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Let $\alpha, \beta, \gamma \in (0, 1)$ such that $\alpha < \beta, \gamma < \beta$. Let $d \in \mathbb{N}, d \geq 4$.

I am interested in bounding from below the spectral radius $\rho(K)$ of the following matrix $K$ of size $d$ (or even better determining it analytically).

$$K = \begin{pmatrix} \sqrt{(1 - \alpha)(1 - \gamma)} & \frac{\sqrt{\alpha \gamma}}{d-1} & \cdots & \cdots & \frac{\sqrt{\alpha \gamma}}{d-1} \\ \frac{\sqrt{\alpha \gamma}}{d-1} & \eta & \frac{\beta}{d-2} & \cdots & \frac{\beta}{d-2} \\ \frac{\sqrt{\alpha \gamma}}{d-1} & \frac{\beta}{d-2} & \eta & \ddots & \vdots \\ \vdots & \vdots & \ddots & \ddots & \frac{\beta}{d-2}\\ \frac{\sqrt{\alpha \gamma}}{d-1} & \frac{\beta}{d-2} & \cdots & \frac{\beta}{d-2} & \eta \\ \end{pmatrix}$$

where $\eta = \sqrt{\left(1 - \beta - \frac{\alpha}{d-1}\right)\left((1 - \beta - \frac{\gamma}{d-1}\right)}$

Notice, as it can only help, that $K$ is symmetric, and has strictly positive entries. So far, have only been able to say that the spectral radius lies (up to reordering of the bounds) in

$$\left[ \sqrt{(1 - \alpha)(1 - \gamma)} + \sqrt{\alpha\gamma}; \frac{\sqrt{\alpha \gamma}}{d-1}+ \eta + \beta \right]$$

since by positivity $\rho(K)$ stands in between the minimum and maximum row sums.

Numerically, though, it seems that I have the following dependency $\rho(K) = 1 - O\left( \frac{f(\alpha, \gamma)}{d} \right)$, and this sort of dependency in $d$ is my goal, but I am a little bit short on tools to confirm it.

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Let $u=(1,0,\ldots,0)^T,\ v=\frac1{\sqrt{d-1}}(0,1,\ldots,1)^T,\ p=\frac{\sqrt{\alpha\gamma}}{d-1},\ q=\frac\beta{d-2}$ and $r=\sqrt{(1-\alpha)(1-\gamma)}$. Then $$ K=(r-\eta+q)uu^T+\sqrt{d-1}\,p(uv^T+vu^T)+(d-1)qvv^T+(\eta-q)I. $$ So, $K-(\eta-q)I$ has rank $\le2$ and its restriction on $\operatorname{span}\{u,v\}$ has a matrix representation $$ A=\pmatrix{r-\eta+q&\sqrt{d-1}\,p\\ \sqrt{d-1}\,p&(d-1)q}. $$ When $d$ is large, $\eta-q>0$ and $A$ is entrywise positive. Hence \begin{align} \rho(K) &=\eta-q+\rho(A)\\ &=\eta-q+ \frac{(r-\eta+dq)+\sqrt{\left[r-\eta-(d-2)q\right]^2+4(d-1)p^2}}{2}\\ &=\frac{(r+\eta+\beta)+\sqrt{(r-\eta-\beta)^2+\frac{4\alpha\gamma}{d-1}}}{2}\\ \end{align} and it is up to you to find an asymptotic bound for it.

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  • $\begingroup$ After computation from your expression, an interesting lower bound then turns out to be $1 - \frac{\alpha + \gamma}{d-1}$. The steps are clear to me except how you determine the rank of $K$. If I manage to express a matrix using a sum of products of $k$ vectors, I get a $k$-rank matrix ? Does this arise directly from the definition of the rank ? $\endgroup$ – ippiki-ookami Aug 6 '18 at 7:46
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    $\begingroup$ @ippiki-ookami I should write "rank $\le2$" instead. Of course, the rank of the sum of $k$ rank-1 matrices is at most $k$, but strict inequality may hold. $\endgroup$ – user1551 Aug 6 '18 at 10:06

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