Originally Posted by

**HappyJoe** **a)**

As for part a), remember what it means for the operator $\displaystyle S^+$ to be linear. You need to check that

$\displaystyle S^+(a+b) = S^+(a)+S^+(b)$

for all infinite row vectors $\displaystyle a,b\in \mathbb{R}^\infty$, and furthermore you need to check for all scalars $\displaystyle c\in\mathbb{R}$ that

$\displaystyle S^+(c a) = cS^+(a).$

Let me get you started on the first part of a), so write

$\displaystyle a=(a_1,a_2,\ldots)$

$\displaystyle b=(b_1,b_2,\ldots)$

Then

$\displaystyle a+b = (a_1+b_1,a_2+b_2,\ldots),$

and hence

$\displaystyle S^+(a+b) = (0,a_1+b_1,a_2+b_2,\ldots).$

On the other hand, we have

$\displaystyle S^+(a) = (0,a_1,a_2,\ldots),$

$\displaystyle S^+(b) = (0,b_1,b_2,\ldots).$

Remember that you want $\displaystyle S^+(a)+S^+(b)$ to be equal to $\displaystyle S^+(a+b)$. Is this clear as of now?

**b)**

For this part, I guess that you just need to realize, what the statement $\displaystyle S^-S^+=I$ really means. To show that two linear operators are equal (which in our case amounts to showing that $\displaystyle S^-S^+$ and $\displaystyle I$ 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 $\displaystyle \mathbb{R}^\infty$. On the right hand side, we have the identity operator on $\displaystyle \mathbb{R}^\infty$, the one that maps an infinite row vector $\displaystyle a=(a_1,a_2,\ldots)$ to itself, so that

$\displaystyle I(a) = a$

for all $\displaystyle a$.

On the left hand side of $\displaystyle S^-S^+=I$, you have the composition of $\displaystyle S^-$ and $\displaystyle S^+$, and the exercise is to prove that if you *first* apply $\displaystyle S^+$ to an infinite row vector $\displaystyle (a_1,a_2,\ldots)$ and *then* apply $\displaystyle S^-$ to whatever the result of this is, then you get the original infinite row vector $\displaystyle (a_1,a_2,\ldots)$ back again.

The calculation goes as follows:

$\displaystyle S^-S^+(a_1,a_2,\ldots) = S^-(0,a_1,a_2,\ldots) = (a_1,a_2,\ldots),$

and this is exactly the identity operator applied to $\displaystyle (a_1,a_2,\ldots)$.

In conclusion, the two operators $\displaystyle S^-S^+$ and $\displaystyle I$ are equal on all vectors $\displaystyle a=(a_1,a_2,\ldots)$, and hence they are equal as operators.

**c)**

Well, there's an easy reason why $\displaystyle S^+$ isn't invertible. The reason is that invertible operators are necessarily bijections, and in the problem description, you have been given that the image $\displaystyle \text{im}S^+$ of $\displaystyle S^+$ is a proper subspace of $\displaystyle \mathbb{R}^\infty$ (which is clear anyway), so the operator $\displaystyle S^+$ is *not* surjective, and hence not bijective.

Another not-as-easy reason is that since $\displaystyle S^-S^+=I$, then *if* $\displaystyle S^+$ had been invertible, it would also have been the case that $\displaystyle S^+S^-=I$.

(Why? If an operator $\displaystyle T$ has both a left inverse $\displaystyle L$, and a right inverse $\displaystyle R$, then the two inverses coincide:

$\displaystyle L = LI = L(TR) = (LT)R = IR = R.$ )

The punchline here is of course that it is not the case that $\displaystyle S^+S^-=I$. Indeed, $\displaystyle S^-$ first maps $\displaystyle (a_1,a_2,\ldots)$ to $\displaystyle (a_2,a_3,\ldots)$, which $\displaystyle S^+$ then maps to $\displaystyle (0,a_2,a_3,\ldots)$. This is not the same as the identity operator $\displaystyle I$ applied to $\displaystyle (a_1,a_2,\ldots)$, and hence $\displaystyle S^+S^-\neq I$.