Turn non-invertable matrix into nilpotent

Question

$A$ is a non-invertable $n$ by $n$ matrix over a field $K$. Show that there are invertable matrices $S,T$ such that $SAT$ is nilpotent.

Answer 1

Suppose $A$ is a $n \times n$ matrix. It is a basic result that we can find invertible $n \times n$ matrices $S'$ and $T'$ such that $$ S' A T' = \begin{pmatrix} 1 & 0 & \cdots & 0 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & 0 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 1 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 & 0 & \cdots & 0 \end{pmatrix}. $$ (To see this take a basis $(b_1, \dotsc, b_k)$ of $\ker(A)$ and extend this to a basis $\mathcal{B} = (b_1, \dotsc, b_k, b_{k+1}, \dotsc, b_n)$ of $K^n$. Then $\{Ab_{k+1}, \dotsc Ab_n\}$ generates $\mathrm{im}(A)$, so we can choose a basis of $\mathrm{im}(A)$ of the form $(Ab_{i_1}, \dotsc, Ab_{i_r})$ for $r = \mathrm{rang}(A)$ and $k+1 \leq i_1 < \dotsb < i_r \leq n$. Extend this to a basis $\mathcal{C} = (Ab_{i_1}, \dotsc, Ab_{i_r},c_{r+1},\dotsc,c_n)$ of $K^n$. If we rearrange $\mathcal{B}$ and $\mathcal{C}$, and take $\mathcal{B}$ as a basis of the domain of the linear map $f \colon K^n \to K^n$, $x \mapsto Ax$ and $\mathcal{C}$ as a basis of its codomain, then with respect to these bases $f$ is represented by the above matrix.)

It sufficies to show the statement for $S' A T'$, so can assume w.l.o.g. that $A$ is of the above form. Then for $r = \mathrm{rang}(A)$ we have $A e_i = e_i$ for every $1 \leq i \leq r$ and $A e_i = 0$ for $r < i \leq n$ (that $r < n$ corresponds to the fact that $A$ is not inverible).

Consider the $n \times n$ permutation matrix $S$ with $Se_i = e_{i+1}$ for every $1 \leq i < n$ and $Se_n = e_1$ and let $T$ be the identity matrix. Then $SAT e_i = e_{i+1}$ for all $1 \leq i \leq r$ and $SAT e_i = 0$ for all $r < i \leq n$, so $SAT$ is nilpotent.