Proving matrix exponent property How can I prove the following equation. I have tried but i couldn't.
$$\exp(A(t_2+t_1))=\exp(At_2)\cdot \exp(At_1)$$
$A$ is a matrix
Will I use state-transition matrix or what ?
Thank you...
 A: For matrix $A$, the definition of exponentiation is 
$$
e^A = \sum_{n\geq 0} \frac{A^n} {n!}
$$
Thus for any two matrices $A,B$
$$
e^{A+B} = \sum_{n\geq 0} \frac{(A+B)^n} {n!}
$$
Now $(A+B)^n = \sum C_i$ where each $C_i$ is a matrix of the form $ABBAABA \cdots BA$ with $n$ factors each of which is either $A$ or $B$. In general, you cannot combine two of the $C_i$, even if they contain the same numbers of $A$ factors, because $AB \neq BA$.
If $A$ commutes with $B$ (that is, if $AB = BA$) then you can move all the $A$'s to the left and group terms with like numbers of $A$ factors, getting
$$
(A+B)^n = \sum \binom{n}{k} A^kB^{n-k}
$$
In that case 
$$
e^{A+B} = \sum_{n\geq 0}  \frac{1}{n!} \sum \binom{n}{k}A^kB^{n-k}
$$
Again using commuting $A,B$ let's multiply
$$
e^A e^B = \sum_{n\geq 0} \frac{A^n} {n!} \sum_{m\geq 0} \frac{B^m} {m!}
$$
In that double sum, the combination $A^pB^q$ appears in the terms with denominator $(p+q)!$ and $\binom{p+q}{p}$ such terms appear. IF we regroup, writing $n=p+q$ and $k=p$, this gives 
$$e^A e^B = \sum_{n\geq 0}  \frac{1}{n!} \sum \binom{n}{k}A^kB^{n-k} = e^{A+B}
$$
Finally, observe that $t_1A$ commutes with $t_2A$ so 
$$
e^{t_1A} e^{t_2A} = e^{t_1A+t_2A} = e^{(t_1+t_2)A}
$$
