Proof of Comparison with the Zero Matrix in the Positive Semidefinite Ordering

Step 1 (Two entrywise identities). Two real $n\times n$ matrices are equal exactly when all their entries agree, so it suffices to compare entries. Let $i$ and $j$ be indices. By the definition of the difference of real matrices, applied twice,

$$\bigl(0_n-(A-B)\bigr)_{ij}=(0_n)_{ij}-(A-B)_{ij}=0-(A_{ij}-B_{ij}).$$

By claim 4 of Additive Cancellation and Elementary Additive Identities in a Field the right-hand side equals $-(A_{ij}-B_{ij})$, and by claim 6 of that lemma this in turn equals $B_{ij}-A_{ij}$, which is the entry $(B-A)_{ij}$. Since $i$ and $j$ were arbitrary,

$$0_n-(A-B)=B-A .$$

Similarly $\bigl((A-B)-0_n\bigr)_{ij}=(A-B)_{ij}-0=(A-B)_{ij}$ by claim 4 of that lemma, so

$$(A-B)-0_n=A-B .$$

Step 2 (Claim 1). Both $A-B$ and $0_n$ lie in $\mathcal{S}(n)$, so The Positive Semidefinite Ordering Compared by Differences applies to them and shows that $A-B\preceq 0_n$ holds if and only if $0_n-(A-B)$ is positive semidefinite. By Step 1 that matrix is $B-A$. Applying The Positive Semidefinite Ordering Compared by Differences to $A$ and $B$ instead, $B-A$ is positive semidefinite if and only if $A\preceq B$. Combining the two equivalences gives claim 1.

Step 3 (Claim 2). By The Positive Semidefinite Ordering Compared by Differences applied to $0_n$ and $A-B$, the relation $0_n\preceq A-B$ holds if and only if $(A-B)-0_n$ is positive semidefinite, and by Step 1 that matrix is $A-B$. Applying The Positive Semidefinite Ordering Compared by Differences to $B$ and $A$ instead, $A-B$ is positive semidefinite if and only if $B\preceq A$. Combining the two equivalences gives claim 2.