Rank + nullity theorem Looking to give an example of a $4×5$ matrix A with $dim(Null(A)) = 3$. My thinking here is that by the $rank+null$ theorem, $rank+null= number of columns$. So the numbers of columns is $5$ so $ 5 = rank + 3 $. I got 3 from $dim(Null(A)) = 3$. So the rank of $A$ would be $2$. So this would just be a $4x5$ matrix with 2 leading 1's. 
Just wondering if my logic was correct. Can I get the null of the matrix from the statement $dim(Null(A)) = 3$ like I did?
 A: $
\DeclareMathOperator{rk}{rk}
\DeclareMathOperator{nul}{nul}
$
$A$ is a map from $V=\mathbb{F}^5$ to $W=\mathbb{F}^4$. The theorem indeed says
$$
\rk(A) + \nul(A) = \dim(V) \iff \\
\rk(A) + 3 = 5 \iff \\
\rk(A) = 2
$$
and
$$
A=
\begin{pmatrix}
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix}
$$
would do the job.
A: Take $A=[\mathbf{e_1}^{t},\mathbf{e_2}^{t},0,0,0]$. 
let $\mathbf{v}=(v_1,..,v_5) \in\mathbb R^5$. Of course $A \mathbf{v} \to (v_1,v_2)$. The kernel of $A$ is the subspace $\mathbb R^3$ generated by vectors of the form $(0,0,\mathbf{e}_3,\mathbf{e}_4,\mathbf{e}_5)$, so it has dimension $3$. Alternatively, we can see that $\dim \mathbb R^5- (\operatorname{rank} A)=\dim (\ker A)$. But the rank is exactly $2$.
The "number of columns" you mention is the dimension (as a vector space) of the domain for $A$ as a linear operator.
The number of linear dependencies give the dimension of the null space, or the things in the kernel of your map. Ultimately, in row echelon form, this is telling you how many columns are really just zeroes.
The number of linear independent vectors tells you the dimension of the image of $A$. These are the leading ones that you were thinking of.
you can ignore the rest if you haven't had any algebra:
From a purely algebraic point of view, this is the first isomorphism theorem:
if $A: V \to W$ is a linear operator, then 
$$A/\ker A \cong \operatorname{Im} A.$$
From this point of view, the rank nullity theorem follows readily, but it is also a useful way to construct such matrices.
A: Yes, you are correct. In general, $A:=\begin{pmatrix}v_1&v_2&v_3&v_4&v_5\end{pmatrix}$ with exactly two of $v_i's $ are linearly independent would follow $\dim(N(A))=3$ 
