Exercise 1286Topology › Topology of Mn(ℝ)L1Added 4 Nov 2026

A non-compact closed subspace: trace-zero matrices

#closed unbounded#trace#non-compactness

Statement

Let $H=\{A\in M_n(\mathbb{R}):\operatorname{tr}(A)=0\}$ . Show that $H$ is closed but not compact.

Hint

$H$ is the kernel of the trace functional; show it is unbounded.

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Closed. The trace is linear, hence continuous; $H=\operatorname{tr}^{-1}(\{0\})$ is closed.

Not compact. The matrices $A_k=k\,E_{12}$ have trace $0$ for every $k$ , but $\|A_k\|_\infty=k\to+\infty$ : unbounded.

Conclusion. $H$ is closed and unbounded, hence not compact.