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 $$\text{Dg}[J_1(0),J_2(0)] = E_{(3,1)}E_{(1,2)}, \label{eq:nilpotent_jkj2_1}$$ and it is immediately apparent that both factors are nilpotent and rank 1 as required.


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. $\square$

### References

[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.