I was trying to solve a rather large system of equalities and inequalities and was stuck, until I realized that converting the inequalities into equalities by adding more variables showed that zero-vector was the only answer. But I'm not so sure if it's correct.
So my question is, it is valid to do so? What I'm trying to do is to find the range of $x$ where the following is true:
$$\mathbb{A} x\leq 0, x\geq 0$$
where $\mathbb{A}$ is a specific $m\times n$ coefficient matrix and $x$ is an $n\times 1$ vector. But is this the same as solving the following:
$$\tilde{\mathbb{A}}\tilde{x}=0, \tilde{x}\geq 0$$ where $\tilde{\mathbb{A}}$ is an $m\times(n+m)$ matrix and $\tilde{x}$ is an $(n+m)\times 1$ vector. By solution, I mean writing the elements of the original $x$ in terms of the elements that are in $\tilde{x}$ but not $x$.
We will show that the two problems are equivalent in the sense they define the same range of solutions.
Define $\tilde{\mathbb{A}} = \begin{bmatrix} \mathbb{A} & \mathbf{I} \end{bmatrix}$ and $\tilde{\mathbf{x}}= \begin{bmatrix} \mathbb{x} \\ \mathbf{t} \end{bmatrix}$.
Then, if $\mathbb{x}$ is a solution of the inequality constrained problem :
$$\mathbb{A}\mathbf{x} \le 0$$
it suffices to take $\mathbf{t}$ as $-\mathbb{A}\mathbf{x}$, which is a positive vector by the inequality above. So we have : $\tilde{\mathbb{A}} \tilde{\mathbf{x}} = \mathbb{A}\mathbf{x}+\mathbf{t}=\mathbb{A}\mathbf{x}-\mathbb{A}\mathbf{x}=0$.
Conversely, if $$\tilde{\mathbb{A}}\tilde{\mathbf{x}} = 0$$
then we have : $\mathbb{A}\mathbf{x}+\mathbf{t} = \tilde{\mathbb{A}}\tilde{\mathbf{x}} = 0$, or equivalently $\mathbb{A}\mathbf{x}=-\mathbf{t}$. Since $\mathbf{t}$ is a positive vector, we can conclude that $\mathbb{A}\mathbf{x} \le 0$.