I would like to check if my proof is correct. I do not have much experience of proving stuff so I would appreciate if anyone could point out and fix some mathematical/proving/wording errors if there some. The statement I am trying to prove is
A polyhedron contains a line if and only if $Ax=0$ has a non-zero solution.
($\Rightarrow$) By contrapositive, assume that $Ax=0$ does not have a non-zero solution. That is $Ax=0$ only if $x$ is a zero-vector. Then, for some $\lambda\in\mathbb{R}$ and $d\in\mathbb{R}$, $A(x+\lambda d)=Ax+\lambda Ad\leq b=0+\lambda\cdot Ad$. This is not always true because for a large $\lambda$, $\lambda\cdot a_i d$ becomes larger than $b_i$ where $a_i,b_i$ are entries of $A$ and $b$ respectively, and $(x+\lambda d)$ is no longer in the polyhedron $P$. Then the polyhedron $P$ does not contain a line.
($\Leftarrow$) By contrapositive, assume that $P$ does not contain a line. Then there exists $\lambda\in\mathbb{R}$ and some constraint $a^T_j \leq b_j$ such that $a^T_j \bar{x}<b_j, a^T_j(\bar{x}+\lambda d)=b_j$. Since $a_jd=0$ and $a_j(\bar{x}+\lambda d)=a_j\bar{x}+\lambda a_j d=a_j\bar{x}=b_j$ for all $j$ in active set of constraints and for some $\lambda\in\mathbb{R}$. Then because $a_j\bar{x}=b_j$, $a_j\bar{x}=0$ only when $\bar{x}=0$. Hence, $Ax=0$ does not have a nonzero solution.
I also need to prove that "$P$ is unbounded if and only if $Ax\leq 0$ has a non-zero solution." However I am struggling how to approach this one. Any hint would also be appreciated.
Thank you in advance