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 $X$.

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.)

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 $X^{\wedge 2n} = 0$. That's because $X^{\wedge 2n}$ is proportional to the Pfaffian of $X$, which is a certain square root of the determinant of $X$.

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 $X$.