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Let $V$ be a finite dimensional vector space and $T$ be linear operator on $V$ such that $rank(T)=rank(T^2)$.Then to prove that the null space and range space of $T $are disjoint, i.e. zero vector is common.

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Let $v\in ker(T)\cap Im(T)$. Then $T(v)=0$ and there is $u\in V$ such that $T(u)=v$. One can see that $T(T(u))=T(v)=0$ Thus, $u\in Ker(T^2)$. Since, $V$ is a finite dimensional vector space, $Ker(T^2)=Ker(T)$. Hence, $u\in Ker(T)$ and so $v=0$, as desired.

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  • $\begingroup$ how $ker(T^2)$= ker (T)? $\endgroup$
    – user123804
    Feb 14, 2014 at 19:43
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    $\begingroup$ As you know, $Ker(T)\subseteq Ker(T^2)$. Since, $rk(T^2)=rk(T)$, one can see $Null(T^2)=Null(T)$. So $Ker(T)= Ker(T^2)$. $\endgroup$
    – Bobby
    Feb 14, 2014 at 19:46
  • $\begingroup$ Sorry it is from Rank nullity theorem ? $\endgroup$
    – user123804
    Feb 14, 2014 at 19:46
  • $\begingroup$ yes, it is from Rank nullity theorem $\endgroup$
    – Bobby
    Feb 14, 2014 at 19:47

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