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Let $A$ be a $*$-algebra.

Let $P(A)=\bigcup_{n=1}^\infty\{p\in M_n(A):\text {$p$ is a projection} \}$. By projection I mean $p=p^*=p^2$.

Define the an equivalence relation on $P(A)$ by $p \sim q \Leftrightarrow $ there is a rectangle matrix $u$ with entries in $A$ such that $p=u^*u$ and $q=uu^*$.

I wonder why this definition coincides with the rank equivalence in the linear algebra, i.e if $A=\mathbb C$, then $p \sim q$ if and only if they have the same rank.

I think the "only if" part is relatively easier, and the answer follows from here. But I have trouble this the other direction: given two rectangle matrices of the same rank, why can they be expressed into $p=u^*u$ and $q=uu^*$? It should follow from a theorem in linear algebra, but what is it? Do we need the hypothesis that $p,q$ are projections?

The source of this question is the Example 7.1.1(page 220) of Murphy's C-algebras and operator theory*.

Thanks in advance!

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    When you write "$-$algebra", with the hyphen inside the math environment, it looks like a minus sign instead of a hyphen. Contrast these: $$ -\text{algebra} \qquad \text{versus}\qquad *\text{-algebra} $$ – Michael Hardy Apr 08 '16 at 16:39
  • Are you sure that this is true? Have you tried some simple examples, say $P$ and $Q$ are projections onto the $x$ and $y$ axis in a 2-dimensional complex vector space? – John Martin Apr 08 '16 at 17:00
  • One thing that you know is if $P:W\to W$ is a projection with $Im(P) = U$ and $Ker(P) = V$ then $W = U \oplus V$ and then you can write $P = AA^$ where $A$ is the inclusion $U \hookrightarrow W$. But I don't see why if $Q$ is another matrix with the same rank it has to be the case that $Q = A^A$... – John Martin Apr 08 '16 at 17:12
  • @JohnMartin For the source of this question, please see Murphy's C*-algebras and operator theory, Example 7.1.1 on page 220 – No One Apr 08 '16 at 18:04

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I owe part of following proof to one of my professors.

Assume rank$(p)$=rank$(q)=k \le$ min$(m,n)$.

Since $p,q$ are projections, there exist unitary matrices $u,v$, such that

$$p=u^* \begin{pmatrix}I_k&O\\O&O\end{pmatrix}_{m\times m} u,$$ and that $$ q=v^* \begin{pmatrix}I_k&O\\O&O\end{pmatrix}_{n\times n} v . $$

Let $w=\begin{pmatrix}I_k&O\\O&O\end{pmatrix}_{m\times n}$, then it follows that $$\begin{pmatrix}I_k&O\\O&O\end{pmatrix}_{m\times m}=ww^*$$

and that

$$\begin{pmatrix}I_k&O\\O&O\end{pmatrix}_{n \times n}=w^*w.$$

Hence, $$p=u^*ww^*u \sim w^*uu^*w=w^*w \sim ww^*=wvv^*w^* \sim v^*w^*wv=q$$

So $p\sim q$ by the transitivity of $\sim$. And I used the condition that $p,q$ are projections.

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