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I read about rank nullity theorem (with proof) but then tried to prove it in different way. Please can you read my proof and tell me if it is correct?

The rank nullity theorem: If $T:V\to W$ is a linear map between finite dimensional vector spaces then $\dim(V) = \dim(\ker(T)) + \dim(\text{im}(T))$.

This is my proof: By induction on $\dim(V)$.

If $\dim(V)=1$ then $T$ is scalar multiplication and either $\ker(T)=\{0\}$ or not. If $\ker(T)=\{0\}$ then $\dim(\text{im}(T)) = 1$. If $\ker(T)\neq\{0\}$ then $\dim(\text{im}(T))= 0$. In both cases $\dim(V) = \dim(\ker(T)) + \dim(\text{im}(T))$.

Assume claim is true if $\dim(V) = n-1$. (induction hypothesis)

Consider the case $\dim(V) = n$. If $\ker(T) = \{0\}$ then $T$ is a bijection $V\to \text{im}(T)$ and the claim is true. If $\ker(T) \neq \{0\}$ then there is $v \notin \ker(T)$ with $v\neq 0$. Let $V' = V \setminus \text{span}(v)$. Apply induction hypothesis to the $V'$:

$$ \dim(V') = n-1=\dim(\ker(T\mid_{V'})) + \dim(\text{im}(T\mid_{V'}))=\dim(\ker(T)) + \dim(\text{im}(T\mid_{V'}))$$

Adding $v$ to $V'$: $$ \dim(V' + \text{span}(v))=\dim(V) = n = \dim(\ker(T)) +\dim(\text{im}(T))$$

Xoque55
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blue
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    Yes, it's correct. Also, it's great to see that you're trying to understand theorems by finding your own proofs; that's a very important aspect of mathematics! (A minor point: the quotient vector space of $V$ by $\text{span}(v)$ is usually denoted by $V/\text{span}(v)$; what you've written as $V'$ is technically a set-theoretic difference rather than a quotient. In particular, in your last line, $V'+\text{span}(v)$ doesn't really make sense unless you're referring to the external direct sum. So, it might be better to write $\text{dim}(V')+\text{dim}(\text{span}(v))=\text{dim}(V)$.) – Amitesh Datta Sep 03 '13 at 09:50
  • @AmiteshDatta Thank you, what about $V' = (V \setminus span(v)) \cup {0}$? Would it be ok instead? Or can I say that because $V' = (V \setminus span(v)) \cup {0}$ and $span(v)$ are subspaces of $V$ their sum is the internal direct product? – blue Sep 03 '13 at 12:46
  • If $V'=(V\setminus \text{span}(v))\cup {0}$, then $V'$ is not a subspace of $V$. (E.g., think about the case of a line in a plane; if I remove the line from the plane, then even retaining the zero vector doesn't result in a vector subspace of the plane.) – Amitesh Datta Sep 03 '13 at 12:53

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