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I am asked to prove this without the rank nullity theorem

My Attempt at a Proof

For the $\implies$;
If $g:V \rightarrow V$ is injective then the dimension of the kernel is 0, and so as ($im$) and ($ker$) are both subspaces of $V$, $dim(im)=dim(v)$ which implies $g$ is surjective.

For the $\impliedby$ ;
If $g:V \rightarrow V$ is surjective, then $dim(v)=dim(im)$ and $dim(ker)$ must be 0, so $g$ is injective.

It seems to me that i have indirectly used the rank nullity theorem, would there be a more suitable proof that doesn't incorporate it?

otupygak
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  • I'm sorry but I don't understand your argument in $\implies$. It appears to me that you're using the rank nullity theorem in $\Longleftarrow$. And from your title I think it sounds that you're only aske to prove $\implies$. – Rudy the Reindeer May 22 '14 at 17:09
  • Oop! I forgot the "dim" before "ker", any linear transformation is injective if and only if the dimension of a kernel is 0, is it not? – otupygak May 22 '14 at 17:11
  • Yes. That's the part I understand. Everything after "and so as..." is what I cannot follow. – Rudy the Reindeer May 22 '14 at 17:14

1 Answers1

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Let $v_1,\ldots,v_n$ be a basis for $V$ (so $V$ has dimension $n$), then since $g$ is injective, $g(v_1),\ldots,g(v_n)$ is linearly independent. However, it's a linearly independent set of $n$ vectors, and $V$ has dimension $n$, so it's a basis, hence the image of $g$ contains the span of $g(v_1),\ldots,g(v_n)$, which being a basis, is all of $V$.

oxeimon
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  • hmm... is $g(v_1),...,g(v_n)$ necessarily independent and not just unique? – otupygak May 22 '14 at 17:37
  • I assume by "unique" you mean "distinct". Anyway, yes, they must be independent. Suppose they satisfy a linear depedence relation $a_1g(v_1) + \cdots + a_ng(v_n) = 0$. But this means $g(a_1v_1 + \cdots + a_nv_n) = 0$. Since $g$ is injective, this means $a_1v_1 + \cdots + a_nv_n = 0$, but since the $v_i$ are linearly independent, implies that the $a_i$'s are all 0. – oxeimon May 22 '14 at 18:06
  • @otupygak hey, if you don't mind please accept this answer if it satisfactorily answered your question. – oxeimon May 14 '15 at 19:42