matrix such that $A^t=-A$, then $A$ is 0.
Proof: Let $J$ be the Jordan form of $A$.
Since $A^t=-A, J^t=-J$.
But $J$ is triangular so that $J^t=-J$ implies that every entry of $J$ is zero.
Since $J=0$ and $A$ is similar to $J$, we see that $A=0$. |
|
