hw3-solutions - MATH 53H, PROBLEM SET #3 SOLUTIONS 6.12....

MATH 53H, PROBLEM SET #3 — SOLUTIONS
6.12. Compute e
A
( e
tA
?), where A is given.
(b) A =
parenleftbigg
2 - 1
1 2
parenrightbigg
. We have p
A
( λ ) = ( λ - 2) 2 + 1, which has roots λ = 2 ± i and corresponding
eigenvectors z
±
= ( ± i, 1) T . Thus A = PDP - 1 where
P =
parenleftbigg
i - i
1 1
parenrightbigg
, D =
parenleftbigg
2 + i 0
0 2 - i
parenrightbigg
, P
- 1
=
1
2
parenleftbigg
- i 1
i 1
parenrightbigg
,
and
e
tA
= Pe
tD
P
- 1
=
e 2 t
2
parenleftbigg
i - i
1 1
parenrightbiggparenleftbigg
e
it
0
0 e
- it
parenrightbiggparenleftbigg
- i 1
i 1
parenrightbigg
=
e
2 t
2
parenleftbigg
e
it
+ e
- it
i ( e
it
- e
- it
)
- i ( e it - e - it ) e it + e - it
parenrightbigg
= e
2 t
parenleftbigg
cos t - sin t
sin t cos t
parenrightbigg
,
using the standard formulas
cos θ = Re e
iθ
=
e iθ + e - iθ
2
and
sin θ = Im e
iθ
=
e
iθ
- e
- iθ
2 i
=
i
2
( e
- iθ
- e
iθ
) .
Alternatively, let B = 2 I and C =
parenleftbigg
0 - 1
1 0
parenrightbigg
, so that A = B + C and BC = CB . Then C
2
= - I
implies
e
tC
=
∞ summationdisplay
j =0
( tC )
j
j !
=


∞ summationdisplay
j =0
( - t
2
)
j
(2 j )!


I +


∞ summationdisplay
j =0
( - 1)
j
t
2 j +1
(2 j + 1)!


C =
parenleftbigg
cos t - sin t
sin t cos t
parenrightbigg
and e
tA
= e
tB
e
tC
= e
2 t
e
tC
as above.
(c) A =
parenleftbigg
2 - 1
0 2
parenrightbigg
. The only eigenvalue of A is 2, so
L =
parenleftbigg
2 0
0 2
parenrightbigg
, N =
parenleftbigg
0 - 1
0 0
parenrightbigg
,
and
e
tA
=
parenleftbigg
e
2 t
-