The problem here is to prove that if $A$ is an $n \times n$ matrix with rank $r$ and is a nilpotent matrix, then $A^{r+1} = 0$.

Solution Outline:

1. Nilpotency: By definition, a matrix $A$ is nilpotent if there exists a positive integer $k$ such that $A^k = 0$. We are given that $A$ is nilpotent, so there exists some integer $m$ for which $A^m = 0$.

2. Rank Argument: The rank of $A$ is $r$, meaning the maximum number of linearly independent columns (or rows) of $A$ is $r$. This also implies that the dimension of the image of $A$ (or the column space) is $r$.

3. General Nilpotency Proof for Rank $r$:

   • In general, if $A$ is an $n \times n$ nilpotent matrix with rank $r$, the power $A^{r+1}$ will map all vectors in the space to zero.
   • This is because each application of $A$ reduces the dimension of the image by at least one until reaching the zero space by the $r+1$-th application, since there are only $r$ independent directions initially.

4. Conclusion:

   • Therefore, for a nilpotent matrix $A$ with rank $r$, $A^{r+1} = 0$.

Final Answer:

Thus, $A^{r+1} = 0$ as required.

Do you want a more detailed step-by-step explanation?

Here are some related questions for further exploration:

1. What defines a nilpotent matrix, and what are some of its properties?
2. How is the rank of a matrix related to its powers?
3. Why does the rank decrease with each power of a nilpotent matrix?
4. How does the Jordan form of a nilpotent matrix provide insight into the nilpotency degree?
5. What is the significance of the minimal polynomial of a matrix in determining nilpotency?

Tip: For any nilpotent matrix, the minimal polynomial's degree provides the smallest power at which the matrix becomes zero.