Page 363 - 35Linear Algebra

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and since ker L = {0 V } we have L = dim ker L = 0, so

dim W = dim V = rank L = dim L(V ).

Since L(V ) is a subspace of W with the same dimension as W, it
must be equal to W. To see why, pick a basis B of L(V ). Each
element of B is a vector in W, so the elements of B form a linearly
independent set in W. Therefore B is a basis of W, since the size
of B is equal to dim W. So L(V ) = span B = W. So L is surjective.

14. (a) F 4 = F 2 + F 3 = 2 + 3 = 5.
(b) The number of pairs of doves in any given year equals the number of
the previous years plus those that hatch and there are as many of them
as pairs of doves in the year before the previous year.

F 1 1 F 2 1
(c) X 1 = = and X 2 = = .
F 0 0 F 1 1

1 1 1 1
MX 1 = = = X 2 .
1 0 0 1
(d) We just need to use the recursion relationship of part (b) in the top
slot of X n+1 :

F n+1 F n + F n−1 1 1 F n
X n+1 = = = = MX n .
F n F n 1 0 F n−1

(e) Notice M is symmetric so this is guaranteed to work.

1 − λ 1 1 2 5
det = λ(λ − 1) − 1 = λ − − ,
1 −λ 2 4
√ !
√ 1± 5
so the eigenvalues are 1± 5 . Hence the eigenvectors are 2 ,
2
1
√ √ √ √
respectively (notice that 1+ 5 + 1 = 1+ 5 1+ 5 and 1− 5 + 1 =
.
√ √ 2 2 2 2
1− 5 1− 5 ). Thus M = PDP −1 with
.
2 2
√ √
! √ !
1+ 5 0 1+ 5 1− 5
D = 2 √ and P = 2 2 .
0 1− 5 1 1
2
n
n
) = PDP
(f) M = (PDP −1 n −1 PDP −1 . . . PDP −1 = PD P −1 .
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