Image and kernel of linear transformation. Conclusions. $\newcommand{\im}{\operatorname{im}}$$X$ is linear space, and $f:X\to X$ is linear transformation.   Then:
a. $\im(f)\subset \im (f\circ f)$
b. $\ker (f)\subset \ker (f\circ f)$
c. $\ker(f)\cap \im(f)=\{0\}$
My approach:
Being honestly I can solve only b.
b.
Consider matrix $M$ for $f$  we have that matrix for $(f\circ f)$ is $M^2$. Therefore if $Ax=0$ then also $A^2x=A0=0$. Hence, $\ker(f)\subset \ker(f\circ f)$.  
I ask for checking my solution for b. and help with rest of question.

$\newcommand{\im}{\operatorname{im}}$Hint. Try to construct linear space $X$ and a (simple) nontrivial linear map $f \colon X → X$ with $\ker f = \im f$. What does that imply for $f ∘ f$ and $\ker f ∩ \im f$? Find such a linear map $f$ in a linear space $X$ of a dimension as small as possible. (Can you go dimension zero or one?)
Further. You can construct a linear transformation of the two-dimensional $ℚ$-space $ℚ^2$ by specifying its action on the standard basis $(1,0)$ and $(0,1)$. Now think of $ℚ^2$ as the Euclidean plane the linear map you want to construct as follows: Project the vertical axis “diagonally” onto the horizontal axis and the horizontal axis to zero, then linearly extend this map. Can you give the matrix for such a map?
