On monomial matrix (generalized permutation matrix ) A matrix $a\in GL_{n}(F)$ is said to be monomial if each row and column has exactly one non-zero entry. Let $N$ denote the set of all monomial matrices. 
I want to prove that following are equivalent 


*

*$A\in N$

*there exist a non singular $D$ (diagonal matrix ) and a permutation matrix $P$ such that $A=DP$

*there exist a non singular $D$ (diagonal matrix ) and a permutation matrix $P$ such that $A=PD$


I have written a small proof. Is it correct ?
Let's say $A\in N$. We know that "If $P$ is a permutation matrix then multiplication by $P$ on right rearranges the columns." So by multiplying suitable permutation matrix say $Q$ on right we get a diagonal matrix (which is obviously non-singular as $A\in N$). 
Precisely $AQ=D$. Multiplying by $Q^{T}$ on both sides we get that $A=DP$
 for some permutation matrix $P$ and diagonal matrix $D$.
Now assume $A=DP$. Then $$DP = PP^{T}DP=PD^{'}$$ where $D^{'}=P^{T}DP$ is again a diagonal matrix, so $A=PD^{'}$.
Now assume $A=PD$. We want to show that $A\in N$, but it is clear as multiplication on left by $P$ will just swap the rows of $D$.
 A: Your proof is alright. You are using the fact about what happens to a matrix when it is multiplied by a permutation matrix. You can change the view point and use the description of what happens to a matrix when it is multiplied by a diagonal matrix.
If $X, D$ are square matrices of the same size with $D$ diagonal,
$XD$ is the matrix obtained from $X$ with its $j$-th column a scalar multiple
of corresponding column of $X$ (the scalar being the $j$-th  entry of the diagonal matrix).
Analogous statement is true for $DX$. So all monomial matrices are obtained from permutation matrices by multiplying with diagonal matrices.
A: It's possible to be much more specific about this. 
If 
$$
\sigma:= 
\begin{pmatrix}
1 & \cdots & n \\
\sigma(1) & \cdots & \sigma(n)
\end{pmatrix} \in S_n,
$$
then the permutation matrix corresponding to $\sigma$ is the $n$-by-$n$ matrix $P_\sigma = \begin{bmatrix} p_{ij} \end{bmatrix}$ defined by 
$$
p_{ij} =
\begin{cases}
1, & j=\sigma(i); \\
0, & \text{otherwise}.
\end{cases}
$$
Suppose that $A=\begin{bmatrix} a_{ij} \end{bmatrix}$ is an $n$-by-$n$ monomial matrix. For every $i \in \{1,\dots,n\}$, let 
$$
\sigma(i):= \min_{1\le j \le n} \{ a_{ij} \mid a_{ij} \ne 0\}. 
$$
Since $A$ is monomial, the function
$$
\sigma:= 
\begin{pmatrix}
1 & \cdots & n \\
\sigma(1) & \cdots & \sigma(n)
\end{pmatrix}
$$
is a permutation (i.e., it belongs to $S_n$).
If $D=\text{diag}(a_{1\sigma(1)},\dots,a_{n\sigma(n)})$, then 
$$
\begin{bmatrix}
DP_\sigma
\end{bmatrix}_{ij}
=
\begin{cases}
a_{i\sigma(i)}, & j=\sigma(i); \\
0, & \text{otherwise}.
\end{cases}
$$
Thus, $A=DP_\sigma$.
With a little care, these steps can be reversed to show that $DP$ is monomial whenever $D$ is a diagonal matrix and $P$ is a permutation matrix.
