I am stuck in the following exercise: (it is one step before describing the Murray-von Neumann equivalence classes in $M_n(\mathbb{C})$.) Here we say that $v$ is a partial isometry when $v=vv^*v$.
Let $p,q\in M_n(\mathbb{C})$ be projections that satisfy $\text{tr}(p)\leq\text{tr}(q)$ (tr denotes the trace function). Show that there exists a partial isometry $v\in M_n(\mathbb{C})$ such that $p=v^*v$ and $vv^*\leq q$.
My work is the following: By the spectral theorem, there exists a diagonal matrix $h$ and a unitary matrix $u$ such that $p=u^*hu$. Since $\text{spec}(p)=\text{spec}(h)=\text{the diagonal entries}$, $h$ has only $0$ and $1$ on the diagonal (and therefore $h=h^*=h^2$). Now we have $p=u^*hu=u^*h^2u=u^*h^*hu=(hu)^*hu$, so $hu$ is a partial isometry (since $(hu)^*hu$ is a projection). What we need to show is that $hu(hu)^*\leq q$. But $hu(hu)^*=huu^*h=h^2=h$, so we need to show that $h\leq q$. I have no idea how to use the trace condition to show this positivity; I know that since $h,q$ are projections, $h\leq q$ is equivalent to $qh=hq=h$. Any ideas?
I'm not sure if you argument is going well. What if $h=p$?
The usual way to do this is as follows:
First show that if $\operatorname{tr}(p)=\operatorname{tr}(q)$, then $p$ and $q$ are equivalent.
Second show that if $\operatorname{tr}(p)\leq \operatorname{tr}(q)$, then $q$ has a subprojection $q'$ with $\operatorname{tr}(q')=\operatorname{tr}(p)$.
Third, use the first two bullets to conclude.
To prove the first part, you look at the ranges of $p$ and $q$, and construct orthonormal bases with the same cardinality. The linear map sending one basis to the other is the partial isometry you need.
You can also prove the second part by looking at orthonormal bases.
Both can also be proven using the spectral theorem (then you deal with diagonals of ones and zeroes instead of orthonormal bases of subspaces). But when you want the infinite-dimensional case, the argument with subspaces goes through untouched.