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Let $V$ be a finite dimensional vector space and let $T:V \to V$ be a linear operator. If the rank of $T^2$ equals then rank of $T$, prove that $(\operatorname{range} T) \cap (\ker T)=\{0\}$

I just want to know whether I'm correct or not. First $T^2$ implies that T is squared, then rank of $T^2=$ rank of $T$ means that $T$ is full rank is that correct? Then the range of $T$ is the space $V$.Is the $\ker T$ means that $\ker T=\{v:T(v)=0\}$? If so range $T \cap \ker T=\{0\}$ right?

Michael Hardy
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user121819
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1 Answers1

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$$T^2(V)\subset T(V)\;,\;\;\text{so}\;\;\dim T^2(V)=\dim T(V)\implies T^2(V)=T(V)$$

Now:

$$x\in\ker T\cap \text{Im}\,T\implies \begin{cases}Tx=0\\{}\\x=Ty\;,\;\;\text{for some}\;\;y\in V\end{cases}$$

but then

$$0=Tx=T^2y\implies y\in \ker T^2=\ker T\;(\text{why this equality? Use the dimensions theorem)}$$

and thus, finally

$$x=Ty=0\implies \ker T\cap \text{Im}\,T=\{0\}$$

DonAntonio
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  • so then, is that correct that rank $T(V) \le$ rank of $T$? – user121819 Jan 18 '14 at 01:21
  • I don't get it, @user121819: rank is usually reserved for operators/matrices, whether dimension's what used for subspaces. Perhaps you meant $;\dim T(V)\le \text{rank},T;$? This, of course, would be absurd as the the left side's the definition of the right one. – DonAntonio Jan 18 '14 at 09:03