Exercise 5.3.22

Answers

With the help of previous exercise, it’s enough to show that $e^{D}$ exist if $P^{- 1} AP = D$ . Since we know that

{(D^{m})}_{ii} = {(D_{ii})}^{m},

we have that ,with the notation the same as the previous exercise,

{(E_{m})}_{ii} = 1 + D_{ii} + \frac{{(D_{ii})}^{2}}{2!} + \dots + \frac{{(D_{ii})}^{m}}{m!},

which tends to $e^{D_{ii}}$ as $m$ tends to infinity. So we know that $e^{D}$ exists.

jephianlin

2011-06-27 00:00