# Bound on ratio between largest eigenvalues of a real symmetric matrix with positive entries

For $$p\in (0,1]$$, denote by $$M_n(p)$$ the set of real symmetric $$n\times n$$ matrices $$M$$ with positive entries, satisfying $$\min_{1\leq i,j,k,l\leq n}\frac{M_{ij}}{M_{kl}}=p.$$ Let $$\{\lambda_j(M)\}^n_{j=1}$$ denote eigenvalues of $$M$$ ordered by absolute value in decreasing order$$|\lambda_1(M)|\geq |\lambda_2(M)|\geq\ldots\geq|\lambda_n(M)|$$. Find $$f(p)\equiv\sup_{M\in M_n(p)}|\lambda_2(M)/\lambda_1(M)|,$$ or at least an upper bound for this quantity (expressed in terms of $$n,p$$).

Note: It is easy to see that $$f(1)=0$$, and $$\lim_{p\to 0} f(p)=1$$ (take the "nearly" identity matrix, for example). Also, consider the matrix $$M=\psi\psi^T$$, where $$\psi^T=(1,q,q^2,\ldots,q^{n-1}),~q=p^{\frac{1}{2n-2}}$$, so that $$M\in M_n(p)$$ and $$|\lambda_2(M)/\lambda_1(M)|=0$$, so the lower bound is trivially zero $$\inf_{M\in M_n(p)}|\lambda_2(M)/\lambda_1(M)|=0.$$ I'm curious if anyone can prove an upper bound that is smaller than 1.

• Perhaps you mean as $p \to 0$, $|\lambda_2/\lambda_1| \to 0$. Otherwise, the matrix $$\pmatrix{1&1\\1&p}$$ disproves this limit since we instead obtain $|\lambda_2/\lambda_1| = \frac 12 (3 + \sqrt{5})$. Commented May 24, 2020 at 22:26
• @Omnomnomnom Thanks. I just deleted that sentence and made the question shorter :) Commented May 24, 2020 at 22:45
• I meant to say as $p\to 1$, but that’s not right either Commented May 24, 2020 at 22:50
• Note that is possible for the matrix to be singular within the constraints of the problem (such as in the case where all matrix entries are identical), which means that there is no possible upper bound. Commented May 24, 2020 at 23:36
• @Omnomnomnom When all entries are identical, then $M$ would be proportional to a rank-1 projector, so $|\lambda_2/\lambda_1|=0$. In general I would agree that $\inf_{M\in M_n(p)} |\lambda_2(M)/\lambda_1(M)|=0$, but my question is about upper bound. Anyway I edited my question to make it more clear. Commented May 25, 2020 at 1:37

Note that the ratio of the two eigenvalues is invariant by scaling $$M$$ by a positive factor. Therefore we may modify the definition of $$M_n(p)$$ to assume that the largest element of every $$A\in M_n(p)$$ is $$1$$ and the smallest element is some $$p\in(0,1]$$. Let $$B$$ be the leading principal $$(n-1)\times(n-1)$$ submatrix of $$A$$. By Cauchy's interlacing inequality, the maximum possible value of $$\frac{\rho(B)}{\rho(A)}$$ serves as an upper bound of $$\frac{|\lambda|_2(A)}{|\lambda|_1(A)}$$.
Now, on one hand, since $$A$$ is nonnegative, for any given $$B$$, $$\rho(A)$$ is always minimised when the entries on the last row and the last column of $$A$$ are minimised, i.e. when the last row and the last column are filled with $$p$$s. On the other hand, when the last row and the last column are filled with $$p$$s, $$\rho(B)$$ is maximised when all entries of $$B$$ are equal to $$1$$.
Therefore, $$\rho(B)/\rho(A)$$ is maximised when $$A=\pmatrix{1&\cdots&1&p\\ \vdots&&\vdots&\vdots\\ 1&\cdots&1&\vdots\\ p&\cdots&\cdots&p} =pee^T+(1-p)vv^T,$$ where $$v=(1,\ldots,1,0)^T$$. For this $$A$$, we have $$\rho(B)=n-1$$ and $$\rho(A)=\frac{np+(n-1)q + \sqrt{\left[np-(n-1)q\right]^2 + 4(n-1)^2pq}}{2},$$ where $$q=1-p$$. It follows that $$\frac{|\lambda|_2(A)}{|\lambda|_1(A)} \le\frac{\rho(B)}{\rho(A)} \le\frac{2(n-1)}{np+(n-1)q + \sqrt{\left[np-(n-1)q\right]^2 + 4(n-1)^2pq}}.\tag{1}$$ Since the RHS of $$(1)$$ is the maximum possible value of $$\rho(B)/\rho(A)$$, it is always $$\le1$$. This bound is quite loose, however, because $$|\lambda|_2(A)$$ can be significantly smaller than $$\rho(B)$$. In particular, when $$p=1$$, we have $$A=ee^T$$ and hence $$|\lambda|_2(A)/|\lambda|_1(A)=0$$, but the upper bound we obtained in the above is $$(n-1)/n$$.
To compensate for the poor performance when $$p$$ is close to $$1$$, we give another upper bound. Let $$A=pee^T+D$$, so that $$D$$ is an entrywise nonnegative symmetric matrix whose maximum element is $$1-p$$. By Weyl's inequalities, \begin{aligned} \lambda_\min(A) &\ge\lambda_\min(pee^T)+\lambda_\min(D)\ge-\rho(D),\\ \lambda_2^\downarrow(A) &\le\lambda_2^\downarrow(pee^T)+\lambda_\max(D)\le\rho(D). \end{aligned} Since the second largest-sized eigenvalue of $$A$$ must lie between $$\lambda_\min(A)$$ and $$\lambda_2^\downarrow(pee^T)$$, the above two inequalities show that its absolute value must be bounded above by $$\rho(D)$$. Therefore $$\frac{|\lambda|_2(A)}{|\lambda|_1(A)} \le\frac{\rho(D)}{\rho(A)} \le\frac{nq}{np} =\frac{q}{p}.\tag{2}$$ So, from $$(1)$$ and $$(2)$$ we obtain $$\frac{|\lambda|_2(A)}{|\lambda|_1(A)} \le\min\left\{ \frac{2(n-1)}{np+(n-1)q + \sqrt{\left[np-(n-1)q\right]^2 + 4(n-1)^2pq}}, \frac{q}{p}\right\}.\tag{3}$$ The bound in $$(3)$$ is now sharp in the limiting case $$p=0$$ (with the bound being $$1$$, which is attained by $$A=I$$) and in also the case $$p=1$$ (with the bound being $$0$$, which is attained by the only member $$A=ee^T$$ of $$M_n(1)$$).