I'm having trouble understanding why, in general, if $\mathbf A, \mathbf B$ are $n \times n$ matrices, it does not hold that $\exp(\mathbf A + \mathbf B) = \exp(\mathbf A)\exp(\mathbf B)$, but holds in general when $\mathbf A\mathbf B = \mathbf B \mathbf A$.

Specificall, I can try to prove this using the definition of matrix exponentials as follows, without using commutativity of the matrices:

Notice that \begin{align*} \exp(\mathbf A + \mathbf B) &\equiv \sum_{i=0}^{\infty} \frac{(\mathbf A + \mathbf B)^i}{i!} \\ &= \sum_{i=0}^{\infty} \sum_{j = 0}^{i} \binom ij \frac{\mathbf A^{i-j}\mathbf B^{j}}{i!} \\ &= \sum_{i=0}^{\infty} \sum_{j = 0}^{i} \frac{\mathbf A^{i-j}}{(i-j)!} \frac{\mathbf B^j}{j!} \end{align*} which does not rely on the commutativity of multiplying $\mathbf A, \mathbf B$, but only relies on the commutativity of matrix addition under absolute convergence.

Meanwhile, \begin{align*} \exp(\mathbf A)\exp(\mathbf B) &\equiv\left(\sum_{k = 0}^{\infty} \frac{\mathbf A^k} {k!} \right)\left(\sum_{l=0}^\infty \frac{\mathbf B^l}{l!}\right)\\ &= \sum_{k=0}^\infty\sum_{l=0}^\infty \frac{\mathbf A^k}{k!} \frac{\mathbf B^l}{l!} \end{align*} which still does not rely on the commutativity of multiplying the two matrices. However, we can see that both sums converge (because the matrix exponential always converges) and also that we can rewrite the first expression to be equal to the second one, because we can set any values for $k, l$ and see that an equivalent term occurs in the first sum, and vice versa.

Why does this fail when $\mathbf A, \mathbf B$ do not commute?


It is not true that $(A+B)^2=A^2+2AB+B^2$. All you can say is that$$(A+B)^2=A^2+AB+BA+B^2.$$

More generally, it is not true that$$(A+B)^n=\sum_{i=0}^n\binom niA^{n-i}B^i.$$

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  • $\begingroup$ Of course, thanks.... I feel stupid now. $\endgroup$ – dks28 Feb 25 at 10:01

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