The automorphism group of the algebra of $n$-dimensional matrices over a field $K$ is $PGL\_n(K)$. The automorphism group of $n-1$-dimensional projective space over $K$ is also $PGL\_n(K)$. Therefore, twists of them over $\bar{K}$ are both classified by the same Galois cohomology group, $H^1(PGL\_n(K))$. 


Twists of $M\_n(K)$ are central simple algebras of rank $n^2$. Twists of $K\mathbb P^{n-1}$ are projective varieties geometrically isomorphic to $\mathbb P^{n-1}$. 


There should be a bijection between these two sets.


How explicit and geometrically/algebraically nice can we make it? 


What properties of one object correspond to properties of the other?



Explicit Bijection between Central Simple Algebras and twists of $\mathbb P^n$
References:


Artin, M.
*Brauer-Severi varieties*. Brauer groups in ring theory and algebraic geometry (Wilrijk, 1981), pp. 194–210, 
Lecture Notes in Math., 917, Springer, Berlin-New York, 1982. 


Knus, Max-Albert; Merkurjev, Alexander; Rost, Markus; Tignol, Jean-Pierre
**The Book of Involutions**. (English summary) 
With a preface in French by J. Tits. American Mathematical Society Colloquium Publications, 44. American Mathematical Society, Providence, RI, 1998


Let $A$ be a central simple $K$-algebra of degree $n$ (i.e., of rank $n^2$). Let $Gr(n, A)$ be the subspace of the grassmannian of rank $n$ dimensional subspaces. Let $X \subset Gr(n, A)$ be the subvariety of subspaces which are also right ideals of $A$. Then $X$ is the corresponding space, called the *Severi-Brauer*, *Brauer-Severi*, or *Chatelet* variety of $A$, and is denoted here by $SB(A)$.


When $A = End\_K(V)$ is trivial, every right ideal of $A$ of dimension n (as a $K$-vector space) is of the form $\operatorname{Hom}\_K(V, U)$, where $U$ is a 1-dimensional vector space. So the space is $\mathbb{P}\_K(V)$.


I am not sure of an explanation in the other direction as nice. It is the case that for a Severi-Brauer variety $SB(A)$, you can consider the following subbundle $\mathcal{I}$ of the constant bundle $SB(A) \times A$. The bundle $\mathcal{I}$ consists of pairs $(I, a)$, where $a \in I$. Then $\mathcal{I}$ is a locally free of rank $n$ and the global endomorphism ring $\operatorname{End}\_{SB(A)}(\mathcal{I})$ is isomorphic to $A$ (or the opposite algebra of $A$, I can't remember which). 


However, this doesn't answer your question of how to start with a twist of $\mathbb{P}^{n-1}$ and nicely produce a central simple algebra of rank $n^2$. I think for that you may have to argue by galois descent.


