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Let $A_{n,k},k=1,\ldots,n$ be a sequence of $n\times n$ upper triangular matrices where $A_{n,1}=I_n$ and $A_{n,k},\quad 2\leq k\leq n$ be a regularly shifted and scaled matrix, with $P_{n,k}$ an $n\times n$ matrix with $0,1$ entries to be specified below. An example illustrating this sequence for $n=7$ is given at the end of the question.

Now define $B_n=A_{n,1}+\cdots+A_{n,n}$ and note that one always obtains an upper triangular matrix with this process.

Informally, initialize the matrix $B_n=[b_{i,j}]$ as the identity. Let $k$ range from $2$ to $n$ and increment each entry $b_{vk-f,vk}$ by $1/k$ for $f=0,\ldots,k-1.$ Do this for $v=1,2,\ldots,\lfloor n/k \rfloor.$

For the case $n=7,$ one obtains the matrix as the sum of the matrices at the end of this question: $$ B_{7}=\begin{bmatrix} 1 & 1/2 &1/3 & 1/4 & 1/5 &1/6 &1/7 \\ 0 & 3/2 &1/3 & 1/4 &1/5 &1/6 &1/7 \\ 0 & 0 & 4/3 & 3/4 &1/5 &1/6 &1/7 \\ 0 & 0 &0 & 7/4 &1/5 &1/2 &1/7 \\ 0 & 0 &0 & 0 &6/5 &1 &1/7 \\ 0 & 0 &0 & 0 &0 &2 & 1/7 \\ 0 & 0 &0 & 0 &0 & 0 & 8/7 \end{bmatrix}. $$ whose eigenvalues are on the diagonal. The nonzero entries of $B_n$ which occur for $j\leq i,$ are given by $$ B_{i,j}= \sum_{d|j} \mathbb{1}\left\{j-i +1 \leq d \right\} d^{-1} \qquad\qquad (1) $$ In fact $d=j$ can also be included giving a single expression for $b_{i,j}$ above. It is also clear that the matrix $B_n$ matches the matrix $B_{n-1,n-1}$ in its upper left $(n-1)\times (n-1)$ submatrix which provides a nice "bordered" recurrence to obtain $B_n$ from $B_{n-1}$ by augmentation. Hence

$$ B_{6}=\begin{bmatrix} 1 & 1/2 &1/3 & 1/4 & 1/5 &1/6 \\ 0 & 3/2 &1/3 & 1/4 &1/5 &1/6 \\ 0 & 0 & 4/3 & 3/4 &1/5 &1/6 \\ 0 & 0 &0 & 7/4 &1/5 &1/2 \\ 0 & 0 &0 & 0 &6/5 &1 \\ 0 & 0 &0 & 0 &0 &2 \end{bmatrix}. $$

Clearly all $B_n$ are positive definite with a unique minimum eigenvalue $1.$ All the entries at and above the main diagonals are nonzero but typically somewhat small.

Question: What can be said about eigenvalues (especially lower bounds) and eigenvectors of the sequence of matrices $C_n:=B_n^T B_n$?

Does the bordered recurrence help in determining bounds on the eigenvalues of $C_n$? Any references to similar problems [structurally w.r.t. the recurrence even if not number theoretic] is appreciated.

Example for $n=7$:

$$A_{7,1}=I_7,$$

$$A_{7,2}=\begin{bmatrix} 0 & 1/2 &0 & 0 &0 & 0 & 0\\ 0 & 1/2 &0 & 0 &0 & 0 & 0\\ 0 & 0 &0 & 1/2 &0 & 0 & 0\\ 0 & 0 &0 & 1/2 &0 & 0 & 0\\ 0 & 0 &0 & 0 &0 & 1/2 & 0\\ 0 & 0 &0 & 0 &0 & 1/2 & 0\\ 0 & 0 &0 & 0 &0 & 0 & 0 \end{bmatrix},$$

$$A_{7,3}=\begin{bmatrix} 0 & 0 &1/3 & 0 &0 & 0 & 0\\ 0 & 0 &1/3 & 0 &0 & 0 & 0\\ 0 & 0 &1/3 & 0 &0 & 0 & 0\\ 0 & 0 &0 & 0 &0 & 1/3 & 0\\ 0 & 0 &0 & 0 &0 & 1/3 & 0\\ 0 & 0 &0 & 0 &0 & 1/3 & 0\\ 0 & 0 &0 & 0 &0 & 0 & 0 \end{bmatrix},$$

$$A_{7,4}=\begin{bmatrix} 0 & 0 &0 & 1/4 &0 & 0 & 0\\ 0 & 0 &0 & 1/4 &0 & 0 & 0\\ 0 & 0 &0 & 1/4 &0 & 0 & 0\\ 0 & 0 &0 & 1/4 &0 & 0 & 0\\ 0 & 0 &0 & 0 &0 & 0 & 0\\ 0 & 0 &0 & 0 &0 & 0 & 0\\ 0 & 0 &0 & 0 &0 & 0 & 0 \end{bmatrix},$$

$$A_{7,5}=\begin{bmatrix} 0 & 0 &0 & 0 &1/5 & 0 & 0\\ 0 & 0 &0 & 0 &1/5 & 0 & 0\\ 0 & 0 &0 & 0 &1/5 & 0 & 0\\ 0 & 0 &0 & 0 &1/5 & 0 & 0\\ 0 & 0 &0 & 0 &1/5 & 0 & 0\\ 0 & 0 &0 & 0 &0 & 0 & 0\\ 0 & 0 &0 & 0 &0 & 0 & 0 \end{bmatrix},$$

$$A_{7,6}=\begin{bmatrix} 0 & 0 &0 & 0& 0 &1/6 & 0 \\ 0 & 0 &0 & 0 &0 &1/6 & 0 \\ 0 & 0 &0 & 0 &0 &1/6 & 0 \\ 0 & 0 &0 & 0 &0 &1/6 & 0 \\ 0 & 0 &0 & 0 &0 &1/6 & 0 \\ 0 & 0 &0 & 0 &0 & 1/6 & 0\\ 0 & 0 &0 & 0 &0 & 0 & 0 \end{bmatrix},$$

and

$$A_{7,7}=\begin{bmatrix} 0 & 0 &0 & 0& 0 &0 &1/7 \\ 0 & 0 &0 & 0 &0 &0 &1/7 \\ 0 & 0 &0 & 0 &0 &0 &1/7 \\ 0 & 0 &0 & 0 &0 &0 &1/7 \\ 0 & 0 &0 & 0 &0 &0 &1/7 \\ 0 & 0 &0 & 0 &0 &0 & 1/7 \\ 0 & 0 &0 & 0 &0 & 0 & 1/7 \end{bmatrix}.$$

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  • $\begingroup$ You won't go far without an acceptable expression (and a motivation) for $B_{i,j}$ and $(B^T B)_{i,j}$. The small eigenvalues of matrices involving $1_{i | j}$ often encode (something related to) the Riemann hypothesis. $\endgroup$ – reuns Dec 2 '17 at 7:17
  • $\begingroup$ @reuns Interesting that you think it might be related to Riemann. I'll write more. Itake it my expression for $B_{i,j}$ is wrong in your opinion. $\endgroup$ – kodlu Dec 2 '17 at 19:39
  • $\begingroup$ @reuns I fixed the expression for $B_{i,j}$ thanks for catching the mistake. $\endgroup$ – kodlu Dec 3 '17 at 16:43
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    $\begingroup$ What is this particular matrix supposed to help for ? $\endgroup$ – reuns Dec 3 '17 at 17:25

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