Partial Solution Set, Leon §4.3 4.3.2 Let

(d) The coordinate vector of p(x) is (a0, a1, a2) T B
with respect to the basis B given by
[1, x, 1 + x2 ]. The nth power of L is given by the nth power of B, which is simple
to compute because of the simple diagonal structure of B; Bn ⎡
0 0 0
= ⎣ 0 1 0
0 0 2n ⎤
⎦ . It
follows that the coordinate vector for L n (p(x)) is B n (a0, a1, a2) T = (0, a1, 2 n a2), so
L n (p(x)) = a1x + 2 n a2(1 + x 2 ).
(e) There isn’t a part (e), but I would like to emphasis how this B = S−1AS helps:
The coordinate vector of p(x) is (a0, a1, a2) T with respect to standard basis. The
nth power of L is given by the nth power of A, which is not so simple to computer
since A is not diagonal as it was in part (d). However, from B = S−1AS we can
find A = SBS −1 , so then A2 = (SBS −1 ) 2 = SBS −1SBS −1 = SB2S −1 . Similarly,
A100 = SB100S−1 , or generally An = SBnS −1 , for all natural numbers n. Thus
⎡
A n = ⎣
1 0 1
0 1 0
0 0 1
⎤ ⎡
⎦ ⎣
0 0 0
0 1 0
0 0 2 n
⎤ ⎡
⎦ ⎣
1 0 1
0 1 0
0 0 1
which is easy to computer as a product of three matrices, instead of multiplying A
by itself n times. It follows that the coordinate vector for L n (p(x)) is A n (a0, a1, a2) T
which is
4.3.8 Suppose that A = SΛS −1 , where Λ is a diagonal matrix with main diagonal λ1, λ2, . . . , λn.
(a) Show that Asi = λisi for each 1 ≤ i ≤ n.
n�
(b) Show that if x = αisi, then Ak n�
x =
i=1
i=1
αiλ k i si.
(c) Suppose that |λi| < 1 for each 1 ≤ i ≤ n. What happens to A k x as k → ∞?
Solution:
(a) For any choice of i, 1 ≤ i ≤ n, we have
Asi = � SΛS −1� =
si
SΛ � S −1 �
si
= SΛei
= S (Λei)
4
= Sλiei
= λiSei
= λisi.
⎤
⎦
−1
,