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If a linear system with the same # of equations as unknowns, then it is a $n \times n$ matrix with full rank. But why does it have an unique solution? What's the logic behind it?

Also:

A linear system with coefficient matrix $A$ has an infinite number of solutions iff $A$ can be row-reduced to an echelon matrix that includes some columns containing no pivots.

I thought this statement is true, since if all columns contain a pivot, then the linear system in matrix form has full rank. So it cannot have an infinite number of solutions. However, the correct answer is false. What would be a counterexample to the statement?

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    $\begingroup$ Beginning not quite true. The system $x+y=3$, $2x+2y=6$ has infinitely many solutions. $\endgroup$ – André Nicolas Feb 25 '14 at 5:24
  • $\begingroup$ If you draw two straight lines in the plane, then one of three conditions holds: (1) they are parallel and do not intersect, (2) they are parallel and are the same or (3) the cross at exactly one point. You can think of each equation as being a line (or plane in higher dimensions). Full rank means not parallel (or suitable generalization in higher dimensions). $\endgroup$ – copper.hat Feb 25 '14 at 5:28
  • $\begingroup$ If the system has a solution, it is unique. Now, no solution has something to do with the matrix. $\endgroup$ – Claude Leibovici Feb 25 '14 at 5:44
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As Andre says this is not true. On the other hand if your question is why is the reason for which a system (of $n$ equation and $n$ unknowns) with complete rank has always a unique solution. In my opinion the better way to see it is by linear maps. Let $n$ be fixed and select a collection of $a_{i,j}\in \mathbb{F}$ for $(i,j) \in \{\ 1, \ldots,n\} \times \{\ 1, \ldots,n\}$. Define the linear map $f$

$$(x_1 \ldots,x_n)\longmapsto \bigg(\sum_{1\le j\le n} a_{1,j}x_j \ldots\sum_{1\le j\le n} a_{n,j}x_j \bigg)$$

When we say that the rank is complete, we say that for all $(c_1 \ldots,c_n)$ always exists a $(x_1 \ldots,x_n)$ such that $f(x_1 \ldots,x_n)=(c_1 \ldots,c_n)$, i.e.,

$$\bigg(\sum_{1\le j\le n} a_{1,j}x_j \ldots\sum_{1\le j\le n} a_{n,j}x_j \bigg)=(c_1 \ldots,c_n)$$

With really means that $\sum_{1\le j\le n} a_{i,j}x_j =c_j$ for our system.

Since the $\text{rank }f=n$. Then by the dimention formula we can conclude that its kernel is trivial, i.e., $\ker f = \{0\}$. Then $f$ is an injective map. So we know that always exists a solution because is surjective (the rank has the same dimension as the target space) and also this solution is unique since is one-to-one.

Using our above notation. What happens when the rank is less than $n$. Then there exists some values in the target space which are "outside" of the image and in these cases there is no solution. Since the rank is less than $n$ so the linear map cannot be surjective and exists vectors in $\mathbb{F}^n \backslash f(\mathbb{F}^n)$.

For the question: when does it have an infinite number of solution? Assuming again that the rank is less than $n$. Let $(c_1, \ldots ,c_n) \in f(\mathbb{F^n})$ (a vector in the image), because is in the image there is a vector in the domain that under $f$ is mapped in $(c_1, \ldots ,c_n)$, let call it $x=(x_1, \ldots, x_n)$. But also we know that the kernel is not trivial, again using the dimension formula we conclude $\dim(\ker f)= n-\text{rank} f>0$. Then there is some vector $(z_1, \ldots, z_n) \not= 0$ such that $f(z_1, \ldots, z_n)= (0, \ldots, 0)$.

Now consider $(x_1,\ldots, x_n)+k(z_1, \ldots, z_n)$, where $k\in \mathbb{F}$ and see what happens under $f$.

\begin{align}f((x_1,\ldots, x_n)+k(z_1, \ldots, z_n))=f(x_1,\ldots, x_n)+kf(z_1, \ldots, z_n)\\ =f(x_1,\ldots, x_n)+0= (c_1 \ldots, c_n) \end{align}

This occurs because $f$ is linear and the vector $(z_1, \ldots, z_n)$ is in the kernel (is zero under $f$). Thus $(x_1,\ldots, x_n)+k(z_1, \ldots, z_n)$ is also a solution. And indeed the set

$$x+\ker f: = \{x+k: x=(x_1,\ldots,x_n)+k, \text{ and } k\in \ker f\}$$

by the same argument as above contain all the solutions. In that case there is a infinite number of solutions.

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