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This is a follow-up to this question: Proving that two systems of linear equations are equivalent if they have the same solutions

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

I found this problem in Hoffman and Kunze's Linear Algebra book. Since row operations or anything like that haven't yet been introduced at this point, I want to solve this without using row operations.

For that reason, this question is not really a duplicate because I'm asking for a different way to solve the problem- something along the lines of the 2nd answer in the linked question, but I have a doubt about that.

Comparing this equation with all the equations of the second system and also utilising the fact given that both the systems have the same solutions

Why? We can only equate coefficients of equations in variables $x_1, x_2$ if they're the same for all values of $x_1, x_2$, and not just specific values for them. Secondly, since we're free to select the scalars $C_1, C_2, \ldots$, that indicates any equation in the second system can be made an arbitrary linear combination of the equations in the first system.

I'm probably missing something obvious but it would be nice to get some clarification on a non-row operation method.

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  • $\begingroup$ what is your definition of equivalent $\endgroup$ – Andres Mejia Apr 20 '18 at 14:30
  • $\begingroup$ If by equivalent you mean that any pair of homogeneous linear equations are the same if they share the same solution, then this is false. To see this, you can fix an arbitrary point in $\mathbb{R}^2$ and consider the infinite pairs of lines intersecting at this point, all different from each other. $\endgroup$ – Alex D Apr 20 '18 at 14:52
  • $\begingroup$ @AndresMejia: Sorry for the late reply. By equivalent, I meant that each equation in one system can be expressed as a linear combination of equations in the other system. $\endgroup$ – Shirish Kulhari Apr 24 '18 at 12:56
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Because there are only two unknowns, say $x$ and $y,$ and the systems are homogeneous, each equation has the form $A_1x + A_2y = 0$ where the coefficients $A_1$ and $A_2$ are scalars. Hence, each equation represents a line through the origin unless $A_1$ and $A_2$ are zero. There are three cases:

The first case is that all the coefficients are zero. Then the systems are trivially equivalent.

The second case is that all the lines are the same so that the solution is the set of points on that line. Then all the equations are scalar multiples of each other, so the systems again are equivalent.

The last case is that at least two lines are different so that the solution is the origin. Let $$\begin{align} A_{11}x + A_{12}y = 0\\ A_{21}x + A_{22}y = 0 \end{align}$$ be two different lines in the first system so that $A_{11}A_{22} - A_{12}A_{21} \ne 0$ and $$B_1x + B_2y = 0$$ be any equation in the second system. We want to show that scalars $c_1$ and $c_2$ exist such that $$B_1x + B_2y = c_1(A_{11}x + A_{12}y) + c_2(A_{21}x + A_{22}y).$$ That equation leads to the system $$\begin{align} A_{11} c_1 + A_{21}c_2 = B_1\\ A_{12} c_1 + A_{22}c_2 = B_2\end{align}$$ which has the solution $$\begin{align} c_1 & = \frac{B_1 A_{22} - B_2 A_{21}}{A_{11}A_{22} - A_{12}A_{21}}\\\\ c_2 & = \frac{A_{11} B_2 - A_{12} B_1}{A_{11}A_{22} - A_{12}A_{21}}.\end{align}$$ Hence, each equation in each system is a linear combination of the equations in the other system. Thus, the systems are equivalent.


My answer is a simpler version of Vikram's, so after you understand mine, try reading his again to clear up your doubts about it.

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