= Solution
Write $y=(y_1,y_2,y_3)^T$. From the coordinate formula for the <cross product>,
$$
\boxed{T=
\begin{pmatrix}
1&y_3&-y_2\\
-y_3&0&y_1\\
y_2&-y_1&0
\end{pmatrix}}.
$$
If $y=c_1e_1$, this becomes
$$
\begin{pmatrix}1&0&0\\0&0&c_1\\0&-c_1&0\end{pmatrix}.
$$
Thus $c_1\ne0$ gives rank three and a zero-dimensional kernel, while $c_1=0$ gives rank one and kernel dimension two.
If $y\mathbin{\cdot}e_1=0$, then $y_1=0$. When $y\ne0$, the equations $Tx=0$ force $x_1=0$ and $y_3x_2-y_2x_3=0$, so
$$
\ker T=\mathbb Ry,
\qquad \operatorname{rank}T=2.
$$
When $y=0$, again $\operatorname{rank}T=1$ and $\dim\ker T=2$. These dimensions also agree with the <rank-nullity theorem>.
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