Where we have used the condition that $ST=TS$, i.e, commutativity?

Question

definition

Let $A$ be an $n\times n$ matrix. Then for $t\in \mathbb R$, $$e^{At}=\sum_{k=0}^\infty \frac{A^kt^k}{k!}\tag{1}$$

Proposition

If $S$ and $T$ are linear transformations on $\mathbb R^n$ which commute, then $e^{S+T}=e^Se^T$.

Proof

By binomial therem, $$(S+T)^n=\sum_{j+k=0}^n n! \frac{S^jT^k}{j!k!}$$

therefore,

$$e^{S+T}=\sum_{n=0}^\infty\sum_{j+k=0}^n \frac{S^jT^k}{j!k!}=\sum_{j=0}^\infty \frac {S^j}{j!}\sum_{k=0}^\infty \frac{T^k}{k!}=e^Se^T$$

In the above steps I have used cauchy product and binomial theorem and the absolute convergence of the two series.

But My doubt is..

Where we have used the condition that $ST=TS$, i.e, commutativity?

Answer 1

You use it to establish binomial theorem. In general $$(A+B)^2=(A+B)(A+B)\neq A^2+2AB+B^2.$$

Answer 2

The binomial theorem step requires commutativity. As a simple example, consider:

$(S+T)^2 = S^2 + ST + TS + T^2$

To write this in the binomial theorem form, we must be able to assert $ST = TS$ so that the middle term is $2\cdot ST$.