# Linear Operators: False for Non-Finite Dimensional Vector Spaces

• sihag
In summary, The conversation discusses the relationship between a linear operator T on a finite dimensional vector space V and its inverse, as well as an example of a linear operator that is surjective but not injective. The example involves the backwards shift on the space of square summable sequences, but can be reduced to a more familiar space of polynomials over a field. The conversation also touches on the concept of linear independence and its relationship to the range of a linear operator. Finally, the conversation mentions a question from a linear algebra text regarding the subspace W spanned by the column vectors of the matrix of T, and clarifies that in the case of linear dependence, W cannot span F^n.
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.

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.

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.

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.

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!

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.

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

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 ?

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.

## 1. What are linear operators?

Linear operators are mathematical functions that map one vector space to another, preserving the basic structure and operations of the original space.

## 2. Why are linear operators false for non-finite dimensional vector spaces?

Linear operators are only defined for finite dimensional vector spaces, meaning that the space has a finite number of basis vectors. In non-finite dimensional spaces, such as infinite dimensional spaces, the concept of linear operators breaks down.

## 3. Can linear operators be applied to all types of vector spaces?

No, linear operators are only applicable to finite dimensional vector spaces. Infinite dimensional spaces, such as function spaces, do not follow the same rules and operations as finite dimensional spaces.

## 4. What is the importance of understanding the limitations of linear operators?

Understanding the limitations of linear operators is important because it allows us to accurately apply and interpret mathematical concepts in the appropriate context. It also helps to avoid errors and misunderstandings when working with vector spaces.

## 5. Are there any exceptions to the rule that linear operators are false for non-finite dimensional vector spaces?

No, the concept of linear operators only applies to finite dimensional vector spaces. However, there may be some special cases where certain properties of linear operators can be extended to non-finite dimensional spaces, but this is not a general rule.

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