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I need some help here...

Let $V$ be a vector space over a field $K$, and $T: V \to W$ a linear transformation. Prove that if $$ dim_KV \lt \infty \ \ \text{and} \ \ \dim_K\operatorname{Im}(T^2) = \dim_K\operatorname{Im}(T)$$ then $$\operatorname{Im}(T) \cap \operatorname{Ker}(T) = \{0\}$$

This is what I made so far.

Let $x \in \operatorname{Im}(T) \cap \operatorname{Ker}(T)$,where T is linear operator. Then $T(x)=0$ and $x=T(v)$ for some $v \in V$

$0 = T(x) = T(T(v)) = T^2(v) \implies v\in \operatorname{Ker}(T^2)$

My intention is to reach $x=0$, but I'm stuck at this point. I also deduced, via dimension theorem, that $\dim_K \operatorname{Ker}(T) = \dim_K \operatorname{Ker}(T^2) $ but I don't know where it can help. Every hint is appreciated.

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  • $\begingroup$ You have $\ker(T) \subset \ker(T^2)$. Now use the dimension. $\endgroup$ – Maik Pickl Apr 4 '16 at 19:49
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The rank-nullity theorem says \begin{align} \dim V&=\dim\operatorname{Im}(T)+\dim\operatorname{Ker}(T) \\ \dim V&=\dim\operatorname{Im}(T^2)+\dim\operatorname{Ker}(T^2) \end{align} Next use the fact that $$ \operatorname{Ker}(T)\subseteq\operatorname{Ker}(T^2) $$ to conclude that $$ \operatorname{Ker}(T)=\operatorname{Ker}(T^2) $$

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  • $\begingroup$ Oh, I did not know that $\operatorname{Ker}(T)\subseteq\operatorname{Ker}(T^2)$. $\endgroup$ – Карпатський Apr 4 '16 at 19:59
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    $\begingroup$ @Карпатський If $T(v)=0$, then $T^2(v)=0$. ;-) $\endgroup$ – egreg Apr 4 '16 at 20:00
  • $\begingroup$ @egreg I haven't done linear algebra for a while, and I'm trying to refresh my memory. How exactly can we conclude that $\operatorname{Ker(T)}=\operatorname{Ker(T^2)}$? Does it follow from the fact that their dimensions are equal and the fact that the one is subset of the other implies that they live in the same space? $\endgroup$ – K.Power Apr 7 '16 at 23:19
  • $\begingroup$ @K.Power Since $\ker(T)\subseteq\ker(T^2)$, when we show the former has the same dimension as the latter, we conclude they're equal (take a basis of $\ker(T)$: it has the number of elements needed to be a basis of $\ker(T^2)$). $\endgroup$ – egreg Apr 7 '16 at 23:44
  • $\begingroup$ @egreg Ok thanks I understand that. $\endgroup$ – K.Power Apr 8 '16 at 15:07

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