I have some questions regarding nilpotent matrices. I know that the trace of an $n\ x\ n$ nilpotent matrix must be zero and that the rank of that matrix must be less than $n$. Thus, the determinant of the matrix will be zero. In searching around online for some examples of nilpotent matrices (other than those obvious ones which have all of their nonzero entries above the main diagonal), I came across this matrix which has a nilpotency index of 3: $$\begin{bmatrix}2&2&-2 \\\\ 5&1&-3 \\\\ 1&5&-3 \end{bmatrix}$$
What I observed is that the entries of the third column of this matrix are $-1/2$ times the sum of the entries in the first and second columns, and I assume this is simply a random option for creating a linear combination of the first two columns so that the matrix isn't full rank. So, I went about trying to create other nilpotent matrices, taking a similar approach. However, this proved to be challenging. But here is one I constructed that is nilpotent (of index 3): $$\begin{bmatrix}-1&-1&-1 \\\\ -1&2&2 \\\\ 2&-1&-1 \end{bmatrix}$$
In this matrix, we have a trace of zero, and the bottom row is the negative of the sum of the first two rows (a different condition from the first matrix listed above). Additionally, the second and third columns are equivalent. Although this turned out to be nilpotent, I went through many trial matrices that weren't nilpotent, but each one had a trace of zero, and I tried to make linear combinations of rows and columns (often both in the same matrix, building off of the selections along the main diagonal). So, my question is basically this: what special properties did these two matrices have that allowed them to be nilpotent that the others lacked (even though I tried constructing them in a very similar manner, even duplicating specific linear combinations with different values)?
I would also like to know this, please. While we can manipulate certain features of a matrix using an LU product (for instance, the product of the entries along the main diagonals of L and U will be the value of the determinant of the LU product, etc.), is there also a way to use an LU product (that is, decomposition in reverse) to force a certain value of the $trace$ of the resulting LU product? In this way, can the LU concept be used to construct a nilpotent matrix? Can we perhaps create an idempotent matrix using a controlled set of L and U matrices? Your insights are greatly appreciated.