Can Linear Surjections Exist with n < m?

[diracdelta]

Homework Statement

Let

be linear surjection. Prove that then n>=m.

Homework Equations

Definition(surjection):

The Attempt at a Solution

Lets assume opposite, n<=m. If that is the case, then for some y from R^m, there is no belonging x from R^n, what is in contradiction with definition where its said for every y.

Is this ok? If no, why?

Thanks! Domagoj

 

[Mark44]

Homework Statement

Let

be linear surjection. Prove that then n>=m.

Homework Equations

Definition(surjection):

The Attempt at a Solution

Lets assume opposite, n<=m.

The opposite would be n < m.

If that is the case, then for some y from R^m, there is no belonging x from R^n

Why is there no x in Rⁿ for that y in R^m? You need to do more than just say the words.

, what is in contradiction with definition where its said for every y.

Is this ok? If no, why?

Thanks! Domagoj

 

[diracdelta]

The opposite would be n < m.
Why is there no x in Rⁿ for that y in R^m?

Ok.

Can one x map to more then one y?

 

[Mark44]

Ok.

Can one x map to more then one y?

A is a linear transformation. Can A(x) map to y₁ and y₂, where y₁ ≠ y₂

 

[diracdelta]

No.
Edit; That is actually what I've been trying to say in first post.

 

[Fredrik]

The Attempt at a Solution

Lets assume opposite, n<=m. If that is the case, then for some y from R^m, there is no belonging x from R^n, what is in contradiction with definition where its said for every y.

This is a good proof strategy, or at least it will be when you replace "n<=m" with "n<m". But you have to explain how you know that what you're saying in the second sentence is true.

Ok.

Can one x map to more then one y?

By definition of "map", no. Note that this has nothing to do with surjectivity or even injectivity. It's just the definition of "map" (="function").

 

[diracdelta]

@Fredrik
Ok.
Lets assume the opposite, n<m. I know from definition of surjection that for every y there has to be x such as f(x)=y.
But since n<m, there exists some element y that won't be mapped. Which is contradiction and not surjective.

 

[Mark44]

@Fredrik
Ok.
Lets assume the opposite, n<m. I know from definition of surjection that for every y there has to be x such as f(x)=y.
But since n<m, there exists some element y that won't be mapped.

Again (as in my post #2) why? Why does it follow that if n < m, then some x element won't be paired with that y?

Which is contradiction and not surjective.

 

[diracdelta]

I would say it like this;
If n <m that means that dimension of domain is lower then dimension of codomain. But definition of surjection , Image of (domain) = codomain.
but dim (domain) = n < dim(codomain)=m, -> contradiction

 

[Fredrik]

I would say it like this;
If n <m that means that dimension of domain is lower then dimension of codomain. But definition of surjection , Image of (domain) = codomain.
but dim (domain) = n < dim(codomain)=m, -> contradiction

It's true that the statements $n<m$ and $A(\mathbb R^n)=\mathbb R^m$ can't both be true, but you can't just say this. You have to prove it.