Can a matrix have the same range and nullspace? If you can pick any $3\times 3$ matrix, is there a matrix that its $R(A) = N(A)$?
If you can pick any $4\times 4$ matrix, is $R(A) = N(A)$ possible?
Here, $R(A)$ is the range of matrix A, and $N(A)$ is the nullspace of matrix A
I thought the answers to both questions would be "no" because
$R(A)$ is obtained from $Ax=b$, where $b \not= 0$
$N(A)$ is obtained from $Ax=0$.
The vectors in both spaces will never be the same. I wonder why the question would be structured in such a way.

Here I assume $R$ means "range", not "row space". 
If $A$ is $2n \times 2n$ then you can do it. Take $n$ independent vectors in $\mathbb{R}^{2n}$, call them $x_1,\dots,x_n$, and another $n$ vectors independent of each other and of the $x_i$, call them $y_1,\dots,y_n$. Now map $x_i$ to $y_i$ and map $y_i$ to zero. If $x_1,\dots,x_n = e_1,\dots,e_n$ and $y_1,\dots,y_n = e_{n+1},\dots,e_{2n}$, then we can write the matrix in block form:
$$\begin{bmatrix} 0 & 0 \\ I_n & 0 \end{bmatrix}$$
If $A$ is not square then this clearly can't happen, because then $R(A)$ and $N(A)$ are subspaces of different spaces. If $A$ is $(2n+1) \times (2n+1)$ then it still can't happen, because the rank-nullity theorem tells you that the dimensions of $R(A)$ and $N(A)$ must be different.
