identity matrices stably equivalent?

Question

I'm reading the K-theory chapter in Murphy's C*-algebras and Operator Theory book. He defines stable equivalence to be $p \approx q$ if there exists positive integer $n$ such that $$ \begin{pmatrix}1_n & 0 \\ 0 & p \end{pmatrix} \sim \begin{pmatrix}1_n & 0 \\ 0 & q \end{pmatrix} $$ where $\sim$ is von Neumann equivalence (there exists partial isometry $u$ such that $u^*u = p, uu^*=q$). He says it is easy to check that $\approx$ is an equivalence relation. I want to check transitivity, so I assume that $p \approx q$ and $q \approx r$ so that $$ \begin{pmatrix}1_n & 0 \\ 0 & p \end{pmatrix} \sim \begin{pmatrix}1_n & 0 \\ 0 & q \end{pmatrix} \quad \text{and} \quad \begin{pmatrix}1_m & 0 \\ 0 & q \end{pmatrix} \sim \begin{pmatrix}1_m & 0 \\ 0 & r \end{pmatrix} $$ for some positive integers $n,m$. Writing this as $1_n \oplus p \sim 1_n \oplus q$ and $1_m \oplus q \sim 1_m \oplus r$, I want to show that $1_n \oplus q \sim 1_m \oplus q$. It would suffice to show $1_n \oplus 1_m$, since $1_n \sim 1_m$ and $q \sim q$ imply $1_n \oplus q \sim 1_m \oplus q$ by his Thm. 7.1.1, but I haven't been able to show this despite fiddling around with potential choices for $u$. I have a feeling I am missing something extremely obvious. I had shown transitivity of $\sim$ by taking the partial isometries $u,v$ involved in $p \sim q$ and $q \sim r$ and letting $w = vu$ be the one to show $p \sim r$, but I knew in that case that the dimensions were compatible to make $vu$. In this case if $n \neq m$ I don't see that as helping. Thanks in advance for any help/hints!

Answer 1

Suppose without loss of generality that $n<m$. From $$\tag1 \begin{bmatrix}1_n & 0 \\ 0 & p \end{bmatrix} \sim \begin{bmatrix}1_n & 0 \\ 0 & q \end{bmatrix} $$ you immediately get that $$\tag2 \begin{bmatrix}1_m & 0 \\ 0 & p \end{bmatrix} \sim \begin{bmatrix}1_m & 0 \\ 0 & q \end{bmatrix}. $$ Indeed, if $u$ is the partial isometry that gives you $(1)$, then $(2)$ is obtained with $$ u'=\begin{bmatrix} 1_{m-n}&0\\0& u\end{bmatrix}, $$ since $$ \begin{bmatrix} 1_m&0\\0&p\end{bmatrix}=\begin{bmatrix} 1_{m-n}&0&0\\0&1_n&0\\0&0&p\end{bmatrix}. $$ Now the transitivity is simply the usual transitivity of Murray-von Neumann equivalence.