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?
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You misread Arturo's answer: the system having more than one solution implies that the rank of $A$ is $\lt n$. – Did Feb 17 '12 at 14:00
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Yes, sorry, this not exactly what I intended to ask. I ll modify the question accordingly. – Ashok Feb 17 '12 at 14:16
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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!
Shaun Ault
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