Skip to main content
edited tags
Link
ABB
  • 4.1k
  • 1
  • 11
  • 19
deleted 1 character in body
Source Link
ABB
  • 4.1k
  • 1
  • 11
  • 19

By the circulant matrix $C$ in $M_n(\mathbb{R})$, we mean that $$C=[e_n|e_1|\cdots|e_{n-1}]$$ where $e_1,\cdots,e_n$ are the standard basis vectors in $\mathbb{R}^n$. It is well-known that
$$C=\mathbf{F}\operatorname{Diag}(1,w,\cdots,w^{n-1})\mathbf{F}^{-1}$$ where $\mathbf{F}$ is just the discrete Fourier matrix and $w=\exp(\frac{2\pi i}{n})$

Let $U=(u_{ij})$ be a upper triangular matrix. We say it is supper upper triangular if $u_{i, i+1}=0$.

Let $U$ be a supersupper upper triangular if whose whose entries are either 0 or 1. Is $C+U$ diagonalizable?

By the circulant matrix $C$ in $M_n(\mathbb{R})$, we mean that $$C=[e_n|e_1|\cdots|e_{n-1}]$$ where $e_1,\cdots,e_n$ are the standard basis vectors in $\mathbb{R}^n$. It is well-known that
$$C=\mathbf{F}\operatorname{Diag}(1,w,\cdots,w^{n-1})\mathbf{F}^{-1}$$ where $\mathbf{F}$ is just the discrete Fourier matrix and $w=\exp(\frac{2\pi i}{n})$

Let $U=(u_{ij})$ be a upper triangular matrix. We say it is supper upper triangular if $u_{i, i+1}=0$.

Let $U$ be a super upper triangular if whose entries are either 0 or 1. Is $C+U$ diagonalizable?

By the circulant matrix $C$ in $M_n(\mathbb{R})$, we mean that $$C=[e_n|e_1|\cdots|e_{n-1}]$$ where $e_1,\cdots,e_n$ are the standard basis vectors in $\mathbb{R}^n$. It is well-known that
$$C=\mathbf{F}\operatorname{Diag}(1,w,\cdots,w^{n-1})\mathbf{F}^{-1}$$ where $\mathbf{F}$ is just the discrete Fourier matrix and $w=\exp(\frac{2\pi i}{n})$

Let $U=(u_{ij})$ be a upper triangular matrix. We say it is supper upper triangular if $u_{i, i+1}=0$.

Let $U$ be a supper upper triangular whose entries are either 0 or 1. Is $C+U$ diagonalizable?

added 106 characters in body; edited title
Source Link
ABB
  • 4.1k
  • 1
  • 11
  • 19

Is the sum of the circulant matrix with a strictlysuper upper triangular matrix diagonalizable?

By the circulant matrix $C$ in $M_n(\mathbb{R})$, we mean that $$C=[e_n|e_1|\cdots|e_{n-1}]$$ where $e_1,\cdots,e_n$ are the standard basis vectors in $\mathbb{R}^n$. It is well-known that
$$C=\mathbf{F}\operatorname{Diag}(1,w,\cdots,w^{n-1})\mathbf{F}^{-1}$$ where $\mathbf{F}$ is just the discrete Fourier matrix and $w=\exp(\frac{2\pi i}{n})$

Let $U=(u_{ij})$ be a strictly upper triangular matrix with $u_{ij}\in \{0,1\}$. We say it is supper upper triangular if $u_{i, i+1}=0$.

Let $U$ be a super upper triangular if whose entries are either 0 or 1. Is Is $C+U$ diagonalizable?

Is the sum of the circulant matrix with a strictly upper triangular matrix diagonalizable?

By the circulant matrix $C$ in $M_n(\mathbb{R})$, we mean that $$C=[e_n|e_1|\cdots|e_{n-1}]$$ where $e_1,\cdots,e_n$ are the standard basis vectors in $\mathbb{R}^n$. It is well-known that
$$C=\mathbf{F}\operatorname{Diag}(1,w,\cdots,w^{n-1})\mathbf{F}^{-1}$$ where $\mathbf{F}$ is just the discrete Fourier matrix and $w=\exp(\frac{2\pi i}{n})$

Let $U=(u_{ij})$ be a strictly upper triangular matrix with $u_{ij}\in \{0,1\}$ . Is $C+U$ diagonalizable?

Is the sum of the circulant matrix with a super upper triangular matrix diagonalizable?

By the circulant matrix $C$ in $M_n(\mathbb{R})$, we mean that $$C=[e_n|e_1|\cdots|e_{n-1}]$$ where $e_1,\cdots,e_n$ are the standard basis vectors in $\mathbb{R}^n$. It is well-known that
$$C=\mathbf{F}\operatorname{Diag}(1,w,\cdots,w^{n-1})\mathbf{F}^{-1}$$ where $\mathbf{F}$ is just the discrete Fourier matrix and $w=\exp(\frac{2\pi i}{n})$

Let $U=(u_{ij})$ be a upper triangular matrix. We say it is supper upper triangular if $u_{i, i+1}=0$.

Let $U$ be a super upper triangular if whose entries are either 0 or 1. Is $C+U$ diagonalizable?

added 27 characters in body
Source Link
ABB
  • 4.1k
  • 1
  • 11
  • 19
Loading
edited title
Source Link
Loading
added 35 characters in body; edited tags
Source Link
ABB
  • 4.1k
  • 1
  • 11
  • 19
Loading
added 35 characters in body; edited tags
Source Link
ABB
  • 4.1k
  • 1
  • 11
  • 19
Loading
Source Link
ABB
  • 4.1k
  • 1
  • 11
  • 19
Loading