Definition of matrix exponential Is there an alternative definition of a matrix exponential so I can use it to prove that $$e^{A}=\sum_{m=0}^{\infty} \frac{1}{m!}(A)^m \;?$$
Thanks a lot in advance!
 A: It's been some time since I've looked at it, but I believe I have occasionally seen the definition
$$ e^A = \lim_{n \to \infty} \left(I + \frac{1}{n} A \right)^n $$
If you're being asked to prove the Taylor series formula, you really ought to consult your textbook to see how your textbook defines the exponential, or other formulas for it that it has proven.
A: No. This is how matrix exponents are defined. In fact, this is how we usually define exponentiation, for example, in $\mathbb{C}$ it's also defined in terms of $e$. Why do you want to anyways?
A: $$\begin{align}&\lim_{n \to \infty} \left(I + \frac{1}{n} A \right)^n = \\ &\lim_{n \to \infty} \left(I^n+(n)I^{n-1}\frac{A}{n}+\frac{(n)(n-1)}{2}I^{n-2}\frac{A^2}{n^2} + \frac{(n)(n-1)(n-2)}{6}I^{n-3}\frac{A^3}{n^3} +\;...\right) = \\ &\lim_{n \to \infty}\left(I + A+\frac{(n)(n-1)}{2n^2}A^2 + \frac{(n)(n-1)(n-2)}{6n^3}A^3 ...\right)= \\ &\lim_{n \to \infty} I + \lim_{n \to \infty}A + \lim_{n \to \infty}\frac{(n)(n-1)}{2n^2}A^2 + \lim_{n \to \infty}\frac{(n)(n-1)(n-2)}{6n^3}A^3 +\; ...\; = \\ &I + \frac{A}{1!} + \frac{A^2}{2!} + \frac{A^3}{3!} +\; ... \;= \\ &\sum_{m=0}^\infty\frac{A^m}{m!} = e^A\end{align}$$
Pretty sure that binomial expansion for matrices is valid when the two matrices commute so this is valid because $IA = AI.$
A: As Stella said, no other way it is defined. That said, there are different ways to calculate it instead of infinite sum.
For instance the exponential of a diagonal matrix can be found by simply exponentiating every diagonal entry.
Going further, for any matrix non-singular $A$ you want to find the exponential of, the best way is to write it as: $$A = P^{-1}DP$$ where $D$ is a diagonal matrix found using eigenvectors/eigenvalues. Then $$e^A = P^{-1}e^{D}P$$ where the exponential of $D$ is just elements of $D$ being exponentiated as it is diagonal so easy to calculate.
There are many other ways including using Jordan forms to calculate exponentials of a matrix.
A: Following on from the remark of user88595, it is useful to know how to compute the exponential of a nilpotent matrix $M$ which has a single Jordan block. So suppose that $M$ is an $n \times n$ matrix with $M^{n} = 0  \neq M^{n-1}.$ Then 
$\{I,M,M^{2},\ldots , M^{n-1} \}$ is linearly independent and $e^{M} = I + M + \frac{M^{2}}{2!} + \ldots + \frac{M^{n-1}}{(n-1)!}.$ More generally, it is easy to see that $e^{(\lambda I + M)} = e^{\lambda}(I + M + \frac{M^{2}}{2!} + \ldots + \frac{M^{n-1}}{(n-1)!}).$ This enables us to compute $e^{J_{r}(\lambda)},$ where 
$J_{r}(\lambda)$ is a Jordan block of size $r$ with unique eigenvalue $\lambda.$ Hence we can calculate $e^{X}$ for any matrix $X$ in Jordan Normal Form.
A: You could define the matrix exponential as the solution of a differential equation, just like you can define the (real or complex) number $e^a$ as $f(1)/f(0)$ where $f$ is any nonzero function defined on an open interval$~I$ containing $[0,1]$ and satisfying the differential equation $f'(x)=af(x)$.
Similarly for any vector $x_0\in V$ there is a vector valued function $f:I\to V$ satisfying the linear differential equation $f'(x)=Af(x)$, and putting $x_1=f(1)$ for this function, the map $V\to V$ sending $x_0\mapsto x_1$ is linear, and equal to $\exp(A)$.
