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I need to prove that a system of linear equations doesn't exist over $\mathbb{R}$ for the solutions set $\{(a, a^2, b)|a,b \in \mathbb{R}\}$. But isn't for example this system an example of when this does happen (for $x = y = 1$): $$\begin{cases} x+y=2 \\ 2x+2y=4 \\ 2x+y=3. \end{cases}$$

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    $\begingroup$ Well, the set you are given is a subset of $\Bbb R^3$, and your system as only two unknowns, so its solution set would be a subset $\Bbb R^2$. Moreover your system has only one solution, and what you are asked is to give a system with an infinite number of solution. $\endgroup$ – Arnaud D. Mar 24 '17 at 9:10
  • $\begingroup$ @ArnaudD. I see. What if I had $\begin{cases} x = a \\ y = a^2 \\ z = b\end{cases}$? $\endgroup$ – Yos Mar 24 '17 at 9:12
  • $\begingroup$ The solution set of your equations (in the post, not as in the comment) in $x,y,z$ is $\{(1,1,b)|b\in\mathbb{R}\}$ and not what is sought. $\endgroup$ – ancientmathematician Mar 24 '17 at 9:12
  • $\begingroup$ The solution set of the system given in your question is $\{\langle 1,1,b\rangle\mid b\in\mathbb R\}$. So not $\{\langle a,a^2,b\rangle\mid a,b\in\mathbb R\}$. $\endgroup$ – drhab Mar 24 '17 at 9:14
  • $\begingroup$ @Yos A system of linear equations for the solutions set $\{(a, a^2, b)|a,b \in \mathbb{R}\}$ means a system such that any solution is of the form $(a,a^2,b)$, and all such vectors are solutions. Your system in your comment has only one solution of this form, but $(a',a'^2,b')$ wouldn't be a solution if $a'\neq a$ or $b'\neq b$. $\endgroup$ – Arnaud D. Mar 24 '17 at 9:17
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A non empty solution set $X \subset \mathbb{R}^n$ to a set of linear equations is a sub-vectorspace of $\mathbb{R}^n$. In the case given let $X = \{(a,a^2,b) \mid a,b \in \mathbb{R} \}$. Consider the element $(1,1,0) \in X$ then if $X$ were a subspace of $\mathbb{R}^3$, certainly $2(1,1,0) \in X$. But if there are $a,b \in \mathbb{R}$ such that $2(1,1,0) = (a,a^2,b)$ then we must have $a = 2$ and $a = \sqrt{2}$ by considering the first two coordinates. So $X$ is not a subspace and so there is no set of linear equations for which $X$ is a solution.

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  • $\begingroup$ @ Ollie: unfortunately we're just in the beginning of the course and spaces and subspaces are beyond the material we've covered so far. $\endgroup$ – Yos Mar 24 '17 at 9:36
  • $\begingroup$ @Yos But vector spaces are really useful for studying systems of linear equations, and should be allowed to be used. $\endgroup$ – Dietrich Burde Mar 24 '17 at 9:38
  • $\begingroup$ @ Dietrich Burde: I'm sure but I can't use this knowledge in the proof anyway. I was hoping there was a simple proof based on $\mathbb{R}$ features. Is it enough to show one example where the solution set doesn't hold in order to prove this? $\endgroup$ – Yos Mar 24 '17 at 9:40
  • $\begingroup$ No, it is not.. $\endgroup$ – Dietrich Burde Mar 24 '17 at 9:42
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Every non-trivial equation ($a_i \ne 0$) of a linear system of three variables describes an affine plane of $\mathbb{R}$: $$ a_{i1} x_1 + a_{i2} x_2 + a_{i3} x_3 = b_i \iff \\ (a_{i1},a_{i2}, a_{i3}) \cdot (x_1,x_2,x_3) = b_i \iff \\ a_i \cdot x = b_i \iff \\ \frac{a_i}{\lVert a_i\rVert} \cdot x = \frac{b_i}{\lVert a_i\rVert} \iff \\ n_i \cdot x = d_i $$ where $n_i$ is a unit normal vector of the $i$-th plane and $d_i$ is the (signed) distance of the plane from the origin.

A solution of a system must be element of the intersection of those planes. So you can get, and thus model, either

  • a plane,
  • a line,
  • a single point or
  • no solution (empty set)

These are the possible intersection sets of a non-empty set of planes in $\mathbb{R}^3$.

Your set \begin{align} S &= \{ (a, a^2, b) \mid a, b \in \mathbb{R} \} \\ &= \{ (x, y, z) \in \mathbb{R}^3 \mid -x^2 + y = 0 \} \end{align} does not fit any of these. Here is a visualization for it:

Graph of the solution set

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  • $\begingroup$ @ mvw: why doesn't the set fit for example single point? $\endgroup$ – Yos Mar 26 '17 at 5:25
  • $\begingroup$ @Yos It contains infinite many points. E.g. vary $b$. Thus not just a single point. I added a visualization. You see it is an infinitely extruded parabola. $\endgroup$ – mvw Mar 26 '17 at 18:29
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This is another answer that was provided in our course. It doesn't involve knowledge of subspaces or any advanced notions at all.

Let $S=\{(a,a^2,b)|a,b \in \mathbb{R}\}$. We have 3 unknowns, le them me $x,y,z$.

Let $kx+ly+mz=r$ (*), $k,l,m,r \in \mathbb{R}$ be an equation in the system. Then for all $a,b, \in \mathbb{R}$: $ka+la^2+mb=r$.

If we choose, for example, $a=0$, $b=1$ then $m=r$.

If we choose $a=0$, $b=2$, then $2m=r$. IT follows then, that $m=r=0$, therefore the the equation (*) is of form $ka+la^2=0$.

If we choose $a=1$, and then $a=-1$, then $k+l=0$ and $-k+l=0$, therefore $k=l=0$.

Therefore all equations in our system are of form $0x+0y+0z=0$ so the set of all solutions is essentially $\mathbb{R^3}$.

But $S \neq \mathbb{R^3}$ because for example (1,2,1) is not of form $(a,a^2,b)$ because $1^2 \neq 2$.

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