Skip to main content
deleted 17 characters in body
Source Link
dave2d
  • 103
  • 5

In one step of solving a difficult problem, I would like to know the largest eigenvalue of a matrix with this pattern: $$A_n = \begin{bmatrix} 0 & 0 & 0 & 0 &\dots & 0 \\ 0 & 1 & 1 & 1&\dots & 1 \\ 0 & 1 &2 &2 &\dots &2\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 1 & 2 & 3 & \dots& n-1 \end{bmatrix}$$

In other words, we can "peel" the "L" shaped layers from upper left corner and all numbers on the layers are consistslayer consist of the same numberentry.

This matrix is symmetric, so all eigenvalues are real. It is of rank $n-1$, and numerical experiment demonstratedemonstrates that it is positive semidefinite. I want to know the largetslargest eigenvalue in terms of $n$, or have a lower bound on the largest eigenvalue in terms of n $n$. I performed numerical experiment on various $n$ and it seems like we can quickly get $\lambda_{max}\approx 0.4n^2$ for large $n$. However, I don't know how to derive this result. I tried to look at the characteristic polynomial, but I could not find a pattern.

So anyone has an idea about deriving a (lower) bound on the largest eigenvalue of this matrix? Thanks!

In one step of solving a difficult problem, I would like to know the largest eigenvalue of a matrix with this pattern: $$A_n = \begin{bmatrix} 0 & 0 & 0 & 0 &\dots & 0 \\ 0 & 1 & 1 & 1&\dots & 1 \\ 0 & 1 &2 &2 &\dots &2\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 1 & 2 & 3 & \dots& n-1 \end{bmatrix}$$

In other words, we can "peel" the "L" shaped layers from upper left corner and all numbers on the layers are consists of the same number.

This matrix is symmetric, so all eigenvalues are real. It is of rank $n-1$, and numerical experiment demonstrate that it is positive semidefinite. I want to know the largets eigenvalue in terms of $n$, or have a lower bound on the largest eigenvalue in terms of n. I performed numerical experiment on various $n$ and it seems like we can quickly get $\lambda_{max}\approx 0.4n^2$ for large $n$. However, I don't know how to derive this result. I tried to look at the characteristic polynomial, but I could not find a pattern.

So anyone has an idea about deriving a (lower) bound on the largest eigenvalue of this matrix? Thanks!

In one step of solving a difficult problem, I would like to know the largest eigenvalue of a matrix with this pattern: $$A_n = \begin{bmatrix} 0 & 0 & 0 & 0 &\dots & 0 \\ 0 & 1 & 1 & 1&\dots & 1 \\ 0 & 1 &2 &2 &\dots &2\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 1 & 2 & 3 & \dots& n-1 \end{bmatrix}$$

In other words, we can "peel" the "L" shaped layers from upper left corner and all layer consist of the same entry.

This matrix is symmetric, so all eigenvalues are real. It is of rank $n-1$, and numerical experiment demonstrates that it is positive semidefinite. I want to know the largest eigenvalue in terms of $n$, or have a lower bound on the largest eigenvalue in terms of $n$. I performed numerical experiment on various $n$ and it seems like we can quickly get $\lambda_{max}\approx 0.4n^2$ for large $n$. However, I don't know how to derive this result. I tried to look at the characteristic polynomial, but I could not find a pattern.

So anyone has an idea about deriving a (lower) bound on the largest eigenvalue of this matrix? Thanks!

deleted 1 character in body
Source Link
dave2d
  • 103
  • 5

In one step of solving a difficult problem, I would like to know the largest eigenvalue of a matrix with this pattern: $$A_n = \begin{bmatrix} 0 & 0 & 0 & 0 &\dots & 0 \\ 0 & 1 & 1 & 1&\dots & 1 \\ 0 & 1 &2 &2 &\dots &2\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 1 & 2 & 3 & \dots& n-1 \end{bmatrix}$$

In other words, we can "peel" the "L" shaped layers from upper rightleft corner and all numbers on the layers are consists of the same number.

This matrix is symmetric, so all eigenvalues are real. It is of rank $n-1$, and numerical experiment demonstrate that it is positive semidefinite. I want to know the largets eigenvalue in terms of $n$, or have a lower bound on the largest eigenvalue in terms of n. I performed numerical experiment on various $n$ and it seems like we can quickly get $\lambda_{max}\approx 0.4n^2$ for large $n$. However, I don't know how to derive this result. I tried to look at the characteristic polynomial, but I could not find a pattern.

So anyone has an idea about deriving a (lower) bound on the largest eigenvalue of this matrix? Thanks!

In one step of solving a difficult problem, I would like to know the largest eigenvalue of a matrix with this pattern: $$A_n = \begin{bmatrix} 0 & 0 & 0 & 0 &\dots & 0 \\ 0 & 1 & 1 & 1&\dots & 1 \\ 0 & 1 &2 &2 &\dots &2\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 1 & 2 & 3 & \dots& n-1 \end{bmatrix}$$

In other words, we can "peel" the "L" shaped layers from upper right corner and all numbers on the layers are consists of the same number.

This matrix is symmetric, so all eigenvalues are real. It is of rank $n-1$, and numerical experiment demonstrate that it is positive semidefinite. I want to know the largets eigenvalue in terms of $n$, or have a lower bound on the largest eigenvalue in terms of n. I performed numerical experiment on various $n$ and it seems like we can quickly get $\lambda_{max}\approx 0.4n^2$ for large $n$. However, I don't know how to derive this result. I tried to look at the characteristic polynomial, but I could not find a pattern.

So anyone has an idea about deriving a (lower) bound on the largest eigenvalue of this matrix? Thanks!

In one step of solving a difficult problem, I would like to know the largest eigenvalue of a matrix with this pattern: $$A_n = \begin{bmatrix} 0 & 0 & 0 & 0 &\dots & 0 \\ 0 & 1 & 1 & 1&\dots & 1 \\ 0 & 1 &2 &2 &\dots &2\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 1 & 2 & 3 & \dots& n-1 \end{bmatrix}$$

In other words, we can "peel" the "L" shaped layers from upper left corner and all numbers on the layers are consists of the same number.

This matrix is symmetric, so all eigenvalues are real. It is of rank $n-1$, and numerical experiment demonstrate that it is positive semidefinite. I want to know the largets eigenvalue in terms of $n$, or have a lower bound on the largest eigenvalue in terms of n. I performed numerical experiment on various $n$ and it seems like we can quickly get $\lambda_{max}\approx 0.4n^2$ for large $n$. However, I don't know how to derive this result. I tried to look at the characteristic polynomial, but I could not find a pattern.

So anyone has an idea about deriving a (lower) bound on the largest eigenvalue of this matrix? Thanks!

Source Link
dave2d
  • 103
  • 5

Largest Eigenvalue of a Matrix with Special Form in terms of n

In one step of solving a difficult problem, I would like to know the largest eigenvalue of a matrix with this pattern: $$A_n = \begin{bmatrix} 0 & 0 & 0 & 0 &\dots & 0 \\ 0 & 1 & 1 & 1&\dots & 1 \\ 0 & 1 &2 &2 &\dots &2\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 1 & 2 & 3 & \dots& n-1 \end{bmatrix}$$

In other words, we can "peel" the "L" shaped layers from upper right corner and all numbers on the layers are consists of the same number.

This matrix is symmetric, so all eigenvalues are real. It is of rank $n-1$, and numerical experiment demonstrate that it is positive semidefinite. I want to know the largets eigenvalue in terms of $n$, or have a lower bound on the largest eigenvalue in terms of n. I performed numerical experiment on various $n$ and it seems like we can quickly get $\lambda_{max}\approx 0.4n^2$ for large $n$. However, I don't know how to derive this result. I tried to look at the characteristic polynomial, but I could not find a pattern.

So anyone has an idea about deriving a (lower) bound on the largest eigenvalue of this matrix? Thanks!