Let $A$ be a unital $C^*$-algebra and $a$ be a positive element in $A$ such that $\|a\|\leq 1$,show that
$p =\left [ \begin{matrix} a & (a-a^2)^{\frac{1}{2}} \\ (a-a^2)^{\frac{1}{2}} & 1-a\\ \end{matrix} \right ]$
is equivalent to $q =\left [ \begin{matrix} 1 & 0 \\ 0 & 0 \\ \end{matrix} \right ].$
My thought :I want to prove that $\|p-q\|<1$,then $p$ is homotopic to $q$ and so $p$ is equivalent to $q$.But I can not show that $\|p-q\|\neq 1$.
Trying to show that $\|p-q\|<1$ won't work, for if $a=0$, we have $p=\begin{bmatrix}0&0\\0&1\end{bmatrix}$, and $\|p-q\|=1$.
Instead, try showing that that the path of projections $(p_t)_{t\in[0,1]}$ given by $$p_t=\begin{bmatrix} t+(1-t)a&((t+(1-t)a)-(t+(1-t)a)^2)^\frac12\\ ((t+(1-t)a)-(t+(1-t)a)^2)^\frac12 &t(1-a) \end{bmatrix}$$ is continuous, for then you have $p\sim_h q$. To show that this path of projections is continuous, it suffices to show that each matrix coefficient is continuous. The only not entirely trivial part of this is showing that the off-diagonal entries are continuous, but this follows from the fact that $t+(1-t)a$ lies in $C^*(1,a)$, is positive, and $\sigma(t+(1-t)a)\subset[0,1]$ for each $t\in[0,1]$.