How does the linear independence or dependence of the set of column vectors of a matrix depend on that of the set of row vectors of the same? 
Let A = ($a_{ij}$) be an mxn matrix. If the set of row vectors of A is linearly independent, is the set column vectors too?  What happens if the row vectors are linearly dependent. Does it affect the linear dependence of the column vectors?

I believe that since the row rank = column rank and m  and  n  are not equal, the linear dependence and independence of the set of row vectors and column vectors should not depend on each other. I'm not sure how to build up a solid argument though.
 A: You can build an intuition for that by using the number of pivots (in other words "rank") 
Consider the following matrix: 
\begin{equation*}
A_{5,4} = 
\begin{pmatrix}
a_{1,1} & a_{1,2} & a_{1,3} & a_{1,4} \\
a_{2,1} & a_{2,2} & \cdots & a_{2,4} \\
\vdots  & \vdots  & \ddots & \vdots  \\
a_{5,1} & a_{5,2} & \cdots & a_{5,4} 
\end{pmatrix}
\end{equation*}
Assuming it can be transformed to reduced row echelon form, we get the following (this is just a concrete example for that sake of understanding):
\begin{equation*}
A'_{5,4} = 
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0  & 0  & 1 & 0  \\
0 & 0 & 0 & 1 \\ 
0 & 0 & 0 & 0 \\ 
\end{pmatrix}
\end{equation*}
we have $4$ pivots, $4$ columns, and $5$ rows, the number of pivots is enough for the columns to be linearly independent but it's not enough for the rows to be so, one row won't have a pivot. 
if in the systems of equations $A\vec x=0$ one row doesn't have a pivot then we don't have a unique solution such that $\vec x=0$ and by definition the rows aren't linearly dependent. 
This applies to any non-square matrix (number of rows $\neq $ number of columns), I chose a $5 X 4$ one just as a concrete example. 
To directly answer your questions: 

What happens if the row vectors are linearly dependent. Does it affect the linear dependence of the column vectors?

No, this is clearly shown in the above example, rows are linearly depedent and columns aren't. 

If the set of row vectors of A is linearly independent, is the set column vectors too?

No, this can be shown in a similar fashion where we have number of columns bigger than the number of rows. 
It's worth noting that if you are dealing with a square matrix the linear dependence/independence of the rows and columns are related (one imply the other) as a pivot for a column will be a pivot for the row. 
A: The determinant of a matrix $A$ is the same as that of its transpose $A^T$. And since invertibility is equivalent to a non zero determinant, the linear independence of the column vectors is equivalent to that of the row vectors.
