# Prove that $V = \text{Im } F \oplus \text{Ker } F$ holds if …

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.

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