Conditions for Surjective and Injective linear maps

  • Thread starter malexj93
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  • #1
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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.
 

Answers and Replies

  • #2
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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.
 
  • #3
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Thanks, that's what I was looking for. I didn't know there was a theorem for that.
 
  • #4
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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.
 

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