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Let $T:V\rightarrow V$ be a linear transformation of vector space with finite dimension.

Prove that the following statements are equivalent: $$1. \ \ V=Ker(T) \oplus Im(T) \\ 2. \ \ Ker(T^2)=Ker(T)$$

Unfortunately, I don't even know where to start.

Please help, thank you!

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First, $\ker (T) \subset \ker (T^2)$ is clear (always). Suppose $V= \ker (T) \oplus Im (T)$ and $v\in \ker( T^2)$.

Write $T(v)=p+q$ with $p\in \ker (T)$, $q\in Im(T)$. Then $0=T^2(v)=T(q)$, so $q\in\ker (T)$. So $q$ belongs to both the kernel and the image. So where does $v$ belong?

Conversely, suppose $\ker(T^2)=\ker(T)$. Then you can see $\ker(T^n)=\ker(T)$. Suppose $v\in \ker(T)\cap Im(T)$. Then $v=T(w)$ for some $w$. But then $0=T(v)=T^{2}(w)$ so $w\in \ker(T)$, which means $v=T(w)=0$. This shows they have trivial intersection. Can you finish it off with the rank-nullity theorem?

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