Exercise 1967Topology › Topology of Mn(ℝ)M1Added 4 Nov 2026

Continuity of the matrix exponential

#matrix exponential#continuity#series

Statement

Show that $\exp:M_n(\mathbb{R})\to M_n(\mathbb{R})$ , $\exp(A)=\sum_{k\ge0}\frac{A^k}{k!}$ , is continuous.

Hint

Bound $\|\exp(A)-\exp(B)\|$ using the series and the inequality $\|A^k-B^k\|\le\|A-B\|\sum_{j=0}^{k-1}\|A\|^j\|B\|^{k-1-j}$ . Or use local boundedness and uniform convergence on bounded sets.

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Solution

Local bound. Fix $R>0$ and $A,B$ with $\|A\|,\|B\|\le R$ . For each $k\ge1$ :
$\|A^k-B^k\|\le\|A-B\|\cdot kR^{k-1}.$
Summing: $\|\exp(A)-\exp(B)\|\le\|A-B\|\sum_{k\ge1}\frac{kR^{k-1}}{k!}=\|A-B\|\,e^R$ .

Conclusion. $\exp$ is locally Lipschitz on $M_n(\mathbb{R})$ , hence continuous.

Continuity of the matrix exponential

Statement

Solution

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