Proving a property of the exponential of square matrices

Question

Given square matrices $A$ and $B$ such that $AB = BA$, and the exponential of a matrix $M$ defined as

$$e^M := \sum_{k=0}^{\infty}\frac{M^k}{k!}$$

prove that $e^{A + B} = e^A e^B$.

I am lost as to how to proceed. I've never seen anything about matrix exponentials before.

Answer 1

$AB=BA$ so we may use the binomial theorem. Then, $e^{A+B} = \lim_{n\to \infty}\sum_{k=0}^n \frac{(A+B)^k}{k!}$. But: $$ \sum_{k=0}^n \frac{(A+B)^k}{k!}=\sum_{k=0}^n\frac{1}{k!}\sum_{i=0}^k \binom{k}{i} A^iB^{k-i}=\sum_{0\leq i\leq k \leq n}\frac{A^i}{i!}\frac{B^{k-i}}{(k-i)!} $$ Now notice how $k-i$ spans all of $\{0,\dots,n\}$. Try and finish the proof.

Answer 2

This is a big overkill, but every pair of commuting complex matrices is the sequential limit of pairs of commuting diagonalisable matrices (see this MO question for instance). So, by passing to limit, you may assume that $A$ and $B$ are simultaneously diagonalisable over $\mathbb C$. This reduces the problem to the scalar case.

Remark. Of course, the above assumes that the truthfulness of the scalar case has been established. Otherwise the above trick is not applicable and you must also prove the scalar case from ground up. This is basically what the other two answers do: repeat the same proof for the scalar case (which can be found in some introductory texts on mathematical analysis, such as baby Rudin), but treat commuting matrices as numbers.

Answer 3

We have

$e^Ae^B= (\sum_{k=0}^{\infty}\frac{A^k}{k!})(\sum_{k=0}^{\infty}\frac{B^k}{k!})$. Since $AB=BA$ you can compute the Cauchy product of the series on the RHS as in the case of real numbers.

The result is $\sum_{k=0}^{\infty} \frac{(A+B)^k}{k!}=e^{A+B}$.