# Linear operators

1. Feb 16, 2008

### sihag

Let T be a linear operator on a finite dimensional vector space V, over the field F.
Suppose TU = I, where U is another linear operator on V, and I is the Identity operator.
It can ofcourse be shown that T is invertible and the invese of T is nothing but U itself.

What I want to know is an example explicitly to show that the above is false if V is not finite dimensional.

Thank You.

2. Feb 16, 2008

### morphism

It suffices to find an operator T that is surjective but not injective.

A standard example of this phenomenon is the backwards shift on $\ell^2$, the space of square summable sequences: T(x_1, x_2, ...) = (x_2, x_3, ...). I'll let you verify this.

3. Feb 16, 2008

### sihag

This space is new to me !
Work me through this example.
Can't we show something on the vector space of R (reals) over rationals Q, or maybe even on vector space of all polynomials over any field.

Thank you.

4. Feb 16, 2008

### morphism

We actually don't really need the space to be l^2, but this is the space this example usually comes up in.

Let's reduce it to a more familiar space: F[x], the space of polynomials over F. Define T:F[x]->F[x] by T(a0 + a1x + ... + anxn) = a1 + a2x + ... + anxn-1. I'll let you verify that this is linear and surjective, but not injective.

5. Feb 16, 2008

### HallsofIvy

To add to what morphism just said, let U(a0+ a1x+ ...+ anxn)= 1+ a0x+ a1x2+ ...+ anxn+1 and it should be clear that TU= I but UT is not equal to I!

6. Feb 16, 2008

### Hurkyl

Staff Emeritus
You've (hopefully) already seen this example, in another context. The construction the others have described precisely corresponds to finding a (set) function from the set of natural numbers to itself that is injective but not surjective.

This is actually a quite general method useful in many contexts -- all you have to do is find a way to relate natural numbers to the kind of structure you're studying.

7. Feb 17, 2008

### sihag

hehe
finally got it !
yes, used similar mappings in ring theory !

8. Feb 17, 2008

### sihag

Well I've come across another question.
It's from Hoffman and Kunze's text on linear algebra.
Let T be a linear operator on F^n, let A be the matrix of T in the standard ordered basis for F^n, and let W be the subspace of F^n spanned by the column vectors of A. What does W have to do with T?

W is simply the space spanned by the range of T, right ? And in case the columns of A are linearly independent, they would form a basis for F^n and so W would be F^n itself.
In this case the transformation T would be necessarily injective, since it is preserving linear independence.
In case the columns of A are linearly dependent, W might or might not span F^n. But the columns of A would in this case not form a basis.

Is there something I've missed here ?

9. Feb 17, 2008

### HallsofIvy

If the columns are not independent, then T does not have an inverse and its range is not all of R^n. Since W does span the range of T, it cannot span F^n. The "might" part of "might or might not" is incorrect.