# When linear homogeneous equations are equivalent?

It is obvious that the following homogeneous linear equations are equivalent $$a_1x_1+\dots+a_nx_n=0$$ $$ka_1x_1+\dots+ka_nx_n=0$$ with $k\in\mathbb{R}\setminus{\{0\}}$. The question is; if two linear homogeneous equations are equivalent, are they multiples? I think I have a proof of this fact, but it seems excessively complicated.

The proof is the following.

Let $A\equiv a_1x_1+\dots+a_nx_n=0$ and $B\equiv b_1x_1+\dots+b_nx_n=0$ be two equivalent equations, lets suppose they are not multiples. Each equation has an associated matrix $\begin{pmatrix}a_1&\dots&a_n\end{pmatrix}$ and $\begin{pmatrix}b_1&\dots&b_n\end{pmatrix}$ respectively. This associated matrices are row-equivalent if and only if the equations are multiples, so they aren't. We know that two matrices are row equivalent if and only if they have the same Hermite normal form, so they don't.

The Hermite normal forms of the associated matrices will be $\begin{pmatrix}1&a_2'&\dots&a_n'\end{pmatrix}$ and $\begin{pmatrix}1&b_2'&\dots&b_n'\end{pmatrix}$ respectively. As they have to be different, lets suppose $a_i'=b_i'$ for all $i\in\{2,\dots,r\}$ and $a_i'\not=b_i'$ for all $i\in\{r+1,\dots,n\}$.

Let $\alpha:=(\alpha_1,\dots,\alpha_n)$ be a solution of $A$ with the $n-r$ last components $\not=0$ (which is always possible). Now, lets insert $\alpha$ in $B\equiv x_1+b_2'x_2+\dots+b_n'x_n=0$, so $\alpha_1+b_2'\alpha_2+\dots+b_n'\alpha_n = \xi$.

Now, lets sum $A\equiv x_1+a_2'x_2+\dots+a_n'x_n=0$ with $B\equiv x_1+b_2'x_2+\dots+b_n'x_n=0$ and plug in $\alpha$, so we obtain $2\alpha_1+2\sum_{i=2}^ra_i'\alpha_i+\sum_{j=r+1}^n(a_j'+b_j')\alpha_j=\xi$

Lets try to find an expression for $\xi$. We know that $2\alpha_1+2\sum_{i=2}^ra_i'\alpha_i+2\sum_{j=r+1}^na_j'\alpha_j=0$, reordering we obtain $2\alpha_1+2\sum_{i=2}^ra_i'\alpha_i=-2\sum_{j=r+1}^na_j'\alpha_j$. Summing up $\sum_{j=r+1}^n(a_j'+b_j')\alpha_j$ in both sides we get $$2\alpha_1+2\sum_{i=2}^ra_i'\alpha_i+\sum_{j=r+1}^n(a_j'+b_j')\alpha_j=-2\sum_{j=r+1}^na_j'\alpha_j+\sum_{j=r+1}^n(a_j'+b_j')\alpha_j=\xi$$ So, reordering $$\xi=\sum_{j=r+1}^n\alpha_j(b_j'-a_j')$$ So, if $\xi=0$ we would have $$\alpha_1+b_2'\alpha_2+\dots+b_n'\alpha_n=\alpha_1+b_2'\alpha_2+\dots+b_n'\alpha_n-\xi=\alpha_1+\sum_{i=2}^ra_i'\alpha_i+\sum_{j=r+1}^nb_j'\alpha_j-\sum_{j=r+1}^n\alpha_j(b_j'-a_j')=\alpha_1+\sum_{k=2}^na_k'\alpha_k=0$$ So, $b_k'=a_k'$ for all $j\in\{r+1,\dots,n\}$, which is a contradiction.

Assume that the solution spaces for $a_1\cdot x_1+...+a_n\cdot x_n=0$ and $b_1\cdot x_1+...+b_n\cdot x_n=0$ are the same. Then two equations together again have the same solution space. Hence the matrix $$\begin{bmatrix}a_1 & a_2 & ... & a_n\\ b_1 & b_2 &...&b_n\end{bmatrix}$$ has rank $1$, which implies that the rows are linearly dependent. Therefore there exists $k$ such that $b_i=k\cdot a_i$ for all $i$.