Show that $\ker(f)\cap \ker(g)=\{0\},\;\ker(f) + \ker(g) = V $

Question

Let $V$ be a vector space of dimension $n$ and $f,g:V\to V$ linear transformations such that $f(f(x))=g(g(x))=0_V$ and $f(g(x))+g(f(x)) = 1_V$ $\forall x\in V$.

Show that $\ker(f)\cap \ker(g)=\{0\}, \;\ker(f) + \ker(g) = V $.

and that $\ker(f)=f(\ker(g)).$

I'm not sure what $1_V$ means, is that the unit vector if so, what does that mean? I know that $0_V$ is the null vector and $0$ is just $0$ but what does it really mean?

My attempt which I think is completely wrong but we see:

Let $x\in \ker(f)$ and $f(f(x)) = 0_V$ but that means $f(0)=0_V$ and that means that $\ker(f) = \{0\}$ same with $\ker(g).$ But that would mean that $\ker(f)+\ker(g)=\{0\}$ which is contradictory with the statement....

UPDATE:

Not sure if I'm right but I think that $Imf\subseteq Kerf$ since $f(f(x))=0_V$, same for $g$. Not sure if that helps neither.

Answer 1

Hints:

• If $\DeclareMathOperator{\Ker}{Ker}x\in\Ker f\cap\Ker g$, what can you deduce for $x$ from the relation $\;f\circ g+g\circ f=\mathbf 1_V$?
• From $\;f\circ f=g\circ g=0$ you can deduce that $\;\operatorname{Im}f\subseteq \Ker f$ and similarly for $g$. Deduce from these inclusions and the rank-nullity theorem a lower bound for the dimensions of $\Ker f$ and $\Ker g$.

Answer 2

Note that if $$ x \in \ker (f) \Rightarrow 0 = f(x) \Rightarrow f(0) = f(f(x)) = 0 \Rightarrow 0 \in \ker(f). $$ And thus also $0 \in \ker (g)$

Assume $\exists x \in \ker (f) \cap \ker (g) \not = 0$. This then implies that

$$ f(g(x)) + g(f(x)) = f(0) + g(0) = 0 \text{ i.e contradiction} $$