# Conditions for Surjective and Injective linear maps

1. Jul 16, 2013

### 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.

2. Jul 16, 2013

### micromass

Staff Emeritus
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.

3. Jul 16, 2013

### malexj93

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

4. Jul 23, 2013

### Theorem.

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