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The below is just me trying to explain my thought process, but it's not necessary to read it to understand my question. My question is just: what does it mean when the rank of a matrix is MORE than the number of variables in the matrix?

Let's say A is an augmented matrix in reduced row echelon form. I know that:

If the rank of A is less than the number of variables in the system of linear equations that A is representing, then there are an infinite amount of solutions to the system (there are x free variables where x = number of variables-rank[A]). The way I justify this is: if there isn't at least one row per variable, at least one variables is "unaccounted for" and therefore can have an infinite amount of values.

If the rank of A is equal to the number of variables (i.e. x = 0 using the above definition of x), then there is a single, unique solution).

But what does it mean when there are MORE non-zero rows than variables? Using the above though process, this would mean that one variable is accounted for more than once. Does that mean there are no solutions to the matrix because one variable is said to have two different values? What if the values agree with each other?

Sorry if this is a stupid question, any help is greatly appreciated!

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  • $\begingroup$ The "rank" in linear algebra usually denotes the number of linearly independent rows. Being non-zero is not enough. This is the relevant notion; the key idea is that, if two linear equations are linearly dependent, then one of them can be eliminated without altering the solution set. This said, having more equations than indeterminates might mean that the system is over-determined; I say "might", because some equations could be eliminable. $\endgroup$ – Giuseppe Negro Nov 1 '18 at 18:56
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I think this is what you mean? https://en.wikipedia.org/wiki/Overdetermined_system

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  • $\begingroup$ Yes, I agree with you, this is the relevant notion to answer this question. $\endgroup$ – Giuseppe Negro Nov 1 '18 at 18:59
  • $\begingroup$ Yes this looks great, thank you!! $\endgroup$ – James Ronald Nov 1 '18 at 20:46
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Rather than the augmented matrix, lets represent our system of linear equations as:

$A\mathbf x = \mathbf b$

$A$ is an $m\times n$ matrix. The rank of $A$ cannot be greater than $\min(m,n)$ It is not possible for $A$ to have a rank greater than the number of variables.

It is possible for $A$ to have more rows than there are variables, in which case the system is over-determined and there is no guarantee that a solution exists. But, the rank of $A$ is still no greater than the number of variables.

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  • $\begingroup$ I think that the OP used "rank" to mean "number of nonzero rows". $\endgroup$ – Giuseppe Negro Nov 1 '18 at 18:58
  • $\begingroup$ Possibly, but "rank" has a meaning, and in this case, I have to trust that people mean what what they say. $\endgroup$ – Doug M Nov 1 '18 at 19:20
  • $\begingroup$ Like someone said, I did mean rank to mean the number of nonzero rows in the reduced row echelon form of the matrix, is that not the correct definition of rank? I apologize if it isn't, could you let me know the correct definition? $\endgroup$ – James Ronald Nov 1 '18 at 20:47
  • $\begingroup$ You definition of rank is fine. If you have a $n\times m$ matrix and you put it into reduced row echelon form, if $m > n$ you will have some number 0 rows, and the number of non-zero rows cannot exceed the number of columns. $\endgroup$ – Doug M Nov 1 '18 at 21:00

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