Self-adjoint order, norm bounds and diagonal tests

A Hermitian matrix $B$ is normal because $B{B}^{\ast }=B^{\ast }B$. By [F2] and the convention transfer in [F1], it has an orthonormal eigenbasis $u_1,\dots ,u_n$. Each eigenvalue ${\lambda }_j$ is real: ${\lambda }_j=⟨u_j,B{u}_j⟩$ is real by [F1]. For $f={\sum }_j{c}_j{u}_j$, orthonormality gives $⟨f,Bf⟩={\sum }_j{\lambda }_j\mathrm{\mid }{c}_j{\mathrm{\mid }}^2$ and $\mathrm{\parallel }Bf{\mathrm{\parallel }}_2^2={\sum }_j{\lambda }_j^2\mathrm{\mid }{c}_j{\mathrm{\mid }}^2$. Thus, when $n>0$, [F3] gives $\mathrm{\parallel }B\mathrm{\parallel }={\max }_j\mathrm{\mid }{\lambda }_j\mathrm{\mid }$: the upper bound follows from the latter sum for unit $f$, and an eigenvector attaining the maximum gives equality.

For each $x\in V$, let $e_x$ be the function equal to $1$ at $x$ and $0$ elsewhere. By [F1], $⟨e_x,e_x⟩=1$ and $⟨e_x,B{e}_x⟩=B_{xx}$. Testing $B-aI$ and $bI-B$ at $e_x$ proves $a\le B_{xx}\le b$.

If $B⪯C$ and $D⪯E$, their defining quadratic forms are nonnegative, and their sum is the quadratic form of $(C+E)-(B+D)$; this proves addition. Restriction follows by testing the original inequality only on vectors $w\in W$.

The vertex space is finite-dimensional by [F1], so [F4] applies to every subspace $W$, with the same orthogonality in the article convention. For $z,t\in {\mathbb{C}}^V$ and scalars $\alpha ,\beta$, the vector $\alpha P_{W}z+\beta P_{W}t$ belongs to $W$, and its difference from $\alpha z+\beta t$ is orthogonal to $W$: pairing with any $w\in W$ gives a linear combination of two zero pairings. Uniqueness in [F4] therefore gives $P_W(\alpha z+\beta t)=\alpha P_{W}z+\beta P_{W}t$, so $P_W$ is linear. For $w\in W$, the orthogonality of $z-P_{W}z$ gives $⟨w,P_{W}z⟩=⟨w,z⟩$.

For real symmetric $B$, [F1] makes $B$ Hermitian. By step 1.1, $\mathrm{\parallel }B\mathrm{\parallel }\le b$ is equivalent, for $n>0$, to $-b\le {\lambda }_j\le b$ for all $j$. The displayed quadratic-form formula in step 1.1 shows that these inequalities imply $⟨f,(B+bI)f⟩\ge 0$ and $⟨f,(bI-B)f⟩\ge 0$ for every $f$. Conversely, testing those two forms at each $u_j$ gives $-b\le {\lambda }_j\le b$. Hence the two matrix-order inequalities are equivalent to the norm bound. If $V$ is empty, the norm is zero by [F3] and both quadratic-form inequalities hold on the zero space, so the equivalence also holds.

For Hermitian $B$ and $u,v\in W$, step 1.4 and conjugate symmetry give $⟨u,P_{W}Bv⟩=⟨u,Bv⟩=⟨Bu,v⟩=⟨P_{W}Bu,v⟩$. Thus the linear compression $w\mapsto P_{W}Bw$ is self-adjoint on $W$. Therefore $⟨w,P_{W}Bw⟩=⟨w,Bw⟩$, and the same holds for $C$. The order of the compressions is consequently equivalent to the restricted quadratic-form inequality in step 1.3, including $W=\{0\}$. These arguments involve only matrices and the stated inner product, so no graph regularity assumption is used. This proves all assertions.