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Prove that if two homogeneous systems of linear equations in two unknowns have the same solutions, then they are equivalent.

*Two systems of linear equations are $\textit{equivalent}$ if each equation in each system is a linear combination of the equations in the other system.

$\textbf{Proof}.$ Let $$a_{11}x+ a_{12}y=0$$ $$a_{21}x+ a_{22}y=0$$ be a homogeneous system of linear equations where $[x,y]$ is a solution. In addition, let $$b_{11}x+ b_{12}y=0$$ $$b_{21}x+ b_{22}y=0$$ be a homogeneous system of linear equations which has the same solution as the above system.

To show that they are equivalent we have to show that each equation in each system is a linear combination of the equations on the other system. For example,

$$a_{11}x+ a_{12}y= c_1 (b_{11}x+ b_{12}y) + c_2 (b_{21}x+ b_{22}y)$$

where $c_1$ and $c_2$ are weights.

The crucial part here is that the two systems have the same solution so that it is possible to write,

$$(c_1 b_{11} + c_2 b_{21} - a_{11})x + (c_1b_{12} + c_2b_{22} - a_{12})y = 0$$.

Now, if $[x,y] \neq [0,0]$ then it would easily follow that each component $a_{ij}$ is a linear combination of the components $b_{ij}$. But, if $[x,y] = [0,0]$ then e.g. $(c_1 b_{11} + c_2 b_{21} - a_{11}) \neq 0$. How should I proceed to prove this part?

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marked as duplicate by Robert Soupe, Claude Leibovici, Arnaud D., JMP, GNUSupporter 8964民主女神 地下教會 Jan 15 '18 at 12:05

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  • $\begingroup$ $[0,0]$ is always a solution to any homogenous system of equations. If there are other solutions, then your calculation will work. But now you need to consider what it tells you about $(a)$ and $(b)$ if $[0,0]$ is the only solution to them. $\endgroup$ – Paul Sinclair Oct 7 '17 at 21:38
  • $\begingroup$ @Paul Sinclair Well, that is exactly my question. $\endgroup$ – mathemania Oct 8 '17 at 1:06
  • $\begingroup$ No. Your question was how to proceed. I am answering that you need to proceed by examing exactly what it says about $a$ that its only .solution is $[0,0]$. And similarly for $b$. And then consider how you can use that information to determine $c$. $\endgroup$ – Paul Sinclair Oct 8 '17 at 1:51
  • $\begingroup$ @Paul Sinclair I'm thinking that since $c_1b_{11} + c_2b_{21} - a_{11} \neq 0$, we have $c_1b_{11} + c_2b_{21} - a_{11} = c$ for some constant $c$. In addition, $c_1b_{12} + c_2b_{22} - a_{12} = d$ for some constant $d$, then we can solve for $c_1$ and $c_2$. But does that really constitute a proof or are there some things that I'm missing? $\endgroup$ – mathemania Oct 8 '17 at 3:51
  • $\begingroup$ I don't see where you are going with that. My hint was that you need to consider what it says about the $a$ system - in particular about the matrix $$\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}$$ if the system $$a_{11}x + a_{12}y = 0\\a_{21}x + a_{22}y = 0$$ has only $[0,0]$ as a solution. (Note that $B = (BA^{-1})A$.) $\endgroup$ – Paul Sinclair Oct 8 '17 at 18:00
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This has occurred to me:

Let $S_{1}\equiv \left\{ \begin{array}{lcc} a_{1}x+b_{1}y=0 \\ \\ c_{1}x+d_{1}y=0 \end{array} \right., A_{1}= \begin{pmatrix} a_{1} & b_{1} \\ c_{1} & d_{1} \end{pmatrix} $ y $S_{2}\equiv \left\{ \begin{array}{lcc} a_{2}x+b_{2}y=0 \\ \\ c_{2}x+d_{2}y=0 \end{array} \right., A_{2}= \begin{pmatrix} a_{2} & b_{2} \\ c_{2} & d_{2} \end{pmatrix}$

Since it is a homogeneous system, the trivial solution $(x, y) = (0,0)$ is always given.

If the common set of solutions were $\mathbb{R}^{2}$ (ie outside $2$ degrees of freedom)

Then, $rank(A_{1})=rank(A_{2})=0\rightarrow A_{1}=A_{2}=\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$ (and it's already)

Let $(x,y)=(0,0)+\mu (r,s)=\mu (r,s), \mu \in \mathbb{R}$ and $rs\neq 0$ be the common solution. Then, with $1$ degree of freedom, $rank \begin{pmatrix} a_{1} & b_{1} \\ c_{1} & d_{1} \end{pmatrix}=rank \begin{pmatrix} a_{2} & b_{2} \\ c_{2} & d_{2} \end{pmatrix}=1$

Let it be for example, $a_{1}\neq 0$ and $a_{2}\neq 0\rightarrow \begin{pmatrix} a_{1} & b_{1} \\ c_{1} & d_{1} \end{pmatrix}\sim \begin{pmatrix} a_{1} & b_{1} \\ a_{2} & b_{2} \end{pmatrix}\sim \begin{pmatrix} ra_{1}+sb_{1} & b_{1} \\ ra_{2}+sb_{2} & b_{2} \end{pmatrix}=\begin{pmatrix} 0 & b_{1} \\ 0 & b_{2} \end{pmatrix}\rightarrow rank \begin{pmatrix} a_{1} & b_{1} \\ a_{2} & b_{2} \end{pmatrix}=1\rightarrow (a_{2},b_{2})=\lambda (a_{1},b_{1})+0(c_{1},d_{1})$

And so the others.

In case of a single (trivial) solution: $rank \begin{pmatrix} a_{1} & b_{1} \\ c_{1} & d_{1} \end{pmatrix}=rank \begin{pmatrix} a_{2} & b_{2} \\ c_{2} & d_{2} \end{pmatrix}=2\rightarrow rank\begin{pmatrix} a_{1} & b_{1} \\ c_{1} & d_{1} \\ a_{2} & b_{2} \\ c_{2} & d_{2} \end{pmatrix}=2$, with $\begin{vmatrix} a_{1} & b_{1} \\ c_{1} & d_{1} \end{vmatrix}\neq 0\rightarrow \left\{ \begin{array}{lcc} F_{3}\equiv (a_{2},b_{2})=\alpha (a_{1},b_{1})+\beta (c_{1},d_{1}) \\ \\ F_{4}\equiv (c_{2},d_{2})=\rho (a_{1},b_{1})+\sigma (c_{1},d_{1}) \end{array} \right.$

(Another idea) Let's go with a geometric answer. Except for the trivial case, a homogeneous system of two equations with two unknowns represents two straight lines that pass through the origin. If the system has infinite solutions, it is the same straight line, which, by matching the solutions, will also be the same straight line of the other system. And if they have a unique solution, both systems are formed by lines that pass through the same point, so that they all belong to the same set of lines, which would be determined, for example, by the first two.

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