The $n×n$ matrix $A_n$ is defined by the elements $a_{ij}=n−|i−j|$. \begin{bmatrix} n & n-1 & n-2 & ... & 1\\ n-1 & n & n-1 & ... & 2\\ n-2 & n-1 & n & ... & 3\\ ... & ... & ... & ... & ...\\ 1 & 2 & 3 & ... & n \end{bmatrix}

The maximal eigenvalue of this Toeplitz matrix seems to be proportional to $n^2$.
Could someone please show me how to write the exact expression of the following limit value?

$\lim_{n \to \infty} \frac{\rho(A)}{n^2}$

Here is my attempt:

Gerschgorin theorem

$\rho(A) \le n+\sum_{i=\left \lceil \frac{n}{2} \right \rceil }^{n} i$

$\rho(A) \le n+\frac{3n^{2}}{4}$

Cauchy-Schwartz inequality

$\langle A\vec{1} ,\vec{1} \rangle \le \Vert A\vec{1} \Vert_2 \cdot \Vert \vec{1} \Vert_2$

Spectral norm

$\|A\|_2 = \sup_{\vec{x} \neq \vec{0}} \frac{\|A\vec{x}\|_2}{\|\vec{x}\|_2}$

$\frac{\|A\vec{1}\|_2}{\|\vec{1}\|_2} \leq \|A\|_2$

$\|A\vec{1}\|_2 \leq \|A\|_2\cdot \|\vec{1}\|_2 $

Real Symmetric

$ \rho(A)=\Vert A \Vert_2$

$\langle A\vec{1} ,\vec{1} \rangle \le \Vert A\vec{1} \Vert_2 \cdot \Vert \vec{1} \Vert_2 \le \|A\|_2\cdot \|\vec{1}\|_2 \cdot \Vert \vec{1} \Vert_2 \le \rho(A) \cdot \sqrt{n} \cdot \sqrt{n} = n \rho(A)$

$\rho(A) \ge \frac{\langle A\vec{1} ,\vec{1} \rangle}{n}=\frac{n^2+\sum_{i=1}^{n-1} 2i^2}{n}=\frac{\frac{n}{3} + \frac{2n^3}{3}}{n}=\frac{2n^2}{3}+\frac{1}{3}$

$\frac{2}{3} \le \frac{\rho(A)}{n^2} \le \frac{3}{4}$

The result obtained by computer calculation for n=10000 is approximately 0.6755169463223237.

I have also attempted to use the inverse matrix to calculate the reciprocals of the eigenvalues. \begin{bmatrix} \frac{1}{2}+\frac{1}{2n+2} & -\frac{1}{2} & & & & \frac{1}{2n+2}\\ -\frac{1}{2} & 1 & -\frac{1}{2} & && \\ & -\frac{1}{2} & 1 & -\frac{1}{2} & &\\ & & -\frac{1}{2} & 1& -\frac{1}{2}&\\ & & & -\frac{1}{2} &1 &-\frac{1}{2}\\ \frac{1}{2n+2} & & & & -\frac{1}{2}& \frac{1}{2}+\frac{1}{2n+2} \end{bmatrix} This inverse matrix appears to asymptotically approach this, a=1 b=$-\frac{1}{2}$

Tridiagonal Matrix $\lambda_k = a + 2b \cos \left( \frac{k\pi}{n} \right), \quad k = 1, 2, 3, \ldots, n.$

\begin{bmatrix} a+b & b & & & & \\ b & a & b & && \\ & b & a & b& &\\ & & b & a& b&\\ & & & b &a &b\\ & & & & b& a+b \end{bmatrix}

However, this is only helpful for finding the limit of the largest eigenvalue of the inverse matrix, which is also the reciprocal of the smallest eigenvalue of the original matrix. Solving for the smallest eigenvalue of the inverse matrix will be significantly affected by the four corners, 1/(2n+2).

The $n×n$ matrix $A_n$ is defined by the elements $a_{ij}=n−|i−j|$. \begin{bmatrix} n & n-1 & n-2 & \cdots & 1\\ n-1 & n & n-1 & \cdots & 2\\ n-2 & n-1 & n & \cdots & 3\\ \vdots & \vdots & \vdots & & \vdots \\ 1 & 2 & 3 & \cdots & n \end{bmatrix}

The maximal eigenvalue of this Toeplitz matrix seems to be proportional to $n^2$.
Could someone please show me how to write the exact expression of the following limit value?

$$\lim_{n \to \infty} \frac{\rho(A)}{n^2}$$

Here is my attempt:

Gerschgorin theorem

$$\rho(A) \le n+\sum_{i=\left \lceil \frac{n}{2} \right \rceil }^n i$$

$$\rho(A) \le n+\frac{3n^{2}}{4}$$

Cauchy-Schwartz inequality

$$\langle A\vec{1} ,\vec{1} \rangle \le \Vert A\vec{1} \Vert_2 \cdot \Vert \vec{1} \Vert_2$$

Spectral norm

$$\|A\|_2 = \sup_{\vec{x} \neq \vec{0}} \frac{\|A\vec{x}\|_2}{\|\vec{x}\|_2}$$

$$\frac{\|A\vec{1}\|_2}{\|\vec{1}\|_2} \leq \|A\|_2$$

$$\|A\vec{1}\|_2 \leq \|A\|_2\cdot \|\vec{1}\|_2 $$

Real Symmetric

$$ \rho(A)=\Vert A \Vert_2$$

$$\langle A\vec{1} ,\vec{1} \rangle \le \Vert A\vec{1} \Vert_2 \cdot \Vert \vec{1} \Vert_2 \le \|A\|_2\cdot \|\vec{1}\|_2 \cdot \Vert \vec{1} \Vert_2 \le \rho(A) \cdot \sqrt{n} \cdot \sqrt{n} = n \rho(A)$$

$$\rho(A) \ge \frac{\langle A\vec{1} ,\vec{1} \rangle}{n} =\frac{n^2+\sum_{i=1}^{n-1} 2i^2}{n} =\frac{\frac{n}{3} + \frac{2n^3}{3}}{n}=\frac{2n^2}{3}+\frac{1}{3}$$

$$\frac{2}{3} \le \frac{\rho(A)}{n^2} \le \frac{3}{4} $$

The result obtained by computer calculation for $n=10000$ is approximately $0.6755169463223237.$

I have also attempted to use the inverse matrix to calculate the reciprocals of the eigenvalues. $$\begin{bmatrix} \frac{1}{2}+\frac{1}{2n+2} & -\frac{1}{2} & & & & \frac{1}{2n+2}\\ -\frac{1}{2} & 1 & -\frac{1}{2} & && \\ & -\frac{1}{2} & 1 & -\frac{1}{2} & &\\ & & -\frac{1}{2} & 1& -\frac{1}{2}&\\ & & & -\frac{1}{2} &1 &-\frac{1}{2}\\ \frac{1}{2n+2} & & & & -\frac{1}{2}& \frac{1}{2}+\frac{1}{2n+2} \end{bmatrix}$$ This inverse matrix appears to asymptotically approach this, $a=1,$ $b=-\frac{1}{2}$

Tridiagonal Matrix $\lambda_k = a + 2b \cos \left( \frac{k\pi}{n} \right), \quad k = 1, 2, 3, \ldots, n.$

$$\begin{bmatrix} a+b & b & & & & \\ b & a & b & && \\ & b & a & b& &\\ & & b & a& b&\\ & & & b &a &b\\ & & & & b& a+b \end{bmatrix}$$

However, this is only helpful for finding the limit of the largest eigenvalue of the inverse matrix, which is also the reciprocal of the smallest eigenvalue of the original matrix. Solving for the smallest eigenvalue of the inverse matrix will be significantly affected by the four corners, $1/(2n+2).$

The $n×n$ matrix $A_n$ is defined by the elements $a_{ij}=n−|i−j|$. \begin{bmatrix} n & n-1 & n-2 & ... & 1\\ n-1 & n & n-1 & ... & 2\\ n-2 & n-1 & n & ... & 3\\ ... & ... & ... & ... & ...\\ 1 & 2 & 3 & ... & n \end{bmatrix}

Max eigenvalue of this Toeplitz matrix seems to be proportional to $n^2$.
Could someone please show me how to write the exact expression of the following limit value?

$\lim_{n \to \infty} \frac{\rho(A)}{n^2}$

Here is my attempt:

Gerschgorin theorem

$\rho(A) \le n+\sum_{i=\left \lceil \frac{n}{2} \right \rceil }^{n} i$

$\rho(A) \le n+\frac{3n^{2}}{4}$

Cauchy-Schwartz inequality

$\langle A\vec{1} ,\vec{1} \rangle \le \Vert A\vec{1} \Vert_2 \cdot \Vert \vec{1} \Vert_2$

Spectral norm

$\|A\|_2 = \sup_{\vec{x} \neq \vec{0}} \frac{\|A\vec{x}\|_2}{\|\vec{x}\|_2}$

$\frac{\|A\vec{1}\|_2}{\|\vec{1}\|_2} \leq \|A\|_2$

$\|A\vec{1}\|_2 \leq \|A\|_2\cdot \|\vec{1}\|_2 $

Real Symmetric

$ \rho(A)=\Vert A \Vert_2$

$\langle A\vec{1} ,\vec{1} \rangle \le \Vert A\vec{1} \Vert_2 \cdot \Vert \vec{1} \Vert_2 \le \|A\|_2\cdot \|\vec{1}\|_2 \cdot \Vert \vec{1} \Vert_2 \le \rho(A) \cdot \sqrt{n} \cdot \sqrt{n} = n \rho(A)$

$\rho(A) \ge \frac{\langle A\vec{1} ,\vec{1} \rangle}{n}=\frac{n^2+\sum_{i=1}^{n-1} 2i^2}{n}=\frac{\frac{n}{3} + \frac{2n^3}{3}}{n}=\frac{2n^2}{3}+\frac{1}{3}$

$\frac{2}{3} \le \frac{\rho(A)}{n^2} \le \frac{3}{4}$

The result obtained by computer calculation for n=10000 is approximately 0.6755169463223237.

I have also attempted to use the inverse matrix to calculate the reciprocals of the eigenvalues. \begin{bmatrix} \frac{1}{2}+\frac{1}{2n+2} & -\frac{1}{2} & & & & \frac{1}{2n+2}\\ -\frac{1}{2} & 1 & -\frac{1}{2} & && \\ & -\frac{1}{2} & 1 & -\frac{1}{2} & &\\ & & -\frac{1}{2} & 1& -\frac{1}{2}&\\ & & & -\frac{1}{2} &1 &-\frac{1}{2}\\ \frac{1}{2n+2} & & & & -\frac{1}{2}& \frac{1}{2}+\frac{1}{2n+2} \end{bmatrix} This inverse matrix appears to asymptotically approach this, a=1 b=$-\frac{1}{2}$

Tridiagonal Matrix $\lambda_k = a + 2b \cos \left( \frac{k\pi}{n} \right), \quad k = 1, 2, 3, \ldots, n.$

\begin{bmatrix} a+b & b & & & & \\ b & a & b & && \\ & b & a & b& &\\ & & b & a& b&\\ & & & b &a &b\\ & & & & b& a+b \end{bmatrix}

However, this is only helpful for finding the limit of the largest eigenvalue of the inverse matrix, which is also the reciprocal of the smallest eigenvalue of the original matrix. Solving for the smallest eigenvalue of the inverse matrix will be significantly affected by the four corners, 1/(2n+2).

The $n×n$ matrix $A_n$ is defined by the elements $a_{ij}=n−|i−j|$. \begin{bmatrix} n & n-1 & n-2 & ... & 1\\ n-1 & n & n-1 & ... & 2\\ n-2 & n-1 & n & ... & 3\\ ... & ... & ... & ... & ...\\ 1 & 2 & 3 & ... & n \end{bmatrix}

The maximal eigenvalue of this Toeplitz matrix seems to be proportional to $n^2$.
Could someone please show me how to write the exact expression of the following limit value?

$\lim_{n \to \infty} \frac{\rho(A)}{n^2}$

Here is my attempt:

Gerschgorin theorem

$\rho(A) \le n+\sum_{i=\left \lceil \frac{n}{2} \right \rceil }^{n} i$

$\rho(A) \le n+\frac{3n^{2}}{4}$

Cauchy-Schwartz inequality

$\langle A\vec{1} ,\vec{1} \rangle \le \Vert A\vec{1} \Vert_2 \cdot \Vert \vec{1} \Vert_2$

Spectral norm

$\|A\|_2 = \sup_{\vec{x} \neq \vec{0}} \frac{\|A\vec{x}\|_2}{\|\vec{x}\|_2}$

$\frac{\|A\vec{1}\|_2}{\|\vec{1}\|_2} \leq \|A\|_2$

$\|A\vec{1}\|_2 \leq \|A\|_2\cdot \|\vec{1}\|_2 $

Real Symmetric

$ \rho(A)=\Vert A \Vert_2$

$\langle A\vec{1} ,\vec{1} \rangle \le \Vert A\vec{1} \Vert_2 \cdot \Vert \vec{1} \Vert_2 \le \|A\|_2\cdot \|\vec{1}\|_2 \cdot \Vert \vec{1} \Vert_2 \le \rho(A) \cdot \sqrt{n} \cdot \sqrt{n} = n \rho(A)$

$\rho(A) \ge \frac{\langle A\vec{1} ,\vec{1} \rangle}{n}=\frac{n^2+\sum_{i=1}^{n-1} 2i^2}{n}=\frac{\frac{n}{3} + \frac{2n^3}{3}}{n}=\frac{2n^2}{3}+\frac{1}{3}$

$\frac{2}{3} \le \frac{\rho(A)}{n^2} \le \frac{3}{4}$

The result obtained by computer calculation for n=10000 is approximately 0.6755169463223237.

I have also attempted to use the inverse matrix to calculate the reciprocals of the eigenvalues. \begin{bmatrix} \frac{1}{2}+\frac{1}{2n+2} & -\frac{1}{2} & & & & \frac{1}{2n+2}\\ -\frac{1}{2} & 1 & -\frac{1}{2} & && \\ & -\frac{1}{2} & 1 & -\frac{1}{2} & &\\ & & -\frac{1}{2} & 1& -\frac{1}{2}&\\ & & & -\frac{1}{2} &1 &-\frac{1}{2}\\ \frac{1}{2n+2} & & & & -\frac{1}{2}& \frac{1}{2}+\frac{1}{2n+2} \end{bmatrix} This inverse matrix appears to asymptotically approach this, a=1 b=$-\frac{1}{2}$

Tridiagonal Matrix $\lambda_k = a + 2b \cos \left( \frac{k\pi}{n} \right), \quad k = 1, 2, 3, \ldots, n.$

\begin{bmatrix} a+b & b & & & & \\ b & a & b & && \\ & b & a & b& &\\ & & b & a& b&\\ & & & b &a &b\\ & & & & b& a+b \end{bmatrix}

However, this is only helpful for finding the limit of the largest eigenvalue of the inverse matrix, which is also the reciprocal of the smallest eigenvalue of the original matrix. Solving for the smallest eigenvalue of the inverse matrix will be significantly affected by the four corners, 1/(2n+2).