I tried to solve the following exercise and would like to know if it is correct and if there is more efficient way to prove it? Any feedback is appreciated.
Chapter 1,
Exercise 6
Prove that if two homogeneous systems of linear equations in two unknowns have the same solutions, then they are equivalent.
Edit
Let $E_1$ denote system $A_{i1}x_1+A_{i2}x_2=0$, where $1\leqslant i \leqslant n$, and
$E_2$ denote system $B_{i1}x_1+B_{i2}x_2=0$, where $1\leqslant i \leqslant m$.
Proof
Let $(a,b)$ be the a solutions of $E_1$ and $E_2$
Then
$\sum_{i=1}^n(a\cdot A_{i1} + b\cdot A_{i2}) =0$,
$\sum_{i=1}^m(a\cdot B_{i1} + b\cdot B_{i2})=0$.
So
$$\sum_{i=1}^{\max(m,n)}(a\cdot A_{i1} + a\cdot B_{i1}) + \sum_{i=1}^{\max(m,n)}(b\cdot A_{i2}+ b\cdot B_{i2})=0$$
Hence
$$\sum_{i=1}^n(a\cdot A_{i1} + b\cdot A_{i2}) = – \sum_{i=1}^m(a\cdot B_{i1} + b\cdot B_{i2})$$
Hence the systems are equivalent.
Best Answer
Per the discussion in the comments, it seems that your solution is incorrect.
Here is one correct approach to the problem. First of all, we consider $3$ separate cases.
Case 1: If every pair of numbers solves $E_1$, then it must hold that equations of $E_1$ are of the form $0x_1 + 0x_2 = 0$. If $E_2$ has the same set of solutions, it must also consist of only zero equations, which means that it is equivalent to $E_1$.
Case 2: Suppose that the only solution to $E_1$ is $(0,0)$. Show that $E_1$ is equivalent to the system of equations $E_0$ defined by $$ 1x_1 + 0x_2 = 0, \qquad 0x_1 + 1x_2 = 0. $$ If $E_1$ and $E_2$ both have $(0,0)$ as their only solution, then we can say that $E_1$ is equivalent to $E_0$, which is equivalent to $E_2$. It follows that $E_1$ is equivalent to $E_2$.
Case 3: Suppose that $E_1$ has one non-zero equation, and also has a non-zero solution $x_1 = a, x_2 = b$. Show that $E_1$ is equivalent to the system of equations $E_0$ whose only equation is $$ -bx_1 + ax_2 = 0. $$ As before: $E_1$ is equivalent to $E_0$, which is equivalent to $E_2$. It follows that $E_1$ is equivalent to $E_2$.
Filling in the blanks:
Useful facts:
This is easy to understand if you think of each solution set to a non-zero equation $Ax + By = 0$ as a line through the origin: any single line contains a point outside the origin, and there is exactly one line between any two points.
A third fact:
Case 3: Because $(a,b)$ is a solution to $E_1$, we have $$ A_{i1}a + A_{i2}b = 0 \implies A_{i1} \cdot a = A_{i2} \cdot (-b). $$ By point 3 above, this means that the $i$th equation is a multiple of $-bx_1 + ax_2 = 0$. In other words, every equation in $E_1$ is a multiple of the only equation in $E_0$. Conversely, we that $E_1$ must have a non-zero equation (say the $i$th equation), and the equation of $E_0$ is a multiple of the $i$th equation of $E_1$. So, $E_0$ and $E_1$ are equivalent.
Case 2: We want to show that every equation in $E_1$ can be written as a linear combination of the equations in $E_0$, and that every equation in $E_0$ can be written as a linear combination of the equations in $E_1$.
The first statement is easy to show (try it, let me know if you're stuck). For the second statement: we know that $E_1$ has $(0,0)$ as its only solution. Using points 1 and 2 above, we argue that $E_1$ contains at least two non-zero equations that are not multiples of each other. In other words, there must exist $i,j$ such that $A_{i1}A_{j2} \neq A_{i2}A_{j1}$ (a convenient way to rewrite $\frac{A_{1i}}{A_{2i}} \neq \frac{A_{j1}}{A_{j2}}$ that still works if one of the denominators is zero).
Let $D = A_{i1}A_{j2} - A_{i2}A_{j1}$. Verify that adding $\frac{A_{j2}}{D}$ times the $i$th equation and $-\frac{A_{i2}}{D}$ of the $j$th equation produces one equation of $E_0$. Adding $-\frac{A_{j1}}{D}$ of the $i$th equation and $\frac{A_{i1}}{D}$ oof the $j$th equation produces the second.
So, $E_1$ is equivalent to $E_0$.