dg.differential geometry - Singular matrix and wedge product

Your condition on $X$ is that it has a kernel, and that by itself does not mean that
$X wedge X$ doesn't have to vanish. For instance in five dimensions, you could have
$$X = begin{pmatrix} 0 & 1 & 0 & 0 & 0 \\ -1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 end{pmatrix}.$$
Then letting $t = (0,0,0,0,1)$, your condition is satisfied, but $X wedge X$ is not zero.

However, it is true in $2n$ dimensions that a non-zero $t$ exists if and only if $X^{wedge n} = 0$. That's because $X^{wedge n}$ is proportional to the Pfaffian of $X$, which is a certain square root of the determinant of $X$. (In odd dimensions, the determinant of an antisymmetric matrix is zero by calculation, while the Pfaffian is set to zero by definition. So $t$ always exists in this case.)