I am practicing Jordan forms and came across the following example: $$A=\left(\begin{array}{rrr} 1&1&1\\-1&-1&-1\\1&1&1\end{array}\right).$$ Jordan canonical form is $J=\begin{pmatrix}0&0&0\\0&0&0\\0&0&1\end{pmatrix}$. My question is shouldn't $J=\begin{pmatrix}0&1&0\\0&0&0\\0&0&1\end{pmatrix}$ be the Jordan form?
I computed the eigenvalues which are 0 (multiplicity 2) and 1 (multiplicity 1). Now because of multiplicity 1 the block containing eigenvalue 1 is $1\times1$ matrix. With 0 we find the eigenvectors to be $E_1=\{\begin{pmatrix}-1\\1\\0\end{pmatrix}, \begin{pmatrix}-1\\0\\1\end{pmatrix}\}$. Since the dimension is 2 my Jordan block will have $2\times2$ block with diagonal entries 0 and thus we get the second form. Where am I making a mistake?