Matrix $p\in M_2(\mathbb C)$ such that $p^2=\bar{p}^t=p$ and $\mathrm{rank}(p)=1$

Question

Let $p= \begin{pmatrix} x & y \\ z & v \end{pmatrix}\in M_2(\mathbb{C})$ such that $p^2=\overline{p}^t=p$ and $\text{rank}(p)=1$.

Why is $p=\begin{pmatrix} t & l\sqrt{t(1-t)} \\ \overline{l}\sqrt{t(1-t)} & 1-t \end{pmatrix}$ for $l\in S^1$ and $t\in [0,1]$?

I'm stuck with my calculations.

Because $p^2=\overline{p}^t=p$ I obtain:

1. $x=\overline{x}=x^2+yz$

2. $v=\overline{v}=v^2+yz$

3. $y=\overline{z}=xy+yv$

4. $z=\overline{y}=zx+vz$.

I.e. $x,v\in\mathbb{R}$ and $y=\overline{z}$. Furthermore we have $\text{rank}(p)=\text{trace}(p)=1\Longrightarrow x+v=1$. So we set $x=t\in\mathbb{R}$ and we get $v=1-t$.

Therefore we obtain: $t=t^2+yz$ and $1-t=(1-t)^2+yz$ but I now stuck with my calculations to determine $y$ and $z$ as above. Can you help me?

Answer 1

Hint: Since $y=\overline{z}$, they must both have the same modulus i.e. $y=rl$ and $z=r\overline{l}$ with $r\in \mathbb{R}$ and $l\in S^1$. But both of the last two equations you wrote out rearrange to $yz=t(1-t)$.

Answer 2

$yz=y{\bar y}=t(t-1)$, then $t\in [0,1]$ , so in $\Bbb{C}$ the solution of $y$ is $l\sqrt{t(1-t)}$ where $|l|=1$.