Matrix units in von Neumann algebras, and $K_0$ groups This question arises from trying to understand the proof of Lemma 3.1.4 in De Commer, Martos, and Nest - Projective representation theory for compact quantum groups and the quantum Baum–Connes assembly map [math.OA].  This Lemma says that when $M$ is a von Neumann algebra, and $(e_{ij})_{i,j=1}^k$ are matrix units in $M$ (so the $e_{ij}$ satisfy the usual rules: $e_{ij}^*=e_{ji}$, $e_{ij} e_{kl} = \delta_{j,k} e_{il}$ and $\sum_i e_{ii}=1$) and also $(f_{ij})_{i,j=1}^k$ are matrix units, then there is a unitary $U\in M$ with $Ue_{ij}U^* = f_{ij}$ for each $i$, $j$.
The proof is very short, but I do not understand it.  Let me say what I do follow:

Note that $e_{11}$ and $f_{11}$ have central support 1

This is just a calculation using a common criterion.

and $k[e_{11}]=[1]=k[f_{11}]$ in $K_0(M)$.

Writing $\sim$ for Murray–von Neumann equivalence, clearly $e_{11} \sim e_{ii}$ for each $i$, so $k[e_{11}] = [e_{11}] + \dotsb + [e_{11}] = \sum_{i=1}^k [e_{ii}]$, but these are now orthogonal projections, so $\sum_{i=1}^k [e_{ii}] = [\sum_{i=1}^k e_{ii}] = [1]$.  (All of this is computed inside $K_0(M)$.)

Hence $e_{11}$ and $f_{11}$ are Murray–von Neumann equivalent by a partial isometry $u$.

This I do not see.  Why is this so?  (The remainder of the proof I do follow: it is just this one, critical, step I am stuck on.)
I am very far from an expert about $K$-theory, but from e.g. Example 3.3.3 in Rordam et al. - Introduction to K-Theory… we can certainly have $K_0(M)=0$, in which case I cannot see how knowing that $k[e_{11}] = k[f_{11}]$ tells us anything.  Am I missing something obvious?
 A: I got a bit confused in trying to fully work out the hints in the comments, and in the end discovered that this is (essentially) a worked example in Kadison and Ringrose, see Exercise 6.9.14.  Note that this does not use K-Theory at all.
Here are the details (following the given answer in Kadison+Ringrose volume 4, but with some alternations as I also couldn't quite follow this).  Fix an arbitrary von Neumann algebra $M$.
Having matrix units $(e_{ij})_{i,j=1}^n$ in $M$ is equivalent to having a family of projections $(e_i)_{i=1}^n$ with $e_1 \sim e_2 \sim \cdots \sim e_n$ and $\sum_i e_i=1$ (which implies that the $(e_i)$ are orthogonal).  In one direction, take $e_i = e_{ii}$ and use that $e_i = e_{ji}^* e_{ji} \sim e_{ji} e_{ji}^* = e_j$.  Conversely, given $(e_i)$, set $e_{ii} = e_i$ and let $e_{ij}$ be the partial isometry implementing that $e_j \sim e_i$.
Claim: Given $(e_i)_{i=1}^n$ and $(f_i)_{i=1}^n$ projections with $e_1\sim\cdots\sim e_j$ and $f_1\sim\cdots\sim f_j$ and $1 = \sum_i e_i = \sum_i f_i$, we have that $e_i\sim f_i$ for each $i$.
Clearly it suffices to show that $e_1 \sim f_1$.  By [KR2, Prop 6.3.7] there is a central projection $z$ with $ze_1$ properly infinite (or $z=0$) and $(1-z)e_1$ finite (or $z=1$).
Suppose $z\not=1$.  By [KR2, Prop 6.2.3] as $z$ is central, $e_1\sim e_i \implies (1-z)e_1 \sim (1-z)e_i$, so by [KR2, Prop 6.3.2] also $(1-z)e_i$ is finite.  Hence the sum is also finite, [KR2, Thm 6.3.8] so $1-z = \sum (1-z)e_i$ is finite.  Finiteness passes to subprojections, [KR2, Prop 6.3.2], so $(1-z)f_i$ is finite.
To ease notation, we work in the von Neumann algebra $(1-z)M$.  So we may suppose that $M$ is finite, and each $e_i, f_i$ is finite.  I want to apply Comparison, but I find the statement of [Tak, Chapter V, Thm 1.8] a bit cleaner to apply.  There is a central projection $w$ with $we_1 \precsim wf_1$ and $(1-w)f_1 \precsim (1-w)e_1$.   By [KR2, Prop 6.2.3] $e_i\sim e_1 \implies we_i \sim we_1$ so $we_i \precsim we_1$; similarly $wf_1 \precsim wf_i$.  As $\precsim$ is a partial order, [KR2, Prop 6.2.5], $we_i \precsim wf_i$ for each $i$.  So for each $i$ there is a projection $g_i$ with
$ we_i \sim g_i \leq wf_i. $
The $(wf_i)$ are orthogonal so the $(g_i)$ are orthogonal.  Summing, [KR2, Prop 6.2.2], gives
$$ w = \sum we_i \sim \sum g_i \leq \sum wf_i = w. $$
As $M$ is finite, $w$ is finite, [KR2, Prop 6.3.2], and so necessarily $\sum g_i = w$, which implies that $g_i = wf_i$ for each $i$.  Hence $we_i \sim wf_i$ for each $i$.  By symmetry of the situation, as $(1-w)f_1 \precsim (1-w)e_1$, the same argument shows that $(1-w)f_i \sim (1-w)e_i$.  Summing shows that $e_i \sim f_i$.
If $z\not=0$, we now look at the von Neumann algebra $zM$; so without loss of generality suppose that $M$ is properly infinite, and that $e_1$ is properly infinite.  As $e_1\sim e_i$, also $e_i$ is properly infinite, [KR2, Prop 6.3.7].  By the lemma below, as $e_2 \sim e_1$ we have that $e_1 \sim e_1+e_2$, and so $e_1+e_2$ is properly infinite.  Then $e_3 \sim e_1 \sim e_1+e_2$, the lemma shows that $e_1 + e_2 \sim e_1 + e_2 + e_3$.  Continue to show that $e_1 \sim e_1 + e_2 \sim \cdots \sim \sum e_i = 1$.  If also $f_1$ is properly infinite, then the same argument shows that $f_1\sim 1$ and so $e_1\sim f_1$.  To show that $f_1$ is properly infinite, let $z_0$ be a central projection with $z_0f_1\not=0$.  If $z_0f_1$ is not infinite, it is finite, so $f_1\sim f_i\implies z_0f_1\sim z_0f_i$ so $z_0f_i$ is finite.  Hence $z_0 = \sum z_0f_i$ is finite.  Thus also $z_0e_i$ is finite for each $i$, but each $e_i$ is properly infinite, so $z_0e_i=0$ for each $i$, so $z_0=0$ contradiction.
So both $(1-z)e_1\sim (1-z)f_1$ and $ze_1\sim zf_1$ so summing gives $e_1\sim f_1$ as required.
Lemma: Let $e,f$ be orthogonal projections with $e\sim f$ and $e$ properly infinite.  Then $e \sim e+f$.

This is a special case of [KR2, Exercise 6.9.4].  By the ``Halving Lemma'' [KR2, Lem 6.3.3], there is $g\leq e$ with $e\sim g \sim e-g$.  Thus $f \sim e-g$.  Summing gives $e+f \sim g+f \sim g+(e-g) = e$.


Notice that the idea is clearly to split into finite and purely infinite parts.  However, I think that we cannot simplify things and "split $M$ into finite and purely infinite parts", but rather only split a projection, here $e_1$, and then work to show that really we have the same splitting for $f_1$.  (The example I have in mind is a sum of $B(H)$ factors.)
[KR2] : Kadison, Ringrose, Fundamentals of the theory of operator algebras. Vol. II
[Tak] : Takesaki, Theory of operator algebras I.
