Linear Systems of Two Variables and Equivalency (1.2.6)

Hoffman and Kunze Linear Algebra, section 1.2, exercise 6:

Let $F_1$ and $F_2$ be homogeneous systems of linear equations in two unknowns. Prove that if their solution sets are identical, then they are equivalent.

Proof: Lemma 1: Admit two linear systems with one equation each, $A : ax_1+bx_2=0$ and $B : cx_1+dx_2=0$. We have $A$ equivalent to $B$ if and only if $A$ and $B$ have the same solution set if and only if $\dfrac{a}{c}=\dfrac{b}{d}$. Proof: (1 ⇒ 2) Theorem 1 ensures this. (2 ⇒ 3) Any solution $(n_1,m_1)$ of $A$ must satisfy $(n_1,-\dfrac{an_1}{b})$, and this is seen to be a valid solution so these tuples form the solution set of $A$. Similarly, the solution set of $B$ is of the form $(n_2,-\dfrac{cn_2}{d})$. Due to these solution sets' equivalence, we have $(1,-\dfrac{c}{d})$ is a solution of $A$, so that $a(1)+b(-\dfrac{c}{d})=0$, entailing (3). (3 ⇒ 1) Let $c_1 = \dfrac{c}{a} = \dfrac{d}{b}$. Evidently $c_1(ax_1+bx_2)=cx_1+dx_2=c_1(0)=0$.$~\square$

Lemma 2: Let $ax_1+bx_2=0$ and $cx_1+dx_2=0$ be two equations that aren't scalar products of each other. We have their shared solution set is $(0,0)$. Proof: The solution sets of both are $(n_1,-\dfrac{an_1}{b})$ and $(n_2,-\dfrac{cn_2}{d})$ respectively. Assume they share a nontrivial solution. Then $n_1=n_2≠0$ and $-\dfrac{an_1}{b}=-\dfrac{cn_2}{d}=-\dfrac{cn_1}{d}$, entailing $\dfrac{a}{c}=\dfrac{b}{d}$ and these two equations are scalar products by the first lemma, a contradiction.$~\square$

Let $F_1$ contain $n$ equations and $F_2$ contain $m$ equations. Assume $\dfrac{A_{j_1 1}}{A_{k_1 1}}=\dfrac{A_{j_1 2}}{A_{k_1 2}}$ and $\dfrac{B_{j_2 1}}{B_{k_2 1}}=\dfrac{B_{j_2 2}}{B_{k_2 2}}$ for all $1 ≤ j_1,k_1 ≤ n$ and $1 ≤ j_2,k_2 ≤ m$, i.e. all the equations of $F_1$ are scalar products of each other, and the same pertaining to $F_2$, as by the first lemma. Now, $(n,-\dfrac{nA_{1 1}}{A_{1 2}})$ is the solution set of $F_1$. Let $B_{v 1}x_1 + B_{v 2}x_2 = 0$ be an arbitrary equation of $F_2$. We have $(1,-\dfrac{A_{1 1}}{A_{1 2}})$ is a solution of this equation, so that $B_{v 1}+B_{v 2}(\dfrac{A_{1 1}}{A_{1 2}})=0$ and now $\dfrac{A_{1 1}}{B_{v 1}}=\dfrac{A_{1 2}}{B_{v 2}}$ so that by the first lemma this arbitrary equation of $F_2$ is a scalar product of the equation from $F_1$, and now $F_2$ is a linear combination of $F_1$. Reversing the argument, we obtain $F_1$ is a linear combination of $F_2$, and now $F_1$ and $F_2$ are equivalent.

So assume that there exist two equations in one of the systems that are not scalar products of each other, and designate their system $F_1$. These equations are $ax_1+bx_2=0$ and $cx_1+dx_2=0$, and evidently at least one of $a$ or $c$ must be nontrivial for their assumption to hold; let it be $c$. Let $nx_1+mx_2=0$ be an arbitrary equation of $F_2$. This equation is a linear combination of the two former by scalars $c_1$ and $c_2$ if (and only if) $c_1a+c_2c=n$ and $c_1b+c_2d=m$; solving these two nonhomogenous equations, we obtain $$c_1=\dfrac{m-\dfrac{dn}{c}}{b-\dfrac{da}{c}}~\text{and}~c_2=\dfrac{n-c_1a}{c}$$ are such solution scalars. These avoid division by zero as $c≠0$ and $b-\dfrac{da}{c}=0 ⇒ \dfrac{a}{c}=\dfrac{b}{d}$, a contradiction by the first lemma and this paragraph's assumption. Thus $F_2$ is a linear combination of $F_1$.

$F_1$ having two mutually non-scalar-product equations implies the same for $F_2$, since by the second lemma the solution set of $F_1$ is $(0,0)$, the same applies for $F_2$, yet the first lemma would entail a nontrivial solution set for $F_2$ if its equations were all scalar products of each other in two unknowns. Thus the argument above can be reversed to obtain $F_1$ is a linear combination of $F_2$ and now $F_1$ is equivalent to $F_2$.$~\square$