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I am reading "Introduction to Linear Algebra"(in Japanese) by Kazuo Matsuzaka.

There is the following problem(Problem 6 on p.224) in this book.

Let $V$ be a vector space.
Let $F$ be a linear map on $V$.
Suppose that $\text{Ker } F^2 = \text{Ker } F$ holds.
Suppose that $\text{Im } F^2 = \text{Im } F$ holds.

Prove that $V = \text{Im } F \oplus \text{Ker } F$ holds.

My attempt is here:

Let $v \in V$.
$F(v) \in \text{Im } F = \text{Im } F^2$.
So, there exists $u \in V$ such that $F(v) = F^2(u)$.
Since $0 = F(v) - F^2(u) = F(v - F(u))$, $v - F(u) \in \text{Ker } F$.
So, $v = F(u) + w$ for some $w \in \text{Ker } F$.
$\therefore$ $V = \text{Im } F + \text{Ker } F$.

If $V$ is finite-dimensional, then $V = \text{Im } F \oplus \text{Ker } F$ because $\dim V = \dim \text{Im } F + \dim \text{Ker } F$.
But the author didn't assume that $V$ is finite-dimensional and I have not used the assumption $\text{Ker } F^2 = \text{Ker } F$ yet.

Please show me an answer for this question.

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If $v\in\operatorname{Im}F\cap\ker F$, then $v=F(u)$, for some $u\in V$, and $F(v)=0$. But then$$0=F(v)=F\bigl(F(u)\bigr)=F^2(u).$$So, $u\in\ker F^2$ and therefore $u\in\ker F$. But then $v=F(u)=0$.

So, $\operatorname{Im}F\cap\ker F=\{0\}$.

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  • $\begingroup$ Jose Carlos Santos, Thank you very much for your answer again! $\endgroup$ – tchappy ha Apr 18 '20 at 11:15
  • $\begingroup$ I'm glad I could help. $\endgroup$ – José Carlos Santos Apr 18 '20 at 11:16

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