Consider the system of linear equations $A\bf{x}=\bf{b}$ where $A$ is an $m\times n$ matrix with $m
Question concerning rank of a matrix
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0Yes, sorry, this not exactly what I intended to ask. I ll modify the question accordingly. – 2012-02-17
1 Answers
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!