I am currently studying material from a text called "Advanced Topics in Linear Algebra" by O'Meara, Clark, and Vinsonhaler. A large topic in the book is the Weyr form of a matrix. The explanation begins with looking at the nilpotent $10 \times 10$ matrix $J$ in Jordan form with Jordan structure (4,4,2), that is for eigenvalues $\lambda_i = 0$ for $i = 1, \dots, 10$, $$ J= \left[\begin{array}{} \left(\begin{array}{} 0 & 1 & 0 & 0\\ & 0 & 1 & 0\\ & & 0 & 1\\ & & & 0 \end{array}\right) & \\ & \left(\begin{array}{} 0 & 1 & 0 & 0\\ & 0 & 1 & 0\\ & & 0 & 1\\ & & & 0 \end{array}\right) &\\ & & \left(\begin{array}{} 0 & 1\\ & 0 \end{array}\right) \end{array}\right] $$ It goes on to describe how we can view $J$ as the linear transformation $T$ of $V = F^{10}$ relative to some (ordered) basis $B = \left\{v_1, v_2, \dots, v_{10} \right\}$, (e.g., the standard basis, in which case T acts under left multiplication by $J$ on column vectors). The text represents the action of $T$ on the basis vectors as: $$ 0 \gets v_1 \gets v_2 \gets v_3 \gets v_4 $$ $$ 0 \gets v_5 \gets v_6 \gets v_7 \gets v_8 $$ $$ 0 \gets v_9 \gets v_{10}. $$ We can view this as the "cyclic" shifting with the 3 Jordan blocks. To then transition toward the definition of the Weyr form they draw our attention to the columns. We see that the first column after the zeros reveals that $\left\{v_1,v_5,v_9 \right\}$ is the basis for the null space of $T$, the next column reveals that $\left\{v_1,v_5,v_9,v_2,v_6,v_{10} \right\}$ is a basis for the null space for $T^2$, etc. If we now reorder the basis vectors of $B$ by column order we get a new basis $$B' = \left\{v_1, v_5, v_9, v_2, v_6, v_{10}, v_3, v_7, v_4, v_8 \right\} = \left\{v'_1, v'_2, \dots, v'_{10} \right\}.$$ We now view $B'$ as a bunch of four groupings corresponding to the columns. We can now represent this as: $$ \begin{matrix} 0 & 0 & 0 \\ \uparrow & \uparrow & \uparrow \\ v'_1 & v'_2 & v'_3 \\ \uparrow & \uparrow & \uparrow \\ v'_4 & v'_5 & v'_6 \\ \uparrow & \uparrow & \\ v'_7 & v'_8 & \\ \uparrow & \uparrow & \\ v'_9 & v'_{10} \end{matrix} $$ Copied and pasted from the text, "This diagram, of course, is just the transpose of the Jordan diagram above. Note with the Jordan view, it was the “within-row” action which was interesting— there was no interaction between rows. With the new basis, it is the “interaction between rows” which strikes one. It is almost as if the various cyclic shiftings of the Jordan matrix on basis vectors within each of its three basic blocks have been replaced by a single cyclic shift on the four subspaces spanned by the rows." The new matrix of $T$ in the new basis $B'$ is: $$ W= \left[ \begin{matrix} 0 & 0 & 0 & 1 & 0 & 0 & & & & \\ 0 & 0 & 0 & 0 & 1 & 0 & & & & \\ 0 & 0 & 0 & 0 & 0 & 1 & & & & \\ & & & 0 & 0 & 0 & 1 & 0 & & \\ & & & 0 & 0 & 0 & 0 & 1 & & \\ & & & 0 & 0 & 0 & 0 & 0 & & \\ & & & & & & 0 & 0 & 1 & 0 \\ & & & & & & 0 & 0 & 0 & 1 \\ & & & & & & & & 0 & 0 \\ & & & & & & & & 0 & 0 \end{matrix} \right] $$ This matrix $W$ will turn out to be the Weyr form of $J$. Since $W$ and $J$ are matrices of the same transformation $T$ under reordering of a basis, we must have $W = P^{-1}JP$ where $P$ is the corresponding permutation matrix.
So, this was a close paraphrase from the text, but what I am seeking is more clarification on how we obtain the new basis, and further where those vectors in $B'$ appear in $W$. Also, this might be trivial but the original vectors in $B$ are the column vectors correct? That is $v_1 = \vec{0}$, $v_5 = \vec{0}$, and $v_9 = \vec{0}$. Any help is greatly appreciated since there is very little info online about the Weyr, at least what I have access to.