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Let $T$ be a linear operator on a finite-dimensional vector space $V$. Let $R$ be the range of $T$ and let $N$ be the null space of $T$. Prove that $R$ and $N$ are independent if and only if $V=R\oplus N$.

My attempt: $(\Rightarrow )$ Suppose $R$ and $N$ are independent. We need to show $V=R+N$. Here is a proof using dimension. Can we prove $V=R+N$ directly from definition? I mean, let $\alpha \in V$. Show $\exists \alpha_1\in R$, $\exists \alpha_2\in N$ such that $\alpha =\alpha_1 + \alpha_2$.

user264745
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    What do you mean by "$R$ and $N$ are independent"? is it $R\cap N={0}$? I don't think this expression is standard. – Anne Bauval Jan 31 '23 at 18:56
  • @AnneBauval yes. $R$ and $N$ are independent$\iff$$R\cap N={0}$. – user264745 Jan 31 '23 at 18:57
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    Why do you want to prove it "directly" rather that using dimension, and what were your attempts to do so? It holds only if $\dim V<\infty$ and you can always "hide" a proof of the rank-nullity theorem inside a proof of this exercise and pretend to have found a "direct" proof, but what for? – Anne Bauval Jan 31 '23 at 18:59
  • @AnneBauval I also think it is very hard write $\alpha =\alpha_1 +\alpha_2$. We don’t have any “nice” structure on $T$, for instance $T^2=T$. – user264745 Jan 31 '23 at 19:01
  • @AnneBauval Thank you for the answer. – user264745 Jan 31 '23 at 19:05
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    @AnneBauval btw Here is another example of showing two subspace are equal using dimension. Without dimension argument it is quiet hard to prove equality. – user264745 Jan 31 '23 at 19:10

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