Map with identical kernel and image Do you know a linear map A on $R^n$ such that $kerA=ImA$?
I may suggest only one, quite strange, example when $n=0$, i.e. $R^0$ and $kerA=ImA=\emptyset$, though I've no idea what A maps there.
But, may be, there is another one on $R^n$ where $n\neq0$? What do you think?
 A: A simple one, in dimension 2, maps $e_1 \mapsto e_2\mapsto 0$, where  $e_1, e_2$ is a basis. 
A: Yes, one exists. I didn’t know one immediately, but I was able to reason one out.
If $\DeclareMathOperator{\ker}{ker} \DeclareMathOperator{\Im}{Im} \ker A = \Im A$, then $\ker A^2=\mathbb{R}^n$, or equivalently $A^2 x = 0$ for all $x\in\mathbb{R}^n$. (This follows immediately from the definition of image and kernel.)
This lead me to try taking $n=2$, and then a solution falls out quite nicely. Consider the map $A:\mathbb{R}^2 \to \mathbb{R}^2$ taking $(x,y) \mapsto (y,0)$.
This is linear:
$$A((x,y)+(x',y')) = A((x+x',y+y')) = (y+y',0) = A((x,y)) + A((x',y')).$$
The image of the map is
$$\Im A = \{(y,0):y\in\mathbb{R}\}$$
and the kernel is
$$\ker A = \{(x,0):x\in\mathbb{R}\} = \Im A,$$
so the condition is satisfied.
Then $A$ is the linear map taking $e_1 \mapsto e_2$ and $e_2 \mapsto 0$ (in the standard basis), whence it has the rather nice matrix representation
$$A = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$$
ETA: I spent a bit of time thinking over this some more, and came up with a slightly more general rule. Suppose $A:\mathbb{R}^{2n} \to \mathbb{R}^{2n}$. We can extend the principle above, as follows, in the standard basis:


*

*If $i=1,\ldots,n$, then $e_i \mapsto e_{i+n}$;

*If $i=n+1,\ldots,2n$, then $e_i \mapsto 0$.


This has
$$\Im A = \{(\underbrace{0,\ldots,0}_n, x_{n+1},\ldots,x_{2n}):x_i \in\mathbb{R}\} = \ker A$$
and is linear by a similar argument to the above. So that gives you one for $\mathbb{R}^{2n}$.
However, you can't find one for odd-dimensional real vector spaces. Why? We use the rank-nullity theorem, which states that
$$r(A)+n(A) = \DeclareMathOperator{\dim}{dim} \dim \Im A + \dim \ker A = \dim \mathbb{R}^n = n$$
But $\Im A = \ker A$, so $\dim \Im A = \dim \ker A$, whence $n$ must be even. So you can't find any linear map like this for odd-dimensional real vector spaces, because the initial premise would never hold.
