For $x\in\Bbb{R}^n$ let supp$(x)=\{1\leq i\leq n:x_i\neq 0\}.$
Let $\mathcal{P}=\mathcal{P}^=(A,b)$ be a polyhedron with $A\in\Bbb{R}^{m\times n},\ b\in\Bbb{R}^m$ and let $x\in\mathcal{P}$.
Show that$$x\text{ is a vertex of }\mathcal{P}\iff \text{rank}(A_{·\text{supp}(x)})=|\text{supp}(x)| $$Here is my attempt. I am very unsure about my answer, so if anyone could point out my mistakes and/or suggest improvements of any kind, I would be very grateful. Thanks.
Because $\mathcal{P}=\mathcal{P}^=(A,b)$ we directly have that $I_x=\{1,\dots, m\}$ and therefore $A_{I_x}=A$ for any point $x\in\mathcal{P}\qquad (1) $
$"\implies":$ First assume that $x$ is a vertex of $\mathcal{P}$.
By definition, we have that, for a point $x\in\mathcal{P}$ and $I_x=\{1\leq i\leq m:\ A_{i·}x=b_i\}\subseteq\{1,\dots,m\}$:$$x\text{ is a vertex of }\mathcal{P}\iff A_{I_x·}x=b_{I_x}\ \land\ \text{matrix }A_{I_x·}\text{ has full rank}$$We can project $\mathcal{P}$ onto $\Bbb{R}^{|\text{supp}(x)|}$.
If $\mathcal{P}$ is a polyhedron, then the projection $\mathcal{P'}$ of $\mathcal{P}$ is also a polyhedron and every vertex of $\mathcal{P'}$ is a projection of a vertex of $\mathcal{P}$. So $x$ projects onto a vertex $x'$ of $\mathcal{P'}$, for which $A_{I_{x'}\text{supp}(x)}$ has full rank. Because of $(1):A_{I_{x'}\text{supp}(x)}=A_{·\text{supp}(x)}$ and because $A_{·\text{supp}(x)}$ has full rank, its rank is equal to the number of columns, which is, by definition, $=|\text{supp}(x)|$. This gives the desired result.
$"\impliedby":$ Now assume that $\text{rank}(A_{·\text{supp}(x)})=|\text{supp}(x)| $.
Let $x'\in\Bbb{R}^{|\text{supp}(x)|}$ be the vector $x$ without its zero rows. Let $A_{·\text{supp}(x)}x'=b'\in\Bbb{R}^m$.
The set of solutions of the set of equations $A_{·\text{supp}(x)}x'=b'$ is a projection of $\mathcal{P}$ onto $\Bbb{R}^{|\text{supp}(x)|} $. Projections of polyhedra are also polyhedra, therefore $\mathcal{P}'=\mathcal{P}'^=(A_{·\text{supp}(x)},b')$ is also a polyhedron.
Let $m\geq k\in\Bbb{N} $ and $n\geq j_0\in\Bbb{N}\ \land\ j_0\notin \text{supp}(x)$. Then $$ \begin{pmatrix} A_{11}&& \dots&& A_{1j_0}&& \dots&& A_{1n}\\ \vdots&&\vdots&& \vdots&& \vdots&& \vdots\\ A_{k1}&& \vdots&& A_{kj_0}&& \vdots&& A_{kn}\\ \vdots&&\vdots&& \vdots&& \vdots&& \vdots\\ A_{m1}&&\dots&& A_{mj_0}&& \dots&& A_{mn} \end{pmatrix}\begin{pmatrix} x_1\\ \vdots\\ x_{j_0}\\ \vdots\\ x_n \end{pmatrix} =\begin{pmatrix} A_{11}x_1+&& \dots&& +A_{1j_0}x_{j_0}+&& \dots&& +A_{1n}x_n\\ \vdots&&\vdots&& \vdots&& \vdots&& \vdots\\ A_{k1}x_1+&& \vdots&& +A_{kj_0}x_{j_0}+&& \vdots&& +A_{kn}x_n\\ \vdots&&\vdots&& \vdots&& \vdots&& \vdots\\ A_{m1}x_1+&&\dots&& +A_{mj_0}x_{j_0}+&& \dots&& +A_{mn}x_n \end{pmatrix}$$$$=\begin{pmatrix} A_{11}x_1+&& \dots&& +A_{1(j_0-1)}x_{(j_0-1)}+A_{1(j_0+1)}x_{(j_0+1)}+&& \dots&& +A_{1n}x_n\\ \vdots&&\vdots&& \vdots&& \vdots&& \vdots\\ A_{k1}x_1+&& \vdots&& +A_{k(j_0-1)}x_{(j_0-1)}+A_{k(j_0+1)}x_{(j_0+1)}+&& \vdots&& +A_{kn}x_n\\ \vdots&&\vdots&& \vdots&& \vdots&& \vdots\\ A_{m1}x_1+&&\dots&& +A_{m(j_0-1)}x_{(j_0-1)}+A_{m(j_0+1)}x_{(j_0+1)}+&& \dots&& +A_{mn}x_n \end{pmatrix} $$Because $x_{j_0}=0$. So for each row $1\leq i\leq m$ of matrices $A$ and $A_{·\text{supp}(x)}$ the multiplication with the vectors $x$ and $x'$, respectively, yields the same result, therefore we get $A_{·\text{supp}(x)}x'=b'=b=Ax$.
By definition the number of columns in $A_{·\text{supp}(x)}$ is always equal to $|\text{supp}(x)| $, therefore the rank of $A_{·\text{supp}(x)}$ is always equal to its number of columns, which means the matrix $A_{·\text{supp}(x)}$ has full rank. With $(1)$ this means that $A_{I_{x'}\text{supp}(x)}$ has full rank. This in turn means, that $x'$ is a vertex of $\mathcal{P}'$, which is a projection of $\mathcal{P}$, and as we proved above, $A_{·\text{supp}(x)}x'=Ax$, so $x$ must be a vertex of $\mathcal{P}$, because $x'$ is a projection of $x$ and vertices of projections are projections of vertices.