Trace, dimension and equivalence in $M_n(\mathbb{C})$

Question

Let $tr: M_n(\mathbb{C}) \to \mathbb{C}$ be the standard trace given by

$tr \begin{pmatrix} x_{11} & x_{12} & x_{13} & \dots & x_{1n} \\ x_{21} & x_{22} & x_{23} & \dots & x_{2n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_{n1} & x_{n2} & x_{d3} & \dots & x_{nn} \end{pmatrix} = \sum_{j=1}^n x_{jj}$

And let p,q be projections in $ M_n(\mathbb{C})$. I want to show the following are equivalent:

(i) $p \sim q$ (where this is the Murray-Von Neumann equivalence)

(ii) Tr(p)=Tr(q)

(iii) $Dim(p(\mathbb{C}^n))=Dim(q(\mathbb{C}^n))$

Idea:

$(i) \to (ii)$

Assume $p \sim q$ the there exist $v \in M_n(\mathbb{C})$ such that $p=v^*v$ and $q=vv^*$. I have from my book that it also means that $v=qv=vp=qvp$. Thus:

$tr(p)=tr(p^*)=tr((v^*v)^*)=tr((v^*vp)^*)=tr((v^*vv^*v)^*)=tr(vv^*vv^*)=tr(qq)=tr(q)$

$(ii) \to (iii)$

This part I struggle more with. How does projection matrices, dimension of their range and the trace of a projection connect to each other?

$(iii) \to (i)$

Again I have some trouble as I am having a hard time seeing how the dimension of projective matrices is connected to the existence of a v such that $p=v^*v$ and $q=vv^*$?

Answer 1

$(ii)\rightarrow (iii)$

The projections in a von Neumann algebra are defined through $p=p^*=p^2$. The only eigenvalues of $p$ are $0$ and $1$. Therefore, $p$ is self-adjoint and can be diagonalized. That is, there exists $u\in M_n(\mathbb{C})$ unitary such that $upu^*=\mathrm{diag}(1,\dots,1_d,0,0,0,0,0)$, where $d$ is the dimension of its range. On the other hand, $d=\mathrm{Tr}(upu^*)=\mathrm{Tr}(pu^*u)=\mathrm{Tr}(p)$. Moreover, if $p\sim q$ holds, $$\mathrm{Tr}(p)=\mathrm{Tr}(v^*v)=\mathrm{Tr}(vv^*)=\mathrm{Tr}(q).$$

$(iii)\rightarrow (i)$

Since $\mathrm{Dim}(p(\mathbb{C}^n))=\mathrm{Dim}(q(\mathbb{C}^n))=d$, one can choose an orthonormal basis $\{x_k\}_{k=1}^d$ in $p(\mathbb{C}^n)$ and an orthonormal basis $\{y_k\}_{k=1}^d$ in $q(\mathbb{C}^n)$ and define $$v=\sum_{k=1}^d y_k\otimes x_k,$$ where $$(y_k \otimes x_k) f=(x_k,f)\,y_k,$$ $(,)$ being the usual inner product on $\mathbb{C}^n$. Show that $p=v^*v$ and $q=vv^*$ and you are ready.