How can I construct a nilpotent matrix of order 100 and index 98?

Question

I know to construct a nilpotent matrix of order $n$ with index of nilpotency $n$, but how to construct a nilpotent matrix of order $n$ but index of nilpotency $(n-2)$? Is there any general rule for the same?

Answer 1

The other answer gives you a very general hint on how to tackle the problem. It can be narrowed to your problem.

Let say we want to construct a matrix of order $m$ and index $n$ ($m\ge n$). The most simple nilpotent matrix having order equal to the index is the one having $1$ on the upper (or under, but this is not the usual form) diagonal, for example with order (and index) $4$ :

$$J = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 &0 &1&0\\ 0&0&0&1\\0&0&0&0\end{bmatrix}$$

Geometrically, if we take the canonical basis $\{e_1,e_2,e_3,e_4\}$, we project $e_4$ into $e_3$, $e_3$ into $e_2$, $e_2$ in to $e_1$ and $e_1$ into $0$. Thus $e_4$ will be projected to $0$ at the $4$-th application of $J$ and is the last one. This kind of matrix is known as Jordan's matrix.

Following the hint given in the other answer, the block matrix $\begin{bmatrix} J &\mathbf{0}\\\mathbf{0}^T&0 \end{bmatrix}$, where $\mathbf{0}$ is a zero vector of compatible size (4 elements in this case) , has the same index than $J$, thus $4$, but has order $5$.

Based on this, we are able to construct a matrix with index $4$ and arbitrary order $\ge4$. Can you generalize the idea ?

Answer 2

Hint: If $A_1$ is a nilpotent matrix of order $n_1$ and index $k_1$, and $A_2$ is a nilpotent matrix of order $n_2$ and index $k_2$, then their direct sum $A_1\oplus A_2$ is a nilpotent matrix of order ? and index ?