[proofplan]
We prove the two implications separately. If the corner is one-dimensional, then every subprojection of $p$ is a scalar multiple of $p$, and the projection equation forces the scalar to be $0$ or $1$. Conversely, assume $p$ is minimal and view $pMp$ as a von Neumann algebra with identity $p$. Spectral projections of any self-adjoint element of $pMp$ would give subprojections of $p$; minimality forces all such spectra to be singletons, so every self-adjoint element is scalar, and then every element is scalar by decomposing into real and imaginary parts.
[/proofplan]
[step:Use the one-dimensional corner condition to rule out proper subprojections]
Assume
\begin{align*}
pMp=\mathbb{C}p.
\end{align*}
Let $q \in M$ be a projection such that $q \le p$. Since $q \le p$ means $pq=q=qp$, we have
\begin{align*}
q=pqp \in pMp.
\end{align*}
Thus there exists $\lambda \in \mathbb{C}$ such that $q=\lambda p$. Since $q$ is a projection,
\begin{align*}
q^2=q.
\end{align*}
Substituting $q=\lambda p$ and using $p^2=p$ gives
\begin{align*}
\lambda^2p=\lambda p.
\end{align*}
Because $p\ne 0$, this implies $\lambda^2=\lambda$, so $\lambda \in \{0,1\}$. Therefore $q=0$ or $q=p$. Hence $p$ is minimal in $M$.
[guided]
Assume first that the corner is exactly the scalar multiples of $p$:
\begin{align*}
pMp=\mathbb{C}p.
\end{align*}
To prove minimality, we must show that no projection strictly between $0$ and $p$ can exist.
Let $q \in M$ be a projection satisfying $q \le p$. For projections, the order relation $q \le p$ is equivalent to the algebraic identities
\begin{align*}
pq=q
\end{align*}
and
\begin{align*}
qp=q.
\end{align*}
Therefore
\begin{align*}
q=pqp.
\end{align*}
This places $q$ inside the corner $pMp$. Since $pMp=\mathbb{C}p$, there is a scalar $\lambda \in \mathbb{C}$ such that
\begin{align*}
q=\lambda p.
\end{align*}
Now we use the fact that $q$ is a projection. The equation $q^2=q$ gives
\begin{align*}
(\lambda p)^2=\lambda p.
\end{align*}
Since $p^2=p$, this becomes
\begin{align*}
\lambda^2p=\lambda p.
\end{align*}
Equivalently,
\begin{align*}
(\lambda^2-\lambda)p=0.
\end{align*}
Because $p\ne 0$, the scalar multiplying $p$ must be zero:
\begin{align*}
\lambda^2-\lambda=0.
\end{align*}
Thus $\lambda \in \{0,1\}$. If $\lambda=0$, then $q=0$; if $\lambda=1$, then $q=p$. Hence the only projections below $p$ are $0$ and $p$, which is exactly minimality of $p$ in $M$.
[/guided]
[/step]
[step:Pass from minimality of $p$ to scalar self-adjoint elements in the corner]
Assume now that $p$ is minimal in $M$. The corner $pMp$ is a von Neumann algebra with identity $p$ under the inherited operations and represented on the [Hilbert space](/page/Hilbert%20Space) $pH$. Let $a \in pMp$ be self-adjoint.
We use the standard spectral-calculus fact for self-adjoint elements in a von Neumann algebra: if $a$ is self-adjoint in the von Neumann algebra $pMp$, then its spectral projections belong to $pMp$, and the spectrum of $a$ is the support of this spectral resolution. Suppose, for contradiction, that $\sigma_{pMp}(a)$ contains two distinct points. Choose $\lambda,\mu \in \sigma_{pMp}(a)$ with $\lambda<\mu$, and choose $t \in \mathbb{R}$ such that $\lambda<t<\mu$. Let
\begin{align*}
e:=\mathbb{1}_{(-\infty,t]}(a)
\end{align*}
be the spectral projection of $a$ in $pMp$ associated to the Borel set $(-\infty,t] \subseteq \mathbb{R}$. Then $e$ is a projection in $pMp$, hence $e \in M$, and because the identity of $pMp$ is $p$, it satisfies $e \le p$.
Since $\lambda$ and $\mu$ lie in the spectrum, the spectral support on both sides of $t$ is nonzero. Thus
\begin{align*}
e\ne 0
\end{align*}
and
\begin{align*}
p-e\ne 0.
\end{align*}
Hence $e$ is a nonzero projection in $M$ with $e<p$, contradicting minimality of $p$. Therefore $\sigma_{pMp}(a)$ is a singleton, say
\begin{align*}
\sigma_{pMp}(a)=\{\alpha\}
\end{align*}
for some $\alpha \in \mathbb{R}$.
Then $a-\alpha p$ is self-adjoint in $pMp$ and has spectrum $\{0\}$. By the spectral-radius formula for normal elements in a $C^*$-algebra,
\begin{align*}
\|a-\alpha p\|_{pMp}=0.
\end{align*}
Thus
\begin{align*}
a=\alpha p.
\end{align*}
So every self-adjoint element of $pMp$ is a real scalar multiple of $p$.
[/step]
[step:Decompose an arbitrary corner element into self-adjoint parts]
Let $x \in pMp$ be arbitrary. Define
\begin{align*}
b:=\frac{x+x^*}{2}
\end{align*}
and
\begin{align*}
c:=\frac{x-x^*}{2i}.
\end{align*}
Since $pMp$ is closed under addition, scalar multiplication, and adjoints, both $b$ and $c$ belong to $pMp$. Moreover,
\begin{align*}
b^*=b
\end{align*}
and
\begin{align*}
c^*=c.
\end{align*}
By the previous step, there exist $\alpha,\beta \in \mathbb{R}$ such that
\begin{align*}
b=\alpha p
\end{align*}
and
\begin{align*}
c=\beta p.
\end{align*}
Therefore
\begin{align*}
x=b+ic=(\alpha+i\beta)p \in \mathbb{C}p.
\end{align*}
This proves
\begin{align*}
pMp \subseteq \mathbb{C}p.
\end{align*}
The reverse inclusion is immediate because $p(\lambda p)p=\lambda p$ for every $\lambda \in \mathbb{C}$. Hence
\begin{align*}
pMp=\mathbb{C}p.
\end{align*}
Combining this with the first step proves the equivalence.
[/step]
