An inequality problem for certain positive-definite matrices

Question

Suppose that $G=(G_{ij})$ is an $n\times n$ positive-definite symmetric matrix with the diagonal entries all equal $1$ and all off-diagonal entries $<0$. Let $a$ be a column $n\times1$ matrix such that $a^\top a=1$. Let $b$ be the column $n\times1$ matrix with entries $b_i:=|(G^{1/2}a)_i|$ and suppose that $(Gb)_i>0$ for all $i$.

Does it then necessarily follow that $b^\top Gb<1$?

Certain numerical experiments suggest that this is true.

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This question is related to two recent questions: this and this.

Answer 1

This is a partial answer for the case where $v\overset{\Delta}=G^{1/2}a \in\mathbb{R}^n_{+}\cup \mathbb{R}^n_{-}$.

Define $\mathcal{S}_G\overset{\Delta}=\left\{v\in\mathbb{R}^n\,:\,(v^\top G^{-1} v=1) \wedge (G\left|v\right|>0)\right\}$.

We can rephrase the question as follows.

Question. Is it true that if $v\in \mathcal{S}_G$, then $\left|v\right|^{\top} G \left|v\right|<1$?

Remark 1. For $n=1$, this is false. For $n=2$, the answer is yes: one can show that $|v|^{\top} G |v|\leq 1-\alpha^2$ whenever $v\in \mathcal{S}_G$, where $-1<\alpha<0$ is the off-diagonal of $G$.

For $n\geq 3$, we have the following.

Claim. If $v\in\left(\mathbb{R}^n_{+}\cup \mathbb{R}^n_{-}\right)\cap \mathcal{S}_G$, then $\left|v\right|^{\top} G \left|v\right|<1$.

Proof. Let $G=I-\overline{G}$. Remark that

$$\left|v\right|^{\top} G \left|v\right|=v^{\top}v-\left|v\right|^{\top}\overline{G}\left|v\right|.\,\,\,(\star)$$

Since $G\succ 0$ and $\overline{G}\geq 0$, then $-1 \prec \overline{G}\prec 1$ (the lemma below renders this step clearer) and we can write $$G^{-1}=I+\sum_{i=1}^\infty \overline{G}^i,$$ which implies $v^\top G^{-1}v=v^{\top}v + v^{\top} \left(\sum_{i=1}^\infty \overline{G}^i\right) v$. Since $v^\top G^{-1}v=1$, we have

$$v^{\top}v = 1- v^{\top} \left(\sum_{i=1}^\infty \overline{G}^i\right) v.\,\,\,(\star\star)$$

In view of equations $(\star)$ and $(\star \star)$, we have

$$\left|v\right|^\top G \left|v\right|<1 \Longleftrightarrow \underbrace{\left|v\right|^\top \overline{G} \left|v\right|}_{>0}+v^{\top} \left(\underbrace{\sum_{i=1}^ \infty\overline{G}^i}_{\geq 0}\right) v>0.$$

Observe that $G\left|v\right|>0\Longrightarrow \left|v\right|>0$ and thus, $\left|v\right|^{\top} \overline{G} \left|v\right|>0$ necessarily.

Therefore, $\left|v\right|^{\top} G \left|v\right|<1$ holds when $v\in\left(\mathbb{R}^n_{+}\cup \mathbb{R}^n_{-}\right)\cap \mathcal{S}_G$. $\,\,\,\,\,\square$

Remark 2. It remains to prove/disprove the claim over the remaining cones/orthants (other than $\mathbb{R}^n_{+}$ or $\mathbb{R}^n_{-}$) in the case $n\geq 3$.

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Lemma. We have $-1\prec \overline{G}\prec 1$.

Proof. That $\overline{G}\prec 1$ is clear. Assume that $\lambda_{\min}(\overline{G})\leq-1$. Then, there is $v\in \mathbb{R}^n$ with $\left\|v\right\|=1$ so that $$v^{\top} \overline{G} v = \sum_{i\sim j} |v_i| |v_j| \overline{G}_{ij}-\sum_{i\nsim j} |v_i| |v_j| \overline{G}_{ij}\leq-1,$$

where $i\sim j$ means that $v_iv_j\geq 0$ and $i\nsim j$ means otherwise $v_iv_j< 0$. That is,

$$\sum_{i\nsim j} |v_i| |v_j| \overline{G}_{ij}\geq 1+\sum_{i\sim j} |v_i| |v_j| \overline{G}_{ij}\geq 1,$$

which implies that we can construct a vector $\widetilde{v}$ with $\|\widetilde{v}\|\leq 1$ so that $\widetilde{v}^{\top} \overline{G} \widetilde{v}\geq 1$, contradicting $\overline{G}\prec 1$.

Answer 2

This question is, essentially, a restatement of the previous question. Now that the previous question is answered, the question on this page can be fully answered as well.

Indeed, let \begin{equation} W:=\sum_i b_i u_i, \end{equation} where $u_i$ is the $i$th column of the matrix $G^{1/2}$. Then \begin{equation} W^\top W=b^\top Gb>0, \end{equation} because the matrix $G$ is positive definite and $(Gb)_i>0$ for all $i$, so that $b\ne0$. So, we can introduce the unit vector \begin{equation} w:=\frac W{(W^\top W)^{1/2}}=\frac W{(b^\top Gb)^{1/2}}, \end{equation} and then we will have $u_j^\top W=(Gb)_j>0$ and hence $u_j^\top w>0$ for all $j$.

Therefore, by the mentioned answer, \begin{equation} S:=\sum_i |u_i^\top a| u_i^\top w<1. \end{equation} On the other hand, \begin{equation} S=\frac1{(b^\top Gb)^{1/2}}\sum_j |u_j^\top a| u_j^\top W =\frac1{(b^\top Gb)^{1/2}}\sum_{j,i} b_j b_i G_{ij}=(b^\top Gb)^{1/2}. \end{equation} Thus, $b^\top Gb<1$. $\quad\Box$