11.5. FUNCTIONS OF MATRICES 303

where ∑∞n=0 anJn

k is an mk×mk matrix of the form given in 11.23 where A = S−1JS andthe Jordan form of A, J is given by 11.22. Therefore, you can define f (A) by the series in11.24.

Here is a simple example.

Example 11.5.2 Find sin(A) where A =

4 1 −1 11 1 0 −10 −1 1 −1−1 2 1 4

 .

In this case, the Jordan canonical form of the matrix is not too hard to find.4 1 −1 11 1 0 −10 −1 1 −1−1 2 1 4

=

2 0 −2 −11 −4 −2 −10 0 −2 1−1 4 4 2

 ·

4 0 0 00 2 1 00 0 2 10 0 0 2



12

12 0 1

218 − 3

8 0 − 18

0 14 − 1

414

0 12

12

12

 .

Then from the above theorem sin(J) is given by

sin

4 0 0 00 2 1 00 0 2 10 0 0 2

=

sin4 0 0 0

0 sin2 cos2 −sin22

0 0 sin2 cos20 0 0 sin2

 .

Therefore, sin(A) =2 0 −2 −11 −4 −2 −10 0 −2 1−1 4 4 2



sin4 0 0 00 sin2 cos2 −sin2

20 0 sin2 cos20 0 0 sin2



12

12 0 1

218 − 3

8 0 − 18

0 14 − 1

414

0 12

12

12

=M where the columns of M are as follows from left to right,

sin412 sin4− 1

2 sin20

− 12 sin4+ 1

2 sin2

 ,

sin4− sin2− cos2

12 sin4+ 3

2 sin2−2cos2−cos2

− 12 sin4− 1

2 sin2+3cos2

 ,

−cos2

sin2sin2− cos2cos2− sin2



sin4− sin2− cos212 sin4+ 1

2 sin2−2cos2−cos2

− 12 sin4+ 1

2 sin2+3cos2

 .