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I'm not very familiar with convex sets and such. Appreciate any feedback for the following proof.

$\textbf{Proposition.}$ Suppose $A$ and $A^{*}$ are convex sets, where a matrix $M$ establishes one-to-one correspondence between the two. Prove that there is also a one-to-one correspondence between the extreme points of $A$ and $A^{*}$.

One-to-One Correspondence (given): $\forall \textbf{a} \in A,~\exists \mathbf{a^{*}} \in A^{*}$ s.t. $~M\mathbf{a} = \mathbf{a^{*}}$,$~$ and $\forall \mathbf{a^{*}} \in A^{*},~ \exists~\text{a single}~\mathbf{a} \in A$ s.t. $~M\mathbf{a} = \mathbf{a^{*}}$.

$\textit{Proof.}$ Let $\xi(A)$ be the set of all extreme points of A. Similarly, let $\xi(A^{*})$ be the set of all extreme points of $A^{*}$. Suppose $x^{*} \in \xi(A^{*})$ is not an extreme point and $x \in \xi(A)$ is an extreme point. Then there exists two distinct $x_1^{*}, x_2^{*} \in A^{*}$ and $\lambda \in (0, 1)$ such that

\begin{equation*} x^{*} = \lambda x_1^{*} + (1 - \lambda)x_2^{*}. \end{equation*}

Hence,

\begin{align*} Mx^{*} &= \lambda Mx_1^{*} + (1 - \lambda)Mx_2^{*} \\ Mx^{*} &= M\left(\lambda x_1^{*} + (1 - \lambda)x_2^{*}\right) \\ x &= \lambda x_1 + (1 - \lambda)x_2, \end{align*}

which is a contradiction, since $x$ is an extreme point.

Now, since $\xi(A^{*}) \subset A^{*}$ and $\xi(A) \subset A$, it directly holds that for every $x^{*} \in \xi(A^{*})$ there exists a single $x \in \xi(A)$ such that $x^{*} = Mx$.

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    I guess it should be $Mx=x^*$ instead of $Mx^*=x$ according to your description of $M$. Otherwise it looks good to me. Also, I'm curious as to the other direction, which I don't think is trivial if, say, $M$ is not invertible. – 2017-01-31
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    You want to go both ways. $x$ an extreme point of $A$ **iff** there do not exist distinct $x_1, x_2 \in A$ and $0 < \lambda < 1$ with $x = \lambda x_1 + (1-\lambda) x_2$ **iff** there do not exist distinct $y_1 = M x_1, y_2 = M x_2 \in A^*$ and $0 < \lambda < 1$ with $M x = \lambda y_1 + (1-\lambda) y_2$ **iff** $Mx$ is an extreme point of $A^*$ – 2017-01-31
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    You're both right, the other direction isn't trivial. I will add it in. – 2017-01-31

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