As for part a), remember what it means for the operator to be linear. You need to check that
for all infinite row vectors , and furthermore you need to check for all scalars that
Let me get you started on the first part of a), so write
On the other hand, we have
Remember that you want to be equal to . Is this clear as of now?
For this part, I guess that you just need to realize, what the statement really means. To show that two linear operators are equal (which in our case amounts to showing that and are equal), you must show that the operators are equal on all vectors, i.e. that the operators are equal when applied to an arbitrary element in your space . On the right hand side, we have the identity operator on , the one that maps an infinite row vector to itself, so that
for all .
On the left hand side of , you have the composition of and , and the exercise is to prove that if you first apply to an infinite row vector and then apply to whatever the result of this is, then you get the original infinite row vector back again.
The calculation goes as follows:
and this is exactly the identity operator applied to .
In conclusion, the two operators and are equal on all vectors , and hence they are equal as operators.
Well, there's an easy reason why isn't invertible. The reason is that invertible operators are necessarily bijections, and in the problem description, you have been given that the image of is a proper subspace of (which is clear anyway), so the operator is not surjective, and hence not bijective.
Another not-as-easy reason is that since , then if had been invertible, it would also have been the case that .
(Why? If an operator has both a left inverse , and a right inverse , then the two inverses coincide:
The punchline here is of course that it is not the case that . Indeed, first maps to , which then maps to . This is not the same as the identity operator applied to , and hence .