Positive radical, corank one, positive definiteness of proper submatrices, and domination exclusions

(The absolute-value inequality.) For $x={\sum }_s{x}_s{e}_s\in V$ put $q(x):=B(x,x)$ and $y:={\sum }_s\mathrm{\mid }{x}_s\mathrm{\mid }{e}_s$. Expanding in the basis $(e_s)$, $q(x)-q(y)=2{\sum }_{\{s,t\}\in \left(\genfrac{}{}{0px}{}{S}{2}\right)}B(e_s,e_t)(x_s{x}_t-\mathrm{\mid }{x}_s\mathrm{\mid }\mathrm{\mid }{x}_t\mathrm{\mid })$, where the sum runs over unordered pairs of distinct indices. Each summand is $\ge 0$ because $B(e_s,e_t)\le 0$ and $x_s{x}_t\le \mathrm{\mid }{x}_s{x}_t\mathrm{\mid }=\mathrm{\mid }{x}_s\mathrm{\mid }\mathrm{\mid }{x}_t\mathrm{\mid }$; hence $q(y)\le q(x)$ [F1].

(The radical of a positive semidefinite form.) If $q(v)=0$ then $v\in \mathrm{rad}(B)$: for every $w\in V$ and every $t\in \mathbb{R}$ one has $0\le q(v+tw)=2tB(v,w)+t^{2}q(w)$. If $q(w)>0$ then choosing $t=-B(v,w)\mathrm{/}q(w)$ gives $-B(v,w{)}^2\mathrm{/}q(w)\ge 0$, so $B(v,w)=0$; if $q(w)=0$ then $2tB(v,w)\ge 0$ for every $t$, so again $B(v,w)=0$.

(Propagation of zeros along the diagram.) Let $y={\sum }_s{y}_s{e}_s\in \mathrm{rad}(B)$ with $y_s\ge 0$ for all $s$, and suppose $y_i=0$ for some $i\in S$. Since $B$ is symmetric and $y$ is radical, $0=B(e_i,y)={\sum }_{t}B(e_i,e_t)y_t$ [F1, F2]. Each summand is $\le 0$: this follows from $y_t\ge 0$ and $B(e_i,e_t)\le 0$ for $t\mathrm{\ne }i$, while the $t=i$ term is $1\cdot 0=0$ [F1]. Hence every summand is $0$; in particular $y_j=0$ for every neighbour $j$ of $i$, because neighbours have $B(e_i,e_j)<0$ by [F1, F4, F7, F12]. Iterating along the connected diagram $\mathrm{\Gamma }$ [F11] gives $y_t=0$ for all $t\in S$.

(Full support of nonzero radical vectors.) Let $x={\sum }_s{x}_s{e}_s\in \mathrm{rad}(B)$, $x\mathrm{\ne }0$, and put $y:={\sum }_s\mathrm{\mid }{x}_s\mathrm{\mid }{e}_s$. By step 1.1, $q(y)\le q(x)=0$, and by positive semidefiniteness $q(y)\ge 0$, so $q(y)=0$ [F3]; by step 1.2, $y\in \mathrm{rad}(B)$, and $y\mathrm{\ne }0$ because $x\mathrm{\ne }0$. If some $x_i$ were $0$, then $y_i=0$ and step 1.3 would give $y=0$, a contradiction. Hence $x_s\mathrm{\ne }0$ for every $s\in S$.

(Existence, uniqueness and full span of the positive radical vector.) Replacing any nonzero $x\in \mathrm{rad}(B)$ by $y={\sum }_s\mathrm{\mid }{x}_s\mathrm{\mid }{e}_s$ gives a nonzero radical vector with nonnegative coordinates; step 1.3 shows every coordinate is positive. Thus the positive radical set is nonempty. If $x,x^{\mathrm{\prime }}$ both belong to it and were linearly independent, let $t_{\ast }:={\min }_s{x}_s\mathrm{/}{x}_s^{\mathrm{\prime }}>0$ and $w:=x-t_{\ast }{x}^{\mathrm{\prime }}$. Bilinearity shows $w\in \mathrm{rad}(B)$; some coordinate of $w$ is $0$, and $w\mathrm{\ne }0$, contradicting step 2.1. Hence any two positive radical vectors are positive scalar multiples. Fix one such vector $\delta$. For an arbitrary $z={\sum }_s{z}_s{e}_s\in \mathrm{rad}(B)$, if some $z_s<0$ put $M:={\sum }_{z_s<0}(-z_s\mathrm{/}{\delta }_s)>0$ and $\epsilon :=1\mathrm{/}(2M)$; otherwise put $\epsilon :=1$. In either case every coordinate of $\delta +\epsilon z$ is positive. Bilinearity puts this vector in $\mathrm{rad}(B)$, so the uniqueness just proved gives $\delta +\epsilon z=t\delta$ for some $t>0$. Therefore $z=((t-1)\mathrm{/}\epsilon )\delta$. This proves $\mathrm{rad}(B)=\mathbb{R}\delta$ and $\dim \mathrm{rad}(B)=1$ by [F8].

(Proper principal submatrices are positive definite.) Let $T⊊S$ and let $u\in {\mathbb{R}}^T$ be nonzero. Pad its coordinates by zero to obtain $\tilde{u}\in V$. Since $B$ is positive semidefinite, $B(\tilde{u},\tilde{u})\ge 0$. If equality held, step 1.2 would put $\tilde{u}$ in $\mathrm{rad}(B)$; it is nonzero and has a zero coordinate outside $T$, contradicting step 2.1. Hence $B(\tilde{u},\tilde{u})>0$ for every nonzero $u$, exactly positive definiteness of the principal submatrix. If $T=\mathrm{\varnothing }$, this condition is vacuous and the zero-dimensional form is positive definite by definition.

(Domination.) Let ${\mathrm{\Gamma }}^{\mathrm{\prime }}$, $T$ and $B_{{\mathrm{\Gamma }}^{\mathrm{\prime }}}$ be as in (3), and write $A^{\mathrm{\prime }}$ for its cosine matrix. Let $A$ be the matrix of $B$; after reordering the vertices so that $T$ comes first, $A^{\mathrm{\prime }}$ is indexed by the same first $\mathrm{\mid }T\mathrm{\mid }$ vertices. For an edge of ${\mathrm{\Gamma }}^{\mathrm{\prime }}$ with label $m^{\mathrm{\prime }}\le m$, the entry is $-\cos (\pi \mathrm{/}{m}^{\mathrm{\prime }})\ge -\cos (\pi \mathrm{/}m)$ by [F7]; if the labels differ, the inequality is strict, including the convention $\pi \mathrm{/}\mathrm{\infty }=0$. For a pair not joined in ${\mathrm{\Gamma }}^{\mathrm{\prime }}$, its entry is $0\ge A_{st}$. Thus $A_{st}\le A_{st}^{\mathrm{\prime }}\le 0$ for all distinct $s,t\in T$, while both diagonals are $1$ [F1, F4]. Suppose $A^{\mathrm{\prime }}$ is not positive definite. By [F3], some nonzero $x\in {\mathbb{R}}^T$ satisfies $x^{\mathsf{T}}{A}^{\mathrm{\prime }}x\le 0$. Pad $z:=(\mathrm{\mid }{x}_s\mathrm{\mid }{)}_{s\in T}$ by zero outside $T$. Then $0\le z^{\mathsf{T}}Az\le {\sum }_{s,t\in T}{A}_{st}^{\mathrm{\prime }}\mathrm{\mid }{x}_s\mathrm{\mid }\mathrm{\mid }{x}_t\mathrm{\mid }\le x^{\mathsf{T}}{A}^{\mathrm{\prime }}x\le 0$. The first inequality is positive semidefiniteness; the second follows termwise from $A_{st}\le A_{st}^{\mathrm{\prime }}$ and nonnegative coordinate products; the third follows termwise because $A_{st}^{\mathrm{\prime }}\le 0$ off the diagonal and $x_s{x}_t\le \mathrm{\mid }{x}_s\mathrm{\mid }\mathrm{\mid }{x}_t\mathrm{\mid }$. Equality throughout gives $B(z,z)=0$, so $z\in \mathrm{rad}(B)$ by step 1.2. Since $z\mathrm{\ne }0$, step 2.1 forces every coordinate of $z$ to be nonzero, hence $T=S$ and every $x_s\mathrm{\ne }0$. Equality in the second inequality then forces $A_{st}=A_{st}^{\mathrm{\prime }}$ for every distinct pair, so the strict monotonicity in [F7] gives the same edges and labels: ${\mathrm{\Gamma }}^{\mathrm{\prime }}={\mathrm{\Gamma }}_T$, contrary to the hypothesis. Therefore $A^{\mathrm{\prime }}$ is positive definite.

(The proper standard parabolics are finite.) Let $T⊊S$. If $T\mathrm{\ne }\mathrm{\varnothing }$, step 3.2 makes its restricted Coxeter form positive definite, and [F9] identifies $(W_T,T)$ as a Coxeter system with that restricted form; [F10] then gives that $W_T$ is finite. If $T=\mathrm{\varnothing }$, [F9] gives $W_T=\{1\}$, also finite.

(The exclusion consequence.) If $A^{\mathrm{\prime }}$ has a nonzero vector $v$ with $B_{{\mathrm{\Gamma }}^{\mathrm{\prime }}}(v,v)\le 0$, then it is not positive definite by definition [F3]. If $T\mathrm{\ne }\mathrm{\varnothing }$ and $\det A^{\mathrm{\prime }}\le 0$, then its last leading principal minor is nonpositive, so $A^{\mathrm{\prime }}$ is not positive definite by Sylvester's criterion [F5]; this also covers the statement's example that additionally assumes a proper principal submatrix is positive definite. By step 3.3, neither obstruction is compatible with strict domination.