Proof of The Positive Semidefinite Ordering is a Partial Order Compatible with the Linear Structure

Each result cited is universally quantified over the data in its own statement.

We work in Routine Facts Used Without Citation: Arithmetic, Maps, Predicativity and Indices. By The Positive Semidefinite Ordering on Symmetric Matrices §order, for $P,Q\in\mathcal{S}(n)$, $P\preceq Q$ means that $z\cdot(Pz)\le z\cdot(Qz)$ for every $z\in\mathbb{R}^{n}$; by Positive Semidefinite and Positive Definite Real Matrices §psd, $P\in\mathcal{S}(n)$ is positive semidefinite when $z\cdot(Pz)\ge0$ for every $z\in\mathbb{R}^{n}$. Every matrix compared below lies in $\mathcal{S}(n)$ by Rules for Symmetric and Positive Semidefinite Matrices: Closure, Self-Adjointness, Polarization, Congruence and Cones §closure.

Quadratic forms of the operations. Let $P,Q\in\mathcal{M}_{n}(\mathbb{R})$, $t\in\mathbb{R}$ and $z\in\mathbb{R}^{n}$.

(a) By Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §matrix-vector, $(P+Q)z=Pz+Qz$, so by The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite §second

$$z\cdot((P+Q)z)=z\cdot(Pz)+z\cdot(Qz).$$

(b) By Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes §matrix-vector, $(tP)z=t(Pz)$, so by The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite §second

$$z\cdot((tP)z)=t\,\bigl(z\cdot(Pz)\bigr).$$

(c) For $i,j\in[n]$, by Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations §operations,

$$(P-Q)_{ij}=P_{ij}-Q_{ij}=P_{ij}+(-1)\,Q_{ij}=\bigl(P+(-1)Q\bigr)_{ij},$$

so $P-Q=P+(-1)Q$ by Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations §equality. By (a) and then (b) with $t=-1$,

$$z\cdot((P-Q)z)=z\cdot(Pz)+(-1)\bigl(z\cdot(Qz)\bigr)=z\cdot(Pz)-z\cdot(Qz).$$

(d) In (d) and in the clauses below, the $0$ in $0z$, in $0\preceq X$, in $X-Y\preceq0$ and in $0\preceq X-Y$ is the zero matrix of Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations §zero; every other $0$, including the scalar in $0P$ and the value $0$ of a dot product, is the real number zero. The zero matrix is the constant map with value $0$, by Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations §zero, so for $i,j\in[n]$ its $(i,j)$ entry is $0=0\cdot P_{ij}=(0P)_{ij}$ by Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations §operations; hence the zero matrix equals the scalar multiple $0P$ by Real Matrices: the Set of Real m x n Matrices with Their Entrywise Operations §equality. By (b) with $t=0$,

$$z\cdot(0z)=z\cdot((0P)z)=0\,\bigl(z\cdot(Pz)\bigr)=0.$$

Clause reflexive. For every $z\in\mathbb{R}^{n}$, $z\cdot(Xz)\le z\cdot(Xz)$; hence $X\preceq X$.

Clause transitive. If $X\preceq Y$ and $Y\preceq Z$, then for every $z\in\mathbb{R}^{n}$, $z\cdot(Xz)\le z\cdot(Yz)\le z\cdot(Zz)$; hence $X\preceq Z$.

Clause antisymmetric. If $X\preceq Y$ and $Y\preceq X$, then $z\cdot(Xz)\le z\cdot(Yz)$ and $z\cdot(Yz)\le z\cdot(Xz)$, so $z\cdot(Xz)=z\cdot(Yz)$, for every $z\in\mathbb{R}^{n}$. By Rules for Symmetric and Positive Semidefinite Matrices: Closure, Self-Adjointness, Polarization, Congruence and Cones §polarization with $A=X$ and $B=Y$, $X=Y$.

Clause translation. Let $z\in\mathbb{R}^{n}$. By (a), $z\cdot((X+Z)z)=z\cdot(Xz)+z\cdot(Zz)$ and $z\cdot((Y+Z)z)=z\cdot(Yz)+z\cdot(Zz)$. Adding or subtracting the real number $z\cdot(Zz)$ on both sides shows that $z\cdot(Xz)\le z\cdot(Yz)$ if and only if $z\cdot((X+Z)z)\le z\cdot((Y+Z)z)$. As this holds for every $z$, $X\preceq Y$ if and only if $X+Z\preceq Y+Z$.

Clause scaling. Assume $X\preceq Y$ and $\mu\ge0$, and let $z\in\mathbb{R}^{n}$. Then $z\cdot(Xz)\le z\cdot(Yz)$, and by (b)

$$z\cdot((\mu X)z)=\mu\,\bigl(z\cdot(Xz)\bigr)\le\mu\,\bigl(z\cdot(Yz)\bigr)=z\cdot((\mu Y)z).$$

Hence $\mu X\preceq\mu Y$.

Clause difference. Let $z\in\mathbb{R}^{n}$. By (c), $z\cdot((Y-X)z)=z\cdot(Yz)-z\cdot(Xz)$, which is $\ge0$ if and only if $z\cdot(Xz)\le z\cdot(Yz)$. As this holds for every $z$, $Y-X$ is positive semidefinite if and only if $X\preceq Y$.

Clause zero. Let $z\in\mathbb{R}^{n}$. By (d), $z\cdot(0z)=0$, so $z\cdot(0z)\le z\cdot(Xz)$ if and only if $z\cdot(Xz)\ge0$. As this holds for every $z$, $0\preceq X$ if and only if $X$ is positive semidefinite.

Clause difference-zero. Let $z\in\mathbb{R}^{n}$. By (d), $z\cdot(0z)=0$, and by (c), $z\cdot((X-Y)z)=z\cdot(Xz)-z\cdot(Yz)$. Hence $z\cdot((X-Y)z)\le z\cdot(0z)$ if and only if $z\cdot(Xz)\le z\cdot(Yz)$; and $z\cdot(0z)\le z\cdot((X-Y)z)$ if and only if $z\cdot(Yz)\le z\cdot(Xz)$. As these hold for every $z$, $X-Y\preceq0$ if and only if $X\preceq Y$, and $0\preceq X-Y$ if and only if $Y\preceq X$.