Question 7.E.3.9: Explain why e^A+B = e^A e^B whenever AB = BA. Give an exampl......
Explain why e^{A+B} = e^A e^B whenever AB = BA. Give an example to show that e^{A+B}, e^A e^B, and e^B e^A all can differ when AB ≠ BA.
Hint: Exercise 7.2.16 can be used for the diagonalizable case. For the general case, consider F(t) = e^{(A+B)t} − e^{At} e^{Bt} and F^′(t).
If A and B are diagonalizable with AB = BA, Exercise 7.2.16 insures A and B can be simultaneously diagonalized. If P^{−1}AP=D_A = diag (λ_1, λ_2, . . . , λ_n) and P^{−1}BP = D_B = diag (μ_1, μ_2, . . . , μ_n), then A + B = P(D_A +D_B) P^{−1}, so
e^{A+B} = P (e^{D_{A} + D_{B}}) P^{-1} = P \begin{pmatrix}e^{λ_{1}+μ_{1}}& 0& · · ·& 0\\ 0 &e^{λ_{2}+μ_{2}}& · · ·& 0\\ \vdots&\vdots&\ddots&\vdots\\ 0 &0 &· · ·& e^{λ_{n}+μ_{n}}\end{pmatrix} P^{-1}
= e^A e^B.
In general, the same brute force multiplication of scalar series that yields
e^{x+y} = \sum\limits_{n=0}^{∞} \frac{(x + y)^n}{n!} = \left(\sum\limits_{n=0}^{∞} \frac{x^n}{n!}\right) \left(\sum\limits_{n=0}^{∞} \frac{y^n}{n!}\right) = e^x e^y
holds for matrix series when AB = BA, but this is quite messy. A more elegant approach is to set F(t) = e^{At+Bt} − e^{At} e^{Bt} and note that F^′(t) = 0 for all t when AB = BA, so F(t) must be a constant matrix for all t. Since F(0) = 0, it follows that e^{(A+B)t} = e^{At} e^{Bt} for all t. To see that e^{A+B}, e^A e^B, and e^B e^A can be different when AB ≠ BA, consider A = \begin{pmatrix}1&0\\0&0\end{pmatrix}and B = \begin{pmatrix}0&1\\1&0\end{pmatrix}.