Math 2270
Quiz 7
1. In the space of all infinite sequences, consider the subspace W of arithmetic sequences, i.e., sequences of the
form
(a, a + k, a + 2k, a + 3k, . . .)
for some constants a and k. Find a basis for W and verify that your basis satisfies all the requirements for a
basis.
If a sequence is in W , then it can be written as
(a, a + k, a + 2k, a + 3k, . . .) = a(1, 1, 1, 1, . . .) + k(0, 1, 2, 3, . . .).
Therefore,
W = span((1, 1, 1, 1, . . .), (0, 1, 2, 3, . . .)).
All we have left to do is show that these two sequences are linearly independent. Suppose c1 and c2 are constants
such that
c1 (1, 1, 1, 1, . . .) + c2 (0, 1, 2, 3, . . .) = (0, 0, 0, 0, . . .).
Then it must be that
c1 + c2 (0) = 0 ⇒ c1 = 0 (first element),
c1 + c2 (1) = 0 ⇒ c2 = 0 (second element).
This shows that B = {(1, 1, 1, 1, . . .), (0, 1, 2, 3, . . .)} is a basis for W .
2. Show that the linear transformation
T (f (t)) = f (2t)
from P2 to P2 (that is T (a + bt + ct2 ) = a + 2bt + 4ct4 ) is an isomorphism.
Since we are given that T is linear, we only have to show that T is one to one and onto. To show that T is onto,
we show that the equation
T (f ) = a + b + ct2
has a solution f ∈ P2 for any choice of a, b, and c. If we take
c
b
f (t) = a + t + t2 ,
2
4
then T (f ) = a + bt + ct2, so T is onto. To show that T is one to one, we can just show that ker(T ) = {0}. Suppose
that
0 = T (a + bt + ct2 ) = a + 2bt + 4ct2 .
Then it must be true that a = b = c = 0 (by the fundamental theorem of algebra). This shows that T is also one
to one, so it is an isomorphism.