Let $A,B \in (X \to X)$ such that $A^2=B^2$ and $ KerA \cap KerB= {\{0}\}$. Show that
$a) A(\mathrm K\mathrm e\mathrm rB) \subseteq \mathrm K\mathrm e\mathrm rA$
$b) \ \mathrm d\mathrm i \mathrm mA(\mathrm K\mathrm e\mathrm rB)=\mathrm d\mathrm i\mathrm m(\mathrm K\mathrm e\mathrm r B) $
$c) \ \mathrm d\mathrm i\mathrm m(\mathrm I\mathrm mA)=\mathrm d\mathrm i\mathrm m(\mathrm I\mathrm mB) $
Can somebody help me with this problem? I was able to show the problem in $a)$.This is what I did
Let $y \in A(\mathrm K\mathrm e\mathrm rB)$ be an arbitrary element. This implies that there exists $x \in \mathrm K\mathrm e\mathrm rB$ such that $A(x)=y$
From $x \in \mathrm K\mathrm e\mathrm rB \Rightarrow B(x)=0 \Rightarrow B^2(x)=0 \Rightarrow A^2(x)=B^2(x)=0 $
$\Rightarrow A(A(x))=0 \Rightarrow A(y)=0 \Rightarrow y \in \mathrm K\mathrm e\mathrm rA $
So we have $A(\mathrm K\mathrm e\mathrm rB) \subseteq \mathrm K\mathrm e\mathrm rA$
But I'm stuck in the problems in $b)$ and $c)$. I tried using the fact that $\mathrm K\mathrm e\mathrm rB \subseteq \mathrm K\mathrm e\mathrm rB^2$ and $\mathrm I \mathrm mB^2 \subseteq \mathrm I \mathrm mB$ but I didn't get anywhere. I would really appreciate some help
For b) we can use that $\text{ker}(A)\cap\text{ker}(B)=\{0\}$. Hence for all non-zero $x\in\ker(B)$ we find that $Ax\neq0$. So $$\dim\text{ker}(B)=\dim A(\text{ker}(B))+\dim\text{ker}(A|_{\text{ker}(B)})=\dim A(\text{ker}(B)).\quad(1)$$
For c) we can use that $$\dim(X)=\dim(\text{ker}(A))+\dim(\text{im}(A))=\dim(\text{ker}(B))+\dim(\text{im}(B)).\quad(2)$$ From a) and b) we know that $$\dim(\text{ker}(B))=\dim(A(\text{ker}(B))\leq\dim(\text{ker}(A))$$ and as $A$ and $B$ are interchangeable we also have $$\dim(\text{ker}(A))=\dim(B(\text{ker}(A))\leq\dim(\text{ker}(B)).$$ Thus $\dim(\text{ker}(A))=\dim(\text{ker}(B))$ and using $(2)$ gives us that $\dim(\text{im}(A))=\dim(\text{im}(B))$.