Proof of minimal eigenvalues possible?

Question

On a German website I discovered a conjecture about a special square matrix. This conjecture contains a statement about the absolute value of the smallest eigenvalue with two formulas in respect to the number of matrix rows/columns (odd/even number).

This can be summarized as follows. Consider a class of $(n+1)\times (n+1)$ matrices with diagonal entries $b^{n} $ with $b\in \mathbb{R}$, $b>1$ and off-diagonal entries for each row the elements of the following set in some order: $$ \left\{ (b-1)b^{n-k} \mid k=1, \dots, n \right\} $$ These matrices have all row sums equal to $2b^{n}-1$ and therefore have the all-ones vector as a positive eigenvector with eigenvalue $2b^{n}-1$, which by Perron-Frobenius is the maximum eigenvalue. Consider a matrix in this class, $A_{\min }$, whose element arrangement leads to the smallest eigenvalue. This smallest eigenvalue is conjectured to be $$ \lambda _{\min }(n)=\left\{ \begin{array}{cc} (b+1)b^{n/2-1}-1, & n~\mathrm{even} \\ 2b^{(n-1)/2}-1, & n~\mathrm{odd} \end{array}\right. $$

My question is whether this assumption can be proven or can it be refuted? Using Gershgorin circles, it can be deduced that the smallest eigenvalue must always be greater than 1 ($\lambda_{\min.} \geq 1$). But the two formulas of the conjecture give the smallest eigenvalue directly, no matter how large the square matrix is. But I can’t find a way to contradict this. Is there any mathematical theorem that confirms or contradicts this hypothesis? Most papers provide statements about the first maximum eigenvalues.

This seems very absurd to me, because as $n$ increases (for larger matrices!!!), the predicted smallest eigenvalue grows exponentially with $n$, while $\lambda_{\min.} \geq 1$ in all cases by using the Gershgorin circles for the rows. Is anyone able to find a proof or method that this conjecture is correct for all $n$ or false for certain values of $n$? — I have no idea, and numerical calculations fail at the size of $n$.

Answer 1

This is a partial answer about generating the matrices with the conjectured minimum eigenvalues, but does not show that they are minimum for these matrices, or that there can't be other matrices with smaller eigenvalues.

Sufficient conditions for the conjectured $\lambda _{\min }$ to be an eigenvalue

For $n$ odd

We first define two sets: $U$ has the diagonal entry $b^{n}$ and the $(n-1)/2 $ smallest entries, and $V$ has the others. $$ U=\left\{ b^{n},(b-1),\ldots (b-1)b^{(n-1)/2-1}\right\} ,~V=\left\{ (b-1)b^{(n-1)/2},\ldots ,(b-1)b^{n-1}\right\}, $$ for which \begin{align} S_{U} &=\sum_{i\in U}i=b^{n}+b^{(n-1)/2}-1 \\ S_{V} &=\sum_{i\in V}i=b^{n}-b^{(n-1)/2} \\ S_{U}-S_{V} &=2b^{(n-1)/2}-1=\lambda _{\min } \end{align}

Let $\alpha $ and $\beta $ be two index sets, disjoint with union $\{1,\ldots ,n+1\}$ and $\left\vert \alpha \right\vert =\left\vert \beta \right\vert =(n+1)/2$. We construct a matrix with an eigenvector for $\lambda _{\min }$ having $\alpha $-indexed entries $+1$ and $\beta $-indexed entries $-1$. For a row in the matrix with row index in $\alpha $ the entries with column indices in $\alpha $ have the entries of $U$ in some order that has $b^{n}$ on the diagonal, and the entries with column indices in $\beta $ have the entries of $V$ in some order. Then multiplying this row with the eigenvector gives $$ S_{U}(+1)+S_{V}(-1)=\lambda _{\min }(+1) $$ which satisfies the eigenvalue/eigenvector requirement.

For a row in the matrix with row index in $\beta $ the entries with column indices in $\beta $ have the entries of $U$ in some order that has $b^{n}$ on the diagonal, and the entries with column indices in $\alpha $ have the entries of $V$ in some order. Then multiplying this row with the eigenvector gives $$ S_{V}(+1)+S_{U}(-1)=\lambda _{\min }(-1) $$ which also satisfies the requirement, and we have found an eigenvector for $\lambda_{\min}$. In other words, we could permute rows and columns to give $2\times 2$ block form with rows from $U$ in the diagonal blocks and rows from $V$ in the off-diagonal blocks; the eigenvector then has its first-half entries $+1$ and second-half entries $-1$ (or vice-versa, which is just a scalar multiple).

(There are some circulant matrices when $\alpha $ has odd indices and $\beta$ has even indices.)

For $n$ even

We define four sets: $W$ has the diagonal entry $b^{n}$ and the $n/2-1$ smallest entries, and $X$ has the $n/2+1$ others; $Y$ has the diagonal entry $b^{n}$ and the $n/2$ smallest entries, and $Z$ has the $n/2$ others: \begin{align} W &=\left\{ b^{n},(b-1),\ldots (b-1)b^{n/2-2}\right\} ,~X=\left\{ (b-1)b^{n/2-1},\ldots ,(b-1)b^{n-1}\right\} \\ Y &=\left\{ b^{n},(b-1),\ldots (b-1)b^{n/2-1}\right\} ,~Z=\left\{ (b-1)b^{n/2},\ldots ,(b-1)b^{n-1}\right\} \end{align} for which \begin{align} S_{W} &=\sum_{i\in W}i=b^{n}+b^{n/2-1}-1 \\ S_{X} &=\sum_{i\in X}i=b^{n}-b^{n/2-1} \\ S_{Y} &=\sum_{i\in Y}i=b^{n}+b^{n/2}-1 \\ S_{Z} &=\sum_{i\in Z}i=b^{n}-b^{n/2} \end{align}

Let $\alpha $ and $\beta $ be two index sets, disjoint with union $\{1,\ldots ,n+1\}$ and $\left\vert \alpha \right\vert =n/2$, $\left\vert \beta \right\vert =n/2+1$. We construct a matrix with an eigenvector for $\lambda _{\min }$ having $\alpha$-indexed entries $(b^{n}-b^{n/2-1})$ and $\beta$-indexed entries $-(b^{n}-b^{n/2})$. For a row in the matrix with row index in $\alpha $ the entries with column indices in $\alpha $ have the entries of $W$ in some order that has $b^{n}$ on the diagonal, and the entries with column indices in $\beta $ have the entries of $X$ in some order. Then multiplying this row with the eigenvector gives $$ S_{W}(b^{n}-b^{n/2-1})+S_{X}(-(b^{n}-b^{n/2}))=\lambda _{\min }(b^{n}-b^{n/2-1}) $$ which satisfies the requirement.

For a row in the matrix with row index in $\beta $ the entries with column indices in $\beta $ have the entries of $Y$ in some order that has $b^{n}$ on the diagonal, and the entries with column indices in $\alpha $ have the entries of $Z$ in some order. Then multiplying this row with the eigenvector gives $$ S_{Z}(b^{n}-b^{n/2-1})+S_{Y}(-(b^{n}-b^{n/2}))=\lambda _{\min }(-(b^{n}-b^{n/2})) $$ which also satisfies the requirement, and we have found an eigenvector for $\lambda_{\min}$.