Is the set of nilpotent $m \times m$ real matrices compact?

I found the proof of this statement, using Heine-Borel theorem on $\mathbb R^n$. Tha'ts quite good.

But, is it possible to prove this statement via the following definition of compactness?

A set is said to be compact if every open cover admits a finite subcover.

Assume the usual topology.

  • $\begingroup$ mmm.. but via opensets how could i start the proof?? $\endgroup$
    – David
    Mar 4 '15 at 3:21
  • 5
    $\begingroup$ It is quite surprising that you found a proof of this fact, as it is false! $\endgroup$ Mar 4 '15 at 3:21
  • $\begingroup$ Yes, well that would be a problem. Yes, this is false. For example, $\begin{pmatrix} 0 & n\\ 0 & 0\end{pmatrix}$ is nilpotent for every $n$, so the set of nilpotent matrices is not bounded. $\endgroup$
    – Moya
    Mar 4 '15 at 3:25
  • 1
    $\begingroup$ Heine Borel says a set is compact in Euclidean space if and only if it is closed and bounded. $\endgroup$
    – Moya
    Mar 4 '15 at 3:29
  • 1
    $\begingroup$ You can exhibit an open covering that has no finite subcovering. Think about trying to find open balls containing matrices of the form I wrote above (in $\mathbb{R}^4$) that are small enough such that they are disjoint. Then you can't refine this cover. $\endgroup$
    – Moya
    Mar 4 '15 at 3:36

The set $N$ of $m \times m$ real nilpotent matrices is compact iff $m = 1$, in which case $N$ is the singleton containing the zero matrix (in particular it is finite, hence compact).

For $m > 1$, consider the norm covering of $M(m, \mathbb{R})$ by open balls, namely the covering $\mathcal{B} := \{B_r : r \in \mathbb{Z}_+\}$, where $$B_r := \{A \in M(m, \mathbb{R} : ||A|| < r \},$$ and $||A||$ is (for example) $\sqrt{\sum_{i, j = 1}^m A_{ij}^2}$.

Now, by construction, $\{B_r \cap N : i \in \mathbb{Z}_+\}$ is an open cover of $N$, and we will show it admits no finite subcover: The sets $B_r$ are nested, that is, $B_1 \subseteq B_2 \subseteq B_3 \subseteq \cdots$, and hence so are the sets $B_r \cap N$; thus, the union of the sets in any finite subcollection $\{B_{r_1} \cap N, \ldots, B_{r_k} \cap N\} \subset \mathcal{B}$ is just $B_R$, where $R := \max\{r_1, \ldots, r_k\}$. On the other hand, the matrix $R_0$ with $(1, 2)$ entry $R$ and all other entries zero is in $N$ (its square is the zero matrix), but $||R_0|| = R$, so $R_0$ is not in $B_R$, so hence arbitrary finite subcollection is not a cover of $N$, that is, $N$ is not compact.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.