# A question on bounded linear operators (Functional Analysis)

1. Jun 7, 2007

### Ricky2357

Suppose T: X -> Y and S: Y -> Z , X,Y,Z normed spaces , are bounded linear operators. Is there an example where T and S are not the zero operators but SoT (composition) is the zero operator?

2. Jun 7, 2007

### matt grime

Yes. Don't try to make S and T very complicated though. This has nothing to do with functional analysis or boundedness of operators at all, really.

3. Jun 7, 2007

### Ricky2357

Okay, can you think of any such simple example?

4. Jun 7, 2007

### Ricky2357

By the way, it has to do with functional analysis since I am asking for an expamle of BOUNDED linear operators and not just linear operators.

5. Jun 7, 2007

### matt grime

Yes. I can. But what I can think of isn't the point. What are some simple examples of normed spaces and (bounded) linear operators? (Hint: you did these very early on in maths.)

6. Jun 7, 2007

### Ricky2357

R with absolute value is simple enough

7. Jun 7, 2007

### Ricky2357

So can you help me out here ?

8. Jun 7, 2007

Note that in a finite dimensional vector space the boundedness condition is trivial.

Thus you would be asking whether it is possible to have a map S which maps the image to T of zero, even though S is not the zero operator.

In other words, you are asking whether there are matrices other than the zero matrix which do not have full rank.

9. Jun 7, 2007

### matt grime

And what are the linear maps from R to R? Just multiplication by a real number, so that won't do. So what's next?

10. Jun 7, 2007

### Ricky2357

Why is that? A linear operator from R into R of the form Tx=k*x is bounded.
(absolute value as norm)

11. Jun 7, 2007

For god's sake! You are asking whether the ring of operators in a vector space V has zero divisors. In the case V=R that ring of operators is a field, so obviously not. However, in the case where V has dimension greater than one, that ring is the ring of matrices.

If you don't know where the ring of matrices has zero divisors you are definitely not ready to do functional analysis.

12. Jun 7, 2007

### Ricky2357

DeadWolfe you have my respect. Your idea was brilliant! It is enough to find two nonzero matrices such that their product is nonzero. Gongratulations man and thank you !

13. Jun 7, 2007

### matt grime

Yes. That was clear from the outset, since bounded linear maps are after all linear maps.

Last edited: Jun 7, 2007
14. Jun 7, 2007

### matt grime

If you don't see that the product of two non-zero real numbers is non-zero, then I echo deadwolfe in saying you shouldn't be doing functional analysis, and instead relearn your first course in algebra. And your first course in linear algebra where the answer for this comes from.

A slicker idea comes from projection onto subspaces (maps XxX-->XxX sending (x,y) to (x,0) and (0,y) would also be trivial examples).

15. Jun 7, 2007

### Ricky2357

Okay something else matt grime. Is it true that the range of a bounded linear operator is always a bounded set?

16. Jun 7, 2007

### matt grime

What is the definition of a bounded linear operator? And range?

17. Jun 7, 2007

No, the identity map on R is a counterexample.

Do you make any effort to answer these questions before you pose them to us?

18. Jun 7, 2007

### mathwonk

come on man, this is trivial from the very definition of "bounded".

i.e.a bounded operaTOR IS ONE SUCH that the image of the unit ball is a bounded set. so hellooo........?

of course i could be wrong and hence appear (correctly) very stupid, but i am just saying look at the definition of bounded operator and answer your own question.

oops i blew it again, as i assumed you meant that the image of a bounded set is bounded, but you said range, well there you go, i am an idiot.

i have noticed that any time I make fun of someone elses ignorance, it is in fact myself who is out of the loop. good lesson as usual. i will probably forget it quickly anyway.

Last edited: Jun 7, 2007
19. Jun 7, 2007

### mathwonk

now that we have settled this issue, i am thinking maybe you meant "compact" operator, but i forget the definition of those. i am guessing though that they have bounded images. i will look it up on wiki stupedia.

wait, how stupid can I be???!!!! any non zero operator has in its image a line which is unbounded!!

duhhh???!!!! ]\\is this correct?? help me out here.

20. Jun 7, 2007