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I was listening to a lecture video my algebra teacher sent us, about systems of equations. As I understood, we can show our system of equations as lines with points at which they cross, where each line presents an equation and and the crossing points present solutions. In pdf it says that, for example:

For a system of equations with 3 equations and 2 unknowns, we can either have 1 point(solution) where they intersect, or infinitely many solutions if they are all overlapping. My question is:

Can't we have 3 solutions for example? Or any other amount more than 1 but not infinite? enter image description here

Javadkhan
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    What kind of equations? In what context? I’ll bet that the context of the quote is that of linear equations with coefficients in the rationals, real, or complex numbers, where the statement is correct: any system of linear equations over any of those types of coefficients has either no solutions, exactly one solution, or infinitely many solutions. If you allow different types of equations, then the situation changes. – Arturo Magidin Apr 17 '21 at 22:14
  • @ArturoMagidin Yes, I forgot to mention that. You are correct, it's about linear equations, but why can't it have 2 or 5 solutions? We can plot the lines and intersecting points like that^ – Javadkhan Apr 17 '21 at 22:17
  • $(x^{2}-1)^{2} + y^{2} = 0$ – Joshua Wang Apr 17 '21 at 22:17
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    No, you can’t have straight lines intersection “like that”: if two straight lines have more than one point in common, then they are the same line and have infinitely many points in common (because a line is completely determined by any two of its points). That is not true for other types of curves. Each equation determines a straight line; so if there is more than one solution, then each pair of lines has more than one point in common, and so they are all the same line. And a line has infinitely many points. – Arturo Magidin Apr 17 '21 at 22:20
  • @ArturoMagidin Thanks, I got it now – Javadkhan Apr 17 '21 at 22:23
  • Maybe not in euclidean space, it would only be that 1, 0, or infinite solutions exist. I guess it could be possible to apply a transformation to the space such that two lines that intersect one in euclidean space would intersect more than once in another space. Though I am not sure if this is possible nor a transformation that would work here – Тyma Gaidash Apr 17 '21 at 22:28
  • @TymaGaidash: No such “transformation” would respect the linearity of the defining equations. – Arturo Magidin Apr 17 '21 at 22:31
  • @ArturoMagidin why would that be the case? – Тyma Gaidash Apr 17 '21 at 22:31
  • Why is there a diagram of squiggly lines, meanwhile everyone is talking about linear equations. A Linear equation is the equation of a straight line. – Adam Rubinson Apr 17 '21 at 22:33
  • @TymaGaldash: Because it is a theorem of linear algebra that a system of linear equations has either no solutions, exactly one solution, or at least as many solutions as there are scalars. If you could apply a “transformation” that suddenly gave you a different number of solutions, then necessarily your system would no longer be a system of linear equations. – Arturo Magidin Apr 17 '21 at 22:33
  • @AdamRubinson: Because the OP messed up and forgot to say “linear equations” and forgot to take into account in his picture. That’s why I asked in the comment if they were linear equations, which the OP confirmed (and the tag [linear-algebra] is there, too). – Arturo Magidin Apr 17 '21 at 22:34

2 Answers2

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Suppose the number of solutions is more than $1$ but less than infinity. Then it is at least $2$. Find $2$ points that are solutions to all 3 equations. Then every equation is a straight line going through these two points. Since there is one unique line going through two distinct points, this means that each of the $3$ equations is the same line. But this means that there are infinitely many points that solve all $3$ equations simultaneously. This contradicts the assumption that “the number of solutions is more than $1$ but less than infinity”.

Adam Rubinson
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  • Why bother with the assumption that there are infinitely many? You never use it until the end when you contradict it in order to conclude it is true. Might as well just start with the assumption “more than two” and end with the conclusion “then infinitely many”. – Arturo Magidin Apr 17 '21 at 22:52
  • @ Arturo I see your point. But...it’s swings and roundabouts imo. OP says he knows you can have $1$ or infinitely many points, so I was starting with that. Also I just sprained my ankle so my thought process might be compromised. – Adam Rubinson Apr 17 '21 at 22:57
  • Nothing personal; it’s a common trope in proof writing, and one of my pet peeves... e.g. here, there, and everywhere. – Arturo Magidin Apr 17 '21 at 23:00
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As noted in the comments (and possibly guess-able from the tags), the quote is about linear equations. This is, of course, key.

The statement holds in any number of variables:

Theorem. A system of $m$ linear equations in $n$ unknowns, $$A\mathbf{x}=\mathbf{b}$$ (where $A$ is an $m\times n$ matrix, $\mathbf{x}=(x_1,\ldots,x_n)^T$, and $\mathbf{b}=(b_1,\ldots,b_m)^T$) has either no solutions, exactly one solution, or at least as many solutions as there are scalars; in particular, if there are infinitely many scalars, then the system has either no solutions, exactly one solution, or infinitely many solutions.

Proof. It is enough to show that if the system has more than one solution, then it has at least as many solutions as there are scalars. Assume that $\mathbf{s} = (s_1,\ldots,s_n)^T$ and $\mathbf{t}=(t_1,\ldots,t_n)^T$ are two distinct solutions. Then $\mathbf{s}-\mathbf{t}\neq\mathbf{0}$.

I claim that for any scalar $\alpha$, $\mathbf{s}_{\alpha}=\mathbf{s}+\alpha(\mathbf{s}-\mathbf{t})$ is also a solution. Note that $\mathbf{s}_{\alpha}=\mathbf{s}_{\beta}$ if and only if $\alpha=\beta$ (precisely because $\mathbf{s}-\mathbf{t}\neq\mathbf{0}$), so that means that there are at least as many solutions as there are scalars.

Indeed, we have that $A\mathbf{s}=A\mathbf{t}=\mathbf{b}$; therefore: $$\begin{align*} A\mathbf{s}_{\alpha} &= A(\mathbf{s}+\alpha(\mathbf{s}-\mathbf{t}))\\ &= A\mathbf{s} + \alpha A\mathbf{s} - \alpha A\mathbf{t}\\ &= \mathbf{b} + \alpha\mathbf{b}-\alpha\mathbf{b}\\ &= \mathbf{b}. \end{align*}$$ Thus, $\mathbf{s}_{\alpha}$ is also a solution, as claimed. $\Box$

Arturo Magidin
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