Prop.: Let T: E -> E be a linear mapping. If T is surjective, then T is injective.
I try to prove using that: If Tx = 0 implies x = 0, then T is injective. But it not work... Any help?
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@ZacharySelk Sorry :( I'll delete this post. But thanks! That solved my problem – Renan R. May 15 '16 at 02:57
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@ZacharySelk I don't know delete the post ;-; – Renan R. May 15 '16 at 02:59
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Trivially if $T$ maps to the same domain ($E \rightarrow E$) and it is surjective (meaning that its imagespace is all of $E$.
By rank nullity theorem (and by extension the dimension theorem) $$dim(Ker(T)) + dim(Im(T)) = dim(E)$$
Thus its kernel space must have dimension of $0$.
q.Then
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1You might want to mention that your answer assumes that $ E $ is finite-dimensional, as this is not mentioned in the question. – Martin Argerami May 15 '16 at 07:12