What is the motivation for Murray-von Neumann equivalence 
Definition: If $p,q$ are projections in a $C^*$-algebra $A$, we say that they are Murray-von Neumann equivalent, and we write $p\sim q$, if there exist $u\in A$ such that $p=u^*u$ and $q=uu^*$.

I wonder what is the motivation of this definition. It turns out that if $A$ is a matrix algebra like $M_n(\mathbb C)$, then this definition is equivalent to saying that $p$ and $q$ have the same rank. But how about the general case? Is Murray-von Neumann equivalence the most natural way to define the equivalence relation on the projections of a $C^*$-algebra in some sense?
 A: Here's one possible take on the question:
$K$-theory for $C^*$-algebras is motivated by topological $K$-theory. Topological $K_0$ for a compact Hausdorff space $X$ is the Grothendieck group of formal differences of isomorphism classes of locally trivial vector bundles over $X$, so we want the operator $K_0$-group of $C(X)$ to match this. 
The Serre-Swan theorem says that every locally trivial vector bundle is complemented, so it corresponds to $P C(X, \mathbb{C}^n)$ for some $n$ and some projection in the $C^*$-algebra $C(X, M_n(\mathbb{C}))$. To think of all of these operators $P$ living in the same $C^*$-algebra, we think of them all as projections in $C(X, \mathcal{K})$, where $\mathcal{K}$ is the compact operators. An isomorphism of vector bundles $P C(X, \mathbb{C}^n), Q C(X, \mathbb{C}^n)$ amounts to a continuous choice of vector-space isomorphisms between their fibres; this is an element $V \in C(X, \mathcal{K})$ such that $V^*V = P$ and $VV^* = Q$. This indicates why Murray-von Neumann equivalence is a better choice than unitary equivalence: once you're sitting in $C(X, \mathcal{K})$, unitary equivalence happens outside your $C^*$-algebra (in its multiplier algebra).
Finally, Murray-von Neumann equivalence plays nicely with the abstract notion of rank given by a trace (when the $C^*$-algebra carries one): if $\tau$ is a trace, and $p$ and $q$ are MvN equivalent, then $\tau(p) = \tau(v^*v) = \tau(vv^*) = \tau(q)$. And the natural alternatives, namely unitary equivalence and homotopy equivalence, turn out to agree with Murray-von Neumann equivalence for projections in $\mathcal{K} \otimes A$.
