Christiaan Hattingh on ResearchGate

It was mentioned before [1] that the right factor on p.229, the last factorization in Lemma 3 by Wu [2], is in fact not nilpotent for certain values of $k$. Explicitly, the given factorization of $\text{Dg}[J_k(0),J_2(0)]$ is invalid for odd $k$, since the matrix \[ \begin{bmatrix} \mathbf{0} & J_2(0) \\ J_k(0) & \mathbf{0} \end{bmatrix} \] is not nilpotent when $k=7$.

Lemma 3 [2] (correction of the erroneous part)

Let $\mathscr{F}$ be an arbitrary field. Let $k$ be a positive, odd integer. Then the matrix $\text{Dg}[J_k(0),J_2(0)]$ is the product of two nilpotent matrices, each with rank equal to the rank of $\text{Dg}[J_k(0),J_2(0)]$.

Proof. First consider the case $k=1$, then \begin{equation} \text{Dg}[J_1(0),J_2(0)] = E_{(3,1)}E_{(1,2)}, \label{eq:nilpotent_jkj2_1} \end{equation} and it is immediately apparent that both factors are nilpotent and rank 1 as required.

Suppose $k \geq 3$. Then \begin{equation} \text{Dg}[J_k(0),J_2(0)] = \left [ {\renewcommand{\arraystretch}{1.2}\begin{array}{c|c} \mathbf{0} & A_1 \\ \hline J_2(0) & \mathbf{0} \end{array}} \right ] \left [ {\renewcommand{\arraystretch}{1.2}\begin{array}{c|c} \mathbf{0} & \begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix} \\ \hline A_2 & \mathbf{0} \end{array}} \right ]=N_1N_2, \label{eq:nilpotent_jkj2_odd} \end{equation} where \[A_1 = [e_2,\mathbf{0},e_4,e_3,e_6,e_5,\ldots,e_{k-1},e_{k-2},e_k]\] and \[A_2 = [e_1,e_4,e_3,e_6,e_5,\ldots,e_{k-1},e_{k-2},e_k,\mathbf{0}].\] Notice in particular that for the case $k=3$ we have $A_1=[e_2,\mathbf{0},e_3]$ and $A_2 = [e_1,e_3,\mathbf{0}]$. Now the factor $N_1$ has rank $k$ which is the same as the rank of $\text{Dg}[J_k(0),J_2(0)]$. Let \begin{equation}Q_1 = [e_1,e_{k+2},e_{k},e_{k-1},e_{k-4},e_{k-5},e_{k-8}\ldots,e_{3-(-1)^{(k-3)/2}},e_{k+1},e_{k-2},e_{k-3},\ldots,e_{3+(-1)^{(k-3)/2}}]. \label{eq:nilpotent_jkj2_odd_q} \end{equation} Then $Q_1$ is a change-of-basis matrix and \[Q_1^{-1}N_1Q_1 = \text{Dg}[J_{k-2 \cdot (\lfloor (k-3)/4 \rfloor)+(-1)^{(k-3)/2}}(0),J_{k-2 \cdot (1+\lfloor (k-3)/4 \rfloor)}(0)],\] so that $N_1$ is nilpotent.

Finally, it is easy to verify that $N_2$ also has rank $k$. Let \[Q_2 = [e_4,e_8,\ldots,e_{k+1},e_1,e_3, \ldots,e_{k},e_2,e_6, \ldots,e_{k-1},e_{k+2} ] \] when $(k-3)/2$ is even, and \[Q_2 = [e_2,e_6, \ldots, e_{k+1},e_1,e_3, \ldots,e_{k},e_4,e_8,\ldots,e_{k-1},e_{k+2}] \] when $(k-3)/2$ is odd. Then \[Q_2^{-1} N_2 Q_2 = \text{Dg} [J_{k-\lfloor (k-3)/4 \rfloor}(0),J_{2+\lfloor (k-3)/4 \rfloor}(0)],\] confirming that $N_2$ is nilpotent, and completing the proof. $\square$


[1] D.K. Bukovsek, T. Kosir, N. Novak, and P. Oblak, Products of commuting nilpotent operators, Electronic Journal of Linear Algebra, 16 (2007) 237 -- 247.

[2] P.Y. Wu, Products of Nilpotent Matrices, Linear Algebra and its Applications, 96 (1987) 227 -- 232.