Question: I came across this question

> 
> Find a matrix $A$ with nullity($A$) = 3 and nullity($A^T$) = 1 which contains no zero elements
> 
> 
> 

and am unable to finish my solution and would appreciate some guidance

My attempted solution utilises that the *maximum* rank of an $n \times m$ matrix is

\begin{equation}
rank(A) \le min(n, m) \tag{1} \label{eq:max-rank}
\end{equation}

Additionally a matrix with $m$ columns satisfies the equality in eq. $\eqref{eq:dimension-theorem}$,

\begin{equation}
rank(A) + nullity(A) = m \tag{2} \label{eq:dimension-theorem}
\end{equation}

Furthermore we have that if $A$ is any matrix, then $rank(A) = rank(A^T)$.

Hence we have that, since $A^T$ has $n$ columns,

\begin{align\*}
nullity(A^T) &= 1 = n - rank(A^T) \textrm{ (from eq. \eqref{eq:dimension-theorem})} \\
&\implies \textrm{ (from eq. \eqref{eq:max-rank})} \\
n - m &= 1
\end{align\*}

(I expect that $rank(A) > 0$ but this might be an unnecessary assumption? If $min(n, m) = n$ then the rank would be $0$)

So we have that $m < n$, specifically $n = m + 1$.

From here I am unable to continue. I do not see how to properly extrapolate, or see what I am missing, to infer proper values for n and m that satisfiy the nullity requirements. Or if I have made a flaw in my reasoning, which is possible since I have been at it for some time and may have mixed something together from my notes.

Answer: 
By equation (2) for $A$, we have 
$$ \def\rk{\operatorname{rank}}\rk A + 3 = n $$
by equation (2) for $A^t$, we have
$$ \rk A + 1 = \rk A^t + 1 = m $$
hence $$n - 3 = m -1 \iff n - m = 2. $$
As we want an easy example, let's try with the smallest possible $m$, namely $m = 1$. If $m = 1$, $A$ were a $(1,3)$-matrix, which can have nullity $3$ only if in consists of zeros, but we do not want zero entries. For $m = 2$, we have a $(2,4)$-matrix $A$. $A$ $(2,4)$-matrix with nullity $3$ must have rank $1$, that is all columns equal. We do not want zeros, hence 
$$ A = \begin{pmatrix} 1 & 1 & 1 & 1\\ 
 1 & 1 & 1 & 1 \end{pmatrix} $$
Start with an invertible (square) matrix without zero entries, e.g. $\begin{pmatrix}1&2\\3&4\end{pmatrix}$. This has both nullities $=0$. Add copies of a column three times to obtain a matrix $\begin{pmatrix}1&2&1&1&1\\3&4&3&3&3\end{pmatrix}$ with nullity $=3$. Then add a copy of a row to ensure nullity $=1$ for the transpose, i.e.,
$$\begin{pmatrix}1&2&1&1&1\\3&4&3&3&3\\3&4&3&3&3\end{pmatrix}.$$
Of course you might as well have started simpler with $\begin{pmatrix}1\end{pmatrix}$ and arrive at
$$\begin{pmatrix}1&1&1&1\\1&1&1&1\end{pmatrix}.$$