Prove the following $e^{iAx} = \cos (x)I + i \sin (x) A$

Question

the question says as following,

Let x be a real number and A a matrix such that $A^2 = I$. Show that the $e^{iAx} = \cos (x)I + i \sin (x) A$.

my problem is that I don't know how to deal with this equation using matrix form, I know how to deal with such a question $e^{ix} = \cos (x) + i \sin (x)$ but how to represent the matrix form in a graph in order to get the values of the exponential function $e^{iAx}$.

I will be very happy if someone explains this equation or any reference will be also good.

Thank you

Answer 1

Hint:

$$e^{iAx}=\sum_{k=0}^\infty\frac{(iAx)^k}{k!}\\=\sum_{k=0,\text{even }k}^\infty\frac{(ix)^k}{k!}A^k+\sum_{k=0,\text{odd }k}^\infty\frac{(ix)^k}{k!}A^k\\=\sum_{k=0,\text{even }k}^\infty\frac{(ix)^k}{k!}I+\sum_{k=0,\text{odd }k}^\infty\frac{(ix)^k}{k!}A.$$

Answer 2

Both sides satisfy the same differential equation and have the same initial conditions, so they agree.

More explicitly, let $f(x)=e^{iAx}$ and let $g(x)=I\cos x + i A \sin x$. These are matrix-valued functions of $x$. Compute the following: $$f'=iAe^{iAx},\quad f''=-e^{iAx},\quad g'=-I\sin x+i A\cos x,\quad g''=-I\cos x -iA\sin x.$$

Therefore $f''=-f$ and $g''=-g$. Moreover $f(0)=g(0)=I$ and $f'(0)=g'(0)=iA$. Thus $f\equiv g$.