Question concerning rank of a matrix Consider the system of linear equations $A\bf{x}=\bf{b}$ where $A$ is an $m\times n$ matrix with $m<n$ and suppose that the system has infinitely many solutions. I want to know under what circumstances the rows of $A$ would be linearly dependent? I know that, if the rank($A$) is less than $m$, it is possible. Is there anything else? 
 A: There is a theorem of linear algebra:  For any $m \times n$ matrix $A$,
$$  \operatorname{rank}(A) + \operatorname{nullity}(A) = n,$$
or in terms of the system of equations $A\mathbf{x} = \mathbf{0}$,
$$ \textrm{(number of independent equations)} + \textrm{(number of independent parameters in the solution)} = \textrm{(number of columns of $A$)} $$
At any rate, if $A\mathbf{x} = \mathbf{b}$ has more than one solution, then
so will the system $A\mathbf{x} = \mathbf{0}$, meaning that $\operatorname{nullity}(A) > 0$.  But this only shows that $\operatorname{rank}(A) < n$.  As @Jyrki Lahtonen stated in his answer, there are no restrictions on $\operatorname{rank}(A)$ with respect to $m$, the number of rows.
[added after question was edited]
For the case $m < n$, we certainly know $\operatorname{rank}(A) \leq m$.  You are asking under what conditions the rows must be dependent.  That condition is exactly equivalent to $\operatorname{rank}(A) < m$.  And we can say more, in terms of the original system. Using the rank formula above, $\operatorname{rank}(A) = n - \operatorname{nullity}(A)$.  So $\operatorname{rank}(A) < m$ forces $\operatorname{nullity}(A) > n - m$.  In practical terms, the condition that $A$ has dependent rows is equivalent to the condition that there are more independent paramaters in the solution than the difference of rows and columns.
Hope this helps!
