Endomorphisms exercise explanation correct? $\DeclareMathOperator{\id}{Id}\DeclareMathOperator{\im}{Im}$I have an exercise from an older exam I‘m trying to solve. But even with solutions I can‘t seem to be able to understand the way the proved them.
Let $V$ be a finite dimensional $\mathbb {R}$-vector space, and let $f$ and $g$ be two endomorphisms of $V$ such that: $f \circ g = g \circ f = 0$
and $f + g = 2\id_V$
$(1)$ Show that $f \circ f = 2  f$
$(2)$ Show that $\im(f)=\ker(g)$
As I already mentioned, I have solutions to both exercises but they seem so short and maybe skipped some intermediate steps, which is why I don‘t really know how they go this far.
Is this "explanation" for $(1)$ correct?
Since $f \circ g = 0$ we have that $f \circ f = f \circ f + 0  = f \circ f + f \circ g =  f \circ(f+g) = 2f$
I‘m not sure why this is the case though, is it just because adding $0$ to $f \circ f$ doesn‘t change the outcome or am I missing something? Also the last step isn't really clear to me. Is
$f \circ(f+g) = 2f$ because of $f + g = 2\id_V$?
Also for $(2)$, I don‘t quite know how this follows from the definitions. I do have the solutions in front of me but I can't really follow it.
 A: Your explanation for $1$ is correct. $f\circ (f+g)=2f\because f+g=2\mathrm{Id}\implies f\circ(f+g)=f\circ(2\mathrm{Id})=2f$. “Adding zero” is shorthand for adding the zero map (a linear operator that sends all inputs to zero) which would not affect the other linear map in the addition; in the same way that zero is an additive identity for numbers, a zero map is an additive identity for linear maps - in fact, functions can be considered as elements of a vector space, and you could consider this as adding the zero vector (and thus not changing anything). $(f+0)(v)=f(v)+0(v)=f(v),\,\forall v\in V$, so the addition is valid and $f=f+0$ - we use the fact that these are linear here.
As for $2$, if $g\circ f=0$, then $g(f(v))=0,\,\forall v\in V$, which means that $g(x)=0,\,\forall x\in\mathrm{Im}(f)$, which means...
And you can consider the kernel of $f$ too; $f\circ g=0\implies\ldots$?
EDIT:
Since this answer is accepted, any future viewers/learners should know that I showed only a partial proof of $2$. One must also show that $\ker g\subseteq\mathrm{Im}\,f$ to complete the proof. This has been done in another answer and in the comments below this post. Bonus: the kernel of $g$ is an eigenspace of $f$, and likewise the kernel of $f$ is an eigenspace of $g$, with eigenvalue $2$ for both.
A: Here is a detailed proof of (2):
First, $g \circ f = 0$ implies that for every $x \in V$, $g(f(x)) = 0$. Therfore, $f(x) \in \mathrm{ker}(g)$ and we obtain $\mathrm{Im}(f) \subseteq \mathrm{ker}(g)$.
For the other direction, use $f + g = 2 \mathrm{Id}_V$. If $x \in \mathrm{ker}(g)$ it is $f(x) = f(x) + g(x) = 2x$. Therfore, by linearity of $f$ we obtain $x = f({x \over 2})$. Thus, $x \in \mathrm{ker}(g)$ implies $x \in \mathrm{Im}(f)$ and this gives the other set inequality.
