**a)**
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

Then

and hence

On the other hand, we have

Remember that you want

to be equal to

. Is this clear as of now?

**b)**
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.

**c)**
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

.