Why is $\sum\limits_{k\ge0}\frac{(-1)^k}{(2k+1)!}t^{2k+1}A^{2k+1}$ absolut convergent?

Question

Why is $\sin(tA)=\sum\limits_{k\ge0}\frac{(-1)^k}{(2k+1)!}t^{2k+1}A^{2k+1}$ absolute convergent ? for a $n\times n$ real matrix $A$ and $t\in \mathbf R$

Which crieterion is to use ?

Answer 1

Hint. One may recall that the power series $$ \sin(x)=\sum\limits_{k\ge0}\frac{(-1)^k}{(2k+1)!}x^{2k+1} $$ has an infinite radius of convergence and use the fact that $$ \left|\left|\frac{(-1)^k}{(2k+1)!}t^{2k+1}A^{2k+1}\right|\right| \leq \frac{1}{(2k+1)!}|t|^{2k+1}\left|\left|A\right|\right|^{2k+1}. $$