Difficulty in proving $\text{Im}f\cap\text{Ker}f\subseteqq\text{Im}(f\circ f-3f)$

Question

I am now having difficulty in trying to prove the relation in linear mapping. $V$ is finite dimension vector space on $\mathbb{C}$, $f:V\to V$ is linear mapping. The given condition is $$\text{dim}\ \text{Im}f\cap\text{Ker}f=\text{dim}\ \text{Im}(f\circ f-3f)=1$$ I was asked to show the including relation $$\text{Im}f\cap\text{Ker}f\subseteqq\text{Im}(f\circ f-3f)$$ First I set the arbitrary element $v\in\text{Im}f\cap\text{Ker}f$, then there exists $u\in V$ such that $f(u)=v$ and $f(v)=0$. It also known that these $u$ belongs to $\text{Ker}f^2$.

Next I want to show there exists $w\in V$ such that $(f\circ f-3f)(w)=v$ so that $v\in\text{Im}(f\circ f-3f)$ could be seen and the relation shall be proved.

Hint. Since you know that $f(u) = v$ and that $f \circ f(u) =0 $, and that $f$ is linear, why don't you try some scalar multiple of $u$? — mildboson, Mar 16 '24 at 10:26
@mildboson It seems just letting $w=-\frac{1}{3}u$ can satisfy it. $(f\circ f-3f)(w)=0+f(u)=v$, very nice hint and my thanks to you. — Cunyi Nan, Mar 16 '24 at 10:31
@Matija By $f\circ f-3f=f\circ(f-3\text{Id})$ could show it belongs to $\text{Im}f$. If I want to show it belongs to $\text{Ker}f$, $(f^3-3f^2)(v)=0$ needs to be shown. — Cunyi Nan, Mar 16 '24 at 10:49

score 0 · Accepted Answer · answered Mar 17 '24 at 03:07

The question is solved and I will post answer here to end up it.

For each vector $v\in\text{Im}f\cap\text{Ker}f$, there exists some $u\in V$ that $f(u)=v$ and $f(v)=f^2(u)=0$. Hence $u\in\text{Ker}f^2$. By letting $w=-\frac{1}{3}u\in V$, then $(f\circ f-3f)(w)=(f\circ f-3f)(-\frac{1}{3}u)=0+f(u)=v$, so the existence of $w$ is proved, hence $\text{Im}f\cap\text{Ker}f\subseteqq\text{Im}(f\circ f-3f)$.

Hinted by @Matija, the opposite can also be shown. It's obvious since the dimension of two spaces are equal, so no doubt they're the same subspace.

Difficulty in proving $\text{Im}f\cap\text{Ker}f\subseteqq\text{Im}(f\circ f-3f)$

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