What is the reasoning behind this theorem? If $T:U \rightarrow V$ is a linear map then $\operatorname{nullity}(T)+\operatorname{rank}(T) =\operatorname{dim}(U)$. I am trying to prove it and looking for hints.
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1$\begingroup$ A good idea would be to point out what $U$, $V$ and $T$ are. I'm assuming vector spaces and a vector space homomorphism? I'm also assuming that the little $u$ should be a capital $U$? If you're working in full generality the first things you want to do is write down the definitions. I would also consider checking if you know anything about quotient spaces. $\endgroup$– DRFMar 30, 2015 at 9:56
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$\begingroup$ Consider general bases $\endgroup$– TrajanMar 30, 2015 at 10:03
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$\begingroup$ This is proved in most texts on linear algebra and most course in linear algebra $\endgroup$– TrajanMar 30, 2015 at 10:03
1 Answer
Assuming $\;U,\,V\;$ are linear spaces and $\;\dim U=n<\infty\;$ , here are some hints for you to work out and prove:
== Choose a basis $\;\{u_1,...,u_k\}\;$ of $\;\ker T\;$ (and thus $\;Null\, T=\dim\ker T=k\;$)
== Complete the above basis to a basis of $\;U\,:\;\;\{u_1,...,u_k,u_{k+1},...,u_n\}\;$ .
== Prove now that $\;\{Tu_{k+1},...,Tu_n\}\;$ is a basis for Im$\,T\;$ (and thus we have that $\;Rank\,T=\dim\text{Im}\,T=n-k\,$)
== Complete, and end, the proof.
Another option, of course, is to look for the proof in thousands of sites and books that deal with basic linear algebra.