# Is the Product of a Nilpotent Matrix and an Invertible Matrix Nilpotent?

## Problem 77

A square matrix $A$ is called nilpotent if there exists a positive integer $k$ such that $A^k=O$, where $O$ is the zero matrix.

(a) If $A$ is a nilpotent $n \times n$ matrix and $B$ is an $n\times n$ matrix such that $AB=BA$. Show that the product $AB$ is nilpotent.

(b) Let $P$ be an invertible $n \times n$ matrix and let $N$ be a nilpotent $n\times n$ matrix. Is the product $PN$ nilpotent? If so, prove it. If not, give a counterexample.

## Hint.

For (b), the statement is false. Try to find a counter example.
A typical nilpotent matrix is an upper triangular matrix whose diagonal entries are all zero.

## Proof.

### (a) Show that $AB$ is nilpotent

Since $A$ is nilpotent, there exists a positive integer $k$ such that $A^k=O$. Then we have
$(AB)^k=(AB)(AB)\cdots (AB)=A^kB^k=OB^k=O.$

Here in the second equality, we used the assumption that $AB=BA$.
Thus we have $(AB)^k=O$, hence the product matrix $AB$ is nilpotent.

### (b) Is $PN$ nilpotent?

In general, the product $PN$ of an invertible matrix $P$ and a nilpotent matrix $N$ is not nilpotent.
Here is a counterexample.
Let
$P=\begin{bmatrix} 1 & 0 & 0 \\ 1 &1 &0 \\ 0 & 0 & 1 \end{bmatrix} \text{ and } N=\begin{bmatrix} 0 & 1 & 1 \\ 0 &0 &1 \\ 0 & 0 & 0 \end{bmatrix}.$

Then the matrix $P$ is invertible since $\det(P)=1$.
(Note that $P$ is a lower triangular matrix. So the determinant is the product of diagonal entries.)

Also, it is easy to see by direct computation that $N^3=O$, hence $N$ is nilpotent. Indeed,
$N^2=\begin{bmatrix} 0 & 0 & 1 \\ 0 &0 &0 \\ 0 & 0 & 0 \end{bmatrix}$ and
$N^3=N^2N=\begin{bmatrix} 0 & 0 & 1 \\ 0 &0 &0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 & 1 \\ 0 &0 &1 \\ 0 & 0 & 0 \end{bmatrix}=O.$

Now the product $PN$ is
$PN=\begin{bmatrix} 0 & 1 & 1 \\ 0 &1 &2 \\ 0 & 0 & 0 \end{bmatrix}.$ We show that $PN$ is not nilpotent.
We have
$(PN)^2=\begin{bmatrix} 0 & 1 & 2 \\ 0 &1 &2 \\ 0 & 0 & 0 \end{bmatrix}$ $(PN)^3=(PN)^2(PN)=\begin{bmatrix} 0 & 1 & 2 \\ 0 &1 &2 \\ 0 & 0 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 & 1 \\ 0 &1 &2 \\ 0 & 0 & 0 \end{bmatrix} =\begin{bmatrix} 0 & 1 & 2 \\ 0 &1 &2 \\ 0 & 0 & 0 \end{bmatrix}.$

This calculation shows that
$(PN)^k=\begin{bmatrix} 0 & 1 & 2 \\ 0 &1 &2 \\ 0 & 0 & 0 \end{bmatrix}\neq O \text{ for all } k \geq 2.$

Thus $PN$ is not nilpotent. In conclusion, the product $PN$ of the invertible matrix $P$ and the nilpotent matrix $N$ is not nilpotent.

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