[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 $M$.[/step]

[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: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]