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I have a system of $L$ linear equations with $N$ unknowns ($N<L$). The equations themselves are not given, but I know that, when I write the equations in matrix format $A x = b$, all $N$ columns in $A$ are linearly independent (I cannot say the same for all rows). I am trying to show that if a solution exists, this will be unique. But I think that this cannot be shown without assuming that the L rows also are linearly independent.

My initial direction was to say that the rank of the matrix is $N$ (since $N < L$, and the columns are linearly independent), and to show that this will equal to the rank of the augmented matrix if there exists a solution. But if there exists rows that are not linearly independent, then the row rank of the matrix can be less than $N$, in which case we will have an underdetermined system of equations —> not necessarily a unique solution.

Can I somehow infer that all rows are linearly independent, if the columns are linearly independent? Alternatively, is there a more direct approach to show that if there exists a solution in this system of equations, this solution is unique.

Thanks

Alok

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If you have more rows that columns, then all rows can NOT be linearly independent — but then, it actually does NOT matter here. You have the right answer in your second paragraph: if you know that all $N$ columns of the coefficient matrix are linearly independent, and if you know that solutions exist, then it has to be the case of a unique solution, since in this case $$\operatorname{rank}(\text{coefficient matrix})=\operatorname{rank}(\text{augmented matrix})=\text{number of variables}.$$

The extra equations in this case don't affect the number of solutions. They just happen to be extraneous, or redundant, equations. They are linearly dependent on other equations, but that in and by itself does NOT create linear dependency among variables or solutions. Here's a quick example: $$\left\{\begin{align} x&=5 \\ 2x&=10\end{align}\right.$$ You can see that in this case: $$\operatorname{rank}(\text{coefficient matrix})=\operatorname{rank}(\text{augmented matrix})=\text{number of variables}=1,$$ so the system has a unique solution, even though $N=1$ and $L=2$. The second equation is just a redundant one, a multiple of the first equation, and it can be eliminated (it will become a row of zeroes when you do Gauss-Jordan reduction).

Also, regarding your statement:

… then the row rank of the matrix can be less than $N$.

The row and column ranks of a matrix are always equal. So if you already know that the column rank of a matrix is $N$, then its row rank can't possibly be less than $N$.

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You do not need the rows to be linearly independent. In fact the $L$ rows can't be linearly independent since $N<L$ but the row rank is equal to $N$.

Suppose $b$ lies in the column space of $A$, then there exists a solution.

Suppose $Ax_1 = b$ and $Ax_2=b$ then we have $A(x_1-x_2)=0$, However, since columns of $A$ are linearly independent, then $x_1=x_2$.

Hence if the solution exists, it is unique.

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