Is a Set of All Nilpotent Matrix a Vector Space?

Nilpotent Matrix Problems and Solutions

Problem 236

Let $V$ denote the vector space of all real $n\times n$ matrices, where $n$ is a positive integer.

Determine whether the set $U$ of all $n\times n$ nilpotent matrices is a subspace of the vector space $V$ or not.

 
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Definition.

An matrix $A$ is a nilpotent matrix if there exists a positive integer $k$ such that $A^k$ is the zero matrix.

Solution.

Case: $n=1$

When $n=1$, the only $1\times 1$ nilpotent matrix is $(0)$. Thus $U=\{(0)\}$ and it is a subspace of $V$.

Case: $n=2$

Let us next consider the case $n=2$.

In this case, we prove that the set $U$ of all $2\times 2$ nilpotent matrices is not a subspace of $V$.
The set $U$ is not a subspace because it is not closed under addition as the following example shows.
Consider the matrices
\[A=\begin{bmatrix}
0 & 1\\
0& 0
\end{bmatrix}
\text{ and }
B=\begin{bmatrix}
0 & 0\\
1& 0
\end{bmatrix}.\] It is easy to compute and see that $A^2$ and $B^2$ are both the zero matrices.
Hence $A$ and $B$ are nilpotent matrices.
However we show that their sum
\[A+B=\begin{bmatrix}
0 & 1\\
1& 0
\end{bmatrix}\] is not a nilpotent matrix. We have
\begin{align*}
(A+B)^k=
\begin{cases}
A+B & \text{ if $k$ is odd}\\
I & \text{ if $k$ is even},
\end{cases}
\end{align*}
where $I$ is $2\times 2$ identity matrix. Thus $(A+B)^k$ is not the zero matrix for any integer $k$, and thus $A+B$ is not a nilpotent matrix.
So we saw that even though $A$ and $B$ are in $U$, their sum $A+B$ is not in $U$.
Namely, the set $U$ is not closed under addition, hence $U$ is not a subspace of $V$.

Case: $n \geq 3$

Finally, we consider the case $n \geq 3$.
We claim that the set $U$ of all $n\times n$ nilpotent matrices is not a subspace of the vector space $V$.

Let $A$ and $B$ be the $2\times 2$ matrices as above. Then consider the $n\times n$ matrices
\[A’=
\left[\begin{array}{c|c}
A & O\\
\hline
O& O
\end{array}
\right] \text{ and }
B’=
\left[\begin{array}{c|c}
B & O\\
\hline
O& O
\end{array}
\right],\] that is, $A’$ is the $n \times n$ matrix whose left top $2\times 2$ part is $A$ and the remaining entries are all $0$. Similarly for $B’$.

Then it is easy to check that $A’^2, B’^2$ are both zero matrices but $(A’+B’)^k$ is not a zero matrix for any integer $k$.
Thus, $A’, B’$ are nilpotent matrices but their sum $A+B$ is not nilpotent.
Hence $U$ is not a subspace of $V$ since it is not closed under addition.

Summary

In summary, the set $U$ of all $n\times n$ nilpotent matrices is a subspace of $V$ if and only if $n=1$.


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