Proofs for right/left shift operators

**Runty** · October 14th 2010, 11:59 AM

This one probably isn't that hard, though right now I'm having some trouble.

Suppose $V$ is a vector space over a field $F$ .

Let $S^+:R^\infty\rightarrow R^\infty$ be the right shift operator and $S^-:R^\infty\rightarrow R^\infty$ be the left shift operator (both as defined below).

Before I go further, I'll need to provide a definition of the right/left shift operators.
Let $V=R^\infty$ be the space of infinite row vectors $(a_1,a_2,...)$
$S^+(a_1,a_2,...)=(0,a_1,a_2,...)$
$S^-(a_1,a_2,a_3,...)=(a_2,a_3,...)$
$\ker S^+$ is the zero space; $\mbox{im}S^+$ is a proper subspace of $V$
$\ker S^-$ is a proper subspace of $V$ ; $\mbox{im}S^-$ is the whole space.

a) Prove that $S^+$ is linear.

b) Prove that $S^-S^+=I$

c) Prove that $S^+$ is not invertible.

This probably isn't that difficult, and I'm just looking at it the wrong way.

**HappyJoe** · October 14th 2010, 12:39 PM

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

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

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

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

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

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

Then

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

and hence

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

On the other hand, we have

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

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

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

$I(a) = a$

for all $a$ .

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

The calculation goes as follows:

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

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

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

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

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

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

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

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

**Runty** · October 15th 2010, 11:56 AM

Originally Posted by HappyJoe

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

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

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

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

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

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

Then

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

and hence

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

On the other hand, we have

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

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

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

$I(a) = a$

for all $a$ .

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

The calculation goes as follows:

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

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

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

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

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

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

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

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

Very helpful stuff, but needs clarifications. (I'm literally trying to kill myself right now due to too many problems)

First, when you say $S^+(a)+S^+(b)$ , would that come out to showing it as $S^+(a)+S^+(b)=(0,a_1,a_2,...)+(0,b_1,b_2,...)=0+a_ 1+a_2+...+0+b_1+b_2+...$ ?

Second, for the third part, don't you need to prove that no transformation $T:R^\infty\rightarrow R^\infty$ exists such that $TS^+=I=S^+T$ ? My Prof. said that doing it for just $S^-$ is not sufficient.

**HappyJoe** · October 16th 2010, 12:54 AM

Originally Posted by Runty

Very helpful stuff, but needs clarifications. (I'm literally trying to kill myself right now due to too many problems)

First, when you say $S^+(a)+S^+(b)$ , would that come out to showing it as $S^+(a)+S^+(b)=(0,a_1,a_2,...)+(0,b_1,b_2,...)=0+a_ 1+a_2+...+0+b_1+b_2+...$ ?

Second, for the third part, don't you need to prove that no transformation $T:R^\infty\rightarrow R^\infty$ exists such that $TS^+=I=S^+T$ ? My Prof. said that doing it for just $S^-$ is not sufficient.

I don't quite understand your first question about $S^+(a)+S^+(b)$ . What are you asking?

For the second question, it is sufficient to check that $S^-$ is not the inverse of $S^+$ . Perhaps either you misunderstood your Prof. or he didn't think it through. Like I said, the deal with this is: Suppose contrariwise that $T\colon \mathbb{R}^\infty\rightarrow\mathbb{R}^\infty$ is an inverse of $S^+$ . In particular, we know that $S^+T = I$ . On the other hand, we know also that $S^-S^+=I$ . Therefore

$S^- = S^-I = S^-(S^+T) = (S^-S^+)T = IT = T.$

In conclusion, if there is an inverse $T$ of $S^+$ , then this inverse $T$ must coincide with $S^-$ , and to check that $S^+$ is not invertible, it thus suffices to check only that $S^-$ is not an inverse.

**Runty** · October 16th 2010, 05:18 PM

Originally Posted by HappyJoe

I don't quite understand your first question about $S^+(a)+S^+(b)$ . What are you asking?

For the second question, it is sufficient to check that $S^-$ is not the inverse of $S^+$ . Perhaps either you misunderstood your Prof. or he didn't think it through. Like I said, the deal with this is: Suppose contrariwise that $T\colon \mathbb{R}^\infty\rightarrow\mathbb{R}^\infty$ is an inverse of $S^+$ . In particular, we know that $S^+T = I$ . On the other hand, we know also that $S^-S^+=I$ . Therefore

$S^- = S^-I = S^-(S^+T) = (S^-S^+)T = IT = T.$

In conclusion, if there is an inverse $T$ of $S^+$ , then this inverse $T$ must coincide with $S^-$ , and to check that $S^+$ is not invertible, it thus suffices to check only that $S^-$ is not an inverse.

Alright, so I've got most of this down. I just need a little help with part a).

The part I'm having slight difficulty with is showing that $S^+(a+b)=S^+(a)+S^+(b)$ . The key thing is trying to show how $S^+(a)+S^+(b)$ , which are $(0,a_1,a_2,...)$ and $(0,b_1,b_2,...)$ , add together to make $(0,a_1+b_1,a_2+b_2,...)$ . I'm guessing that I'm not viewing it in the correct fashion.

Here's what I have written out right now, though I'm sure I'm missing some detail.
$S^+(a)+S^+(b)=(0,a_1,a_2,...)+(0,b_1,b_2,...)$
$=0+a_1+a_2+...+0+b_1+b_2+...$
$=0+a_1+b_1+a_2+b_2+...$
$=(0,a_1+b_1,a_2+b_2,...)=S^+(a+b)$

Can you point out where I've made a mistake, if any?

**HappyJoe** · October 17th 2010, 01:47 AM

Originally Posted by Runty

Alright, so I've got most of this down. I just need a little help with part a).

The part I'm having slight difficulty with is showing that $S^+(a+b)=S^+(a)+S^+(b)$ . The key thing is trying to show how $S^+(a)+S^+(b)$ , which are $(0,a_1,a_2,...)$ and $(0,b_1,b_2,...)$ , add together to make $(0,a_1+b_1,a_2+b_2,...)$ . I'm guessing that I'm not viewing it in the correct fashion.

Here's what I have written out right now, though I'm sure I'm missing some detail.
$S^+(a)+S^+(b)=(0,a_1,a_2,...)+(0,b_1,b_2,...)$
$=0+a_1+a_2+...+0+b_1+b_2+...$
$=0+a_1+b_1+a_2+b_2+...$
$=(0,a_1+b_1,a_2+b_2,...)=S^+(a+b)$

Can you point out where I've made a mistake, if any?

There's a part of what you write that doesn't make sense, namely

$(0,a_1,a_2,...)+(0,b_1,b_2,...) = 0+a_1+a_2+...+0+b_1+b_2+...$

What happened here? First, the left hand side is the sum of two infinite row vectors, and as such, should be another infinite row vector, but your right hand side is an infinite sum of real numbers, so the two objects (the sum of two infinite row vectors vs. an infinite sum of real numbers) are incompatible and cannot be equal - they live in different worlds. Second (and less important in this case), you may not know, if the infinite sum of real numbers on the right hand side converges, but this is irrelevant anyway, because we are not going to use that sum.

So, remember that in a vector space, there has been defined an addition between pairs of elements of the vector space. In the vector space $\mathbb{R}^\infty$ , the vectors are infinite row vectors $(x_1,x_2,\ldots)$ , and given two infinite row vectors $(x_1,x_2,\ldots)$ and $(y_1,y_2,\ldots)$ , there is a definition of adding them, such that this addition produces a new vector in your vector space, i.e. the result of this addition is also an infinite row vector.

If you have not been told in the problem description, or in your text book, what the addition on vectors in $\mathbb{R}^\infty$ is, then you have to pick the "natural" addition. If you don't know, what the "natural" addition is, don't worry, you'll know in future scenarios, when you're more experienced. So, the addition of two vectors in our space is defined like:

$(x_1,x_2,\ldots) + (y_1,y_2,\ldots) = (x_1+y_1,x_2+y_2,\ldots).$

This is what is meant, when adding two infinite row vectors. The result is another infinite row vector, whose coordinates are the sum of the coordinates of the two vectors we added to begin with.

Try using this to see what

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

is, when written as a single infinite row vector. Compare this to $S^+(a+b)$ .

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