$t\mapsto\sin(tA)$ is continuous

Question

How to show that $t\mapsto\sin(tA)$ is continuous for a real matrix $A\in Mat(n,n)$

Can I use trigonometric identity,

$\sin y-\sin x=2\cos\left(\frac{x+y}{2}\right)\sin(y-x)$ but this holds only for real numbers, does it matter ?

Answer 1

By definition, $\sin tA=\frac1{2i}(e^{itA}-e^{-itA})$. So you need only to show that $$t\mapsto e^{itA},\quad t\in\Bbb R$$ is continuous. (For the $e^{-itA}$ part, just note that it is the inverse of $e^{itA}$ and that taking inverse is a continuous map in the invertible matrix space. [Or rather, we don't even need to bother with taking inverse, thanks to @Dr.MV's comment, we only need to note that it's a composition of $t\mapsto -t\mapsto e^{-itA}$]).

Now, for $x,y\in\Bbb R$, we see that $$\|e^{ixA}-e^{iyA}\|\le\|e^{ixA}\|\|I-e^{i(y-x)A}\|\le\|e^{ixA}\| \sum_{n=1}^\infty \frac1{n!}(|y-x|\|A\|)^n.$$ So it reduces to show the following limit $$\lim_{s\to 0^+}(e^{s}-1)=0. $$