Conditions for Surjective and Injective linear maps

[malexj93]

Hello,

I'm not sure if this should go under the HW/CW section, since it's not really a homework question, just a curiosity about certain kinds of functions. My specific question is this:

If M: U→V is injective and dim(U)=dim(V), does that imply that M is surjective (and therefore bijective) as well? And why is that?

I believe it does, but I can't seem to justify it.

 

[micromass]

Hello,

I'm not sure if this should go under the HW/CW section, since it's not really a homework question, just a curiosity about certain kinds of functions. My specific question is this:

If M: U→V is injective and dim(U)=dim(V), does that imply that M is surjective (and therefore bijective) as well? And why is that?

I believe it does, but I can't seem to justify it.

For finite-dimensional spaces, it is true. It is called the Fredholm alternative: http://en.wikipedia.org/wiki/Fredholm_alternative

Do you know the rank-nullity theorem? It follows easily from there.

 

[malexj93]

Thanks, that's what I was looking for. I didn't know there was a theorem for that.

 

[Theorem.]

Think about it intuitively too. Say [itex]V=\{v_1,v_2,...,v_k\}[/itex] is a basis for the Domain space. What is the image of [itex]V[/itex]? are the elements of [itex] M(V)[/itex] Linearly independent? It's a good exercise to figure this out. If they are, by the dimension condition you have found a basis.

 

[blue_raver22]

I can provide an answer to your question about surjective and injective linear maps. First, let's define what it means for a linear map to be injective and surjective.

A linear map M: U→V is injective if for any two elements u1 and u2 in U, M(u1) = M(u2) implies that u1 = u2. This means that each element in U has a unique corresponding image in V under M.

On the other hand, a linear map M: U→V is surjective if for every element v in V, there exists an element u in U such that M(u) = v. In other words, every element in V has at least one pre-image in U under M.

Now, to answer your question, if M is both injective and has the same dimension as U and V, then M must also be surjective. This is because if U and V have the same dimension, then they have the same number of elements. And if M is injective, each element in U has a unique image in V. Therefore, there cannot be any elements in V that do not have a corresponding pre-image in U, making M surjective.

To visualize this, imagine U and V as two different sized buckets, with U having the same number of balls as V. If M is injective, each ball in U has a unique spot in V. And if U and V have the same number of balls, there cannot be any empty spots in V, making M surjective.

I hope this explanation helps. If you have any further questions, please don't hesitate to ask. As scientists, it's important to always question and seek understanding.