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Let $T:E\rightarrow E$, be a linear transformation. Prove that $E=\ker(T)\oplus \text{Im}(T)$ if and only if $\ker(T)=\ker(T^2)$

I've come across this problem, and I am attempting to show that $Ker(T)\subset Ker(T^2)$ and $Ker(T^2)\subset Ker(T)$ for the first part and only I proved that $Ker(T)\subset Ker(T^2)$.

for the second part I'm still trying to interpret the implications of $Ker(T)=Ker(T^2)$.

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  • $\begingroup$ Try to prove that $\ker T=\ker T^2\iff \ker T\cap\operatorname{im}T=\{0\}$. Observe that $\ker T\subseteq \ker T^2$ is always true. $\endgroup$ – user228113 Nov 27 '16 at 4:33
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Note that $\ker T$ and $\text{Image} T$ are both subspaces of $E$ and hence $\text{Image} T+\ker T$ is a subspace of $E$.

Now ,$\ker T=\ker T^2\iff \ker T\cap \text{Image} T=\{0\}$.

Also by Rank-Nullity Theorem $\dim \ker T+\dim \text{Image} T=\dim E$.

So,

$\dim (\text{Image} T+\ker T)=\dim \ker T+\dim \text{Image} T+\dim (\ker T\cap \text{Image} T)=\dim \ker T+\dim \text{Image} T=\dim E\implies E=\ker T+ \text{Image} T$.

Moreover,$\ker T\cap \text{Image} T=\{0\}\implies E=\ker T\oplus \text{Image} T$.

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  • $\begingroup$ From what you are saying its clear that $kerT=kerT^2\Rightarrow kerT\cap ImT={0}$. But I don't think it's so clear the other way, $kerT\cap ImT={0}\Rightarrow kerT=kerT^2$. $\endgroup$ – César Rosendo Nov 28 '16 at 4:38
  • $\begingroup$ Yes its quite clear but you need to work it out first $\endgroup$ – Learnmore Nov 28 '16 at 4:57

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