Question 4.2.3: Let V be the space of all infinite sequences of real numbers......
Let V be the space of all infinite sequences of real numbers. Consider the transformation
T (x_0, x_1, x_2, . . .) = (x_1, x_2, x_3, . . .)
from V to V. (We drop the first term, x_0, of the sequence.)
a. Show that T is a linear transformation.
b. Find the kernel of T .
c. Is the sequence (1, 2, 3, . . .) in the image of T ?
d. Find the image of T .
a. T((x_0, x_1, x_2, . . .) + (y_0, y_1, y_2, . . .))=T (x_0 + y_0, x_1 + y_1, x_2 + y_2, . . .)= (x_1 + y_1, x_2 + y_2, x_3 + y_3, . . .)
equals
T((x_0, x_1, x_2, . . .) +T (y_0, y_1, y_2, . . .))= (x_1 , x_2, x_3, . . .)+ ( y_1, y_2, y_3, . . .)= (x_1+y_1,x_2+y_2,x_3+y_3, . . . ).We leave it to the reader to verify the constant-multiple rule.
b. The kernel consists of everything that is transformed to zero, that is, all sequences (x_0, x_1, x_2, . . .) such that
T (x_0, x_1, x_2, . . .) = (x_1, x_2, x_3, . . .) = (0, 0, 0, . . .).
This means that entries x_1, x_2, x_3,… all have to be zero, while x_0 is arbitrary. Thus, ker(T ) consists of all sequences of the form (x_0, 0, 0, . . .), where x_0 is arbitrary. The kernel of T is one-dimensional, with basis (1, 0, 0, 0, . . .). The nullity of T is 1.
c. We need to find a sequence (x_0, x_1, x_2, . . .) such that
T (x_0, x_1, x_2, . . .) = (x_1, x_2, x_3, . . .) = (1, 2, 3, . . .).
It is required that x_1 = 1, x_2 = 2, x_3 = 3, … , and we can choose any value for x_0, for example, x_0 = 0. Thus,
(1, 2, 3, . . .) = T (0, 1, 2, 3, . . .)
is indeed in the image of T .
d. Mimicking our solution in part (c), we can write any sequence (b_0, b_1, b_2, . . .) as
(b_0, b_1, b_2, . . .) = T (0, b_0, b_1, b_2, . . .),
so that im(T ) = V.