Exercise 5.3.21

Answers

We set

B_{m} = I + A + \frac{A^{2}}{2!} + \dots + \frac{A^{m}}{m!}

and

E_{m} = I + D + \frac{D^{2}}{2!} + \dots + \frac{A^{m}}{m!} .

We may observe that $B_{m} = P E_{m} P^{- 1}$ for all $m$ . So by the definition of exponential of matrix and Theorem 5.12, we have

e^{A} = \lim_{m \to \infty} B_{m} = \lim_{m \to \infty} (P E_{m} P^{- 1}) = P (\lim_{m \to \infty} E_{m}) P^{- 1} = P e^{D} P^{- 1} .

jephianlin

2011-06-27 00:00