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Alright, I'm very new to Mathematica and I have yet to learn a lot so please bear with me (sorry if this is a duplicate!):

I want to create an $n\times n$ matrix $C$ with elements $c_{i,j}$ with $i,j \in {1, \ldots n}$ which I would do like this

n = 4;
matC = Array[Subscript[c, #1, #2] &, {n, n}]

Now I want to define each $c_{i,j}=\{0,1\}$ but I have no idea how to effectively delay the assignment! I have tried using the Notation package but I just can't put my mind around it. I would go about it like this:

Do[Subscript[c,i,j] := {0,1}, {i,1,n}, {j,1,n}]

Finally I would like to calculate $det(tI_n - C) = 0$ and return all possible $C$ which I would do as follows:

Solve[Det[t*IdentityMatrix[n] - matC] == 0, matC]

Any help would be much appreciated!

I forgot to mention that I would actually like to find those $C$ where $det(tI_n - C) = a_nt^n - a_{n-1}t^{n-1} - \ldots - a_1t^1 - a_0 = a_nt^n - \sum_{i=0}^{n-1} a_it^i = 0$ for every $a_i = 1$.

As an example if $n = 3$ then

$C = \left\{ \begin{pmatrix} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 1 & 0 & 1 \end{pmatrix} \right\} $

Actually, I just found out, that the general solution for any $n$ will be

$C = \begin{pmatrix} 1 & 1 & 0 & \ldots & 0\\ 1 & 0 & 1 & \ldots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \vdots & \vdots & \vdots & & 1 \\ 1 & 0 & 0 & \ldots & 0 \end{pmatrix} $

and every shift along the diagonal, but I guess I should rather not post this here.

Alright, I'm very new to Mathematica and I have yet to learn a lot so please bear with me (sorry if this is a duplicate!):

I want to create an $n\times n$ matrix $C$ with elements $c_{i,j}$ with $i,j \in {1, \ldots n}$ which I would do like this

n = 4;
matC = Array[Subscript[c, #1, #2] &, {n, n}]

Now I want to define each $c_{i,j}=\{0,1\}$ but I have no idea how to effectively delay the assignment! I have tried using the Notation package but I just can't put my mind around it. I would go about it like this:

Do[Subscript[c,i,j] := {0,1}, {i,1,n}, {j,1,n}]

Finally I would like to calculate $det(tI_n - C) = 0$ and return all possible $C$ which I would do as follows:

Solve[Det[t*IdentityMatrix[n] - matC] == 0, matC]

Any help would be much appreciated!

I forgot to mention that I would actually like to find those $C$ where $det(tI_n - C) = a_nt^n - a_{n-1}t^{n-1} - \ldots - a_1t^1 - a_0 = a_nt^n - \sum_{i=0}^{n-1} a_it^i = 0$ for every $a_i = 1$.

As an example if $n = 3$ then

$C = \left\{ \begin{pmatrix} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 1 & 0 & 1 \end{pmatrix} \right\} $

Actually, I just found out, that the general solution for any $n$ will be

$C = \begin{pmatrix} 1 & 1 & 0 & \ldots & 0\\ 1 & 0 & 1 & \ldots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \vdots & \vdots & \vdots & & 1 \\ 1 & 0 & 0 & \ldots & 0 \end{pmatrix} $

and every shift along the diagonal, but I guess I should rather post this somewhere here.

Alright, I'm very new to Mathematica and I have yet to learn a lot so please bear with me (sorry if this is a duplicate!):

I want to create an $n\times n$ matrix $C$ with elements $c_{i,j}$ with $i,j \in {1, \ldots n}$ which I would do like this

n = 4;
matC = Array[Subscript[c, #1, #2] &, {n, n}]

Now I want to define each $c_{i,j}=\{0,1\}$ but I have no idea how to effectively delay the assignment! I have tried using the Notation package but I just can't put my mind around it. I would go about it like this:

Do[Subscript[c,i,j] := {0,1}, {i,1,n}, {j,1,n}]

Finally I would like to calculate $det(tI_n - C) = 0$ and return all possible $C$ which I would do as follows:

Solve[Det[t*IdentityMatrix[n] - matC] == 0, matC]

Any help would be much appreciated!

I forgot to mention that I would actually like to find those $C$ where $det(tI_n - C) = a_nt^n - a_{n-1}t^{n-1} - \ldots - a_1t^1 - a_0 = a_nt^n - \sum_{i=0}^{n-1} a_it^i = 0$ for every $a_i = 1$.

As an example if $n = 3$ then

$C = \left\{ \begin{pmatrix} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 1 & 0 & 1 \end{pmatrix} \right\} $

Alright, I'm very new to Mathematica and I have yet to learn a lot so please bear with me (sorry if this is a duplicate!):

I want to create an $n\times n$ matrix $C$ with elements $c_{i,j}$ with $i,j \in {1, \ldots n}$ which I would do like this

n = 4;
matC = Array[Subscript[c, #1, #2] &, {n, n}]

Now I want to define each $c_{i,j}=\{0,1\}$ but I have no idea how to effectively delay the assignment! I have tried using the Notation package but I just can't put my mind around it. I would go about it like this:

Do[Subscript[c,i,j] := {0,1}, {i,1,n}, {j,1,n}]

Finally I would like to calculate $det(tI_n - C) = 0$ and return all possible $C$ which I would do as follows:

Solve[Det[t*IdentityMatrix[n] - matC] == 0, matC]

Any help would be much appreciated!

I forgot to mention that I would actually like to find those $C$ where $det(tI_n - C) = a_nt^n - a_{n-1}t^{n-1} - \ldots - a_1t^1 - a_0 = a_nt^n - \sum_{i=0}^{n-1} a_it^i = 0$ for every $a_i = 1$.

As an example if $n = 3$ then

$C = \left\{ \begin{pmatrix} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 1 & 0 & 1 \end{pmatrix} \right\} $

Actually, I just found out, that the general solution for any $n$ will be

$C = \begin{pmatrix} 1 & 1 & 0 & \ldots & 0\\ 1 & 0 & 1 & \ldots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \vdots & \vdots & \vdots & & 1 \\ 1 & 0 & 0 & \ldots & 0 \end{pmatrix} $

and every shift along the diagonal, but I guess I should rather not post this here.

Alright, I'm very new to Mathematica and I have yet to learn a lot so please bear with me (sorry if this is a duplicate!):

I want to create an $n\times n$ matrix $C$ with elements $c_{i,j}$ with $i,j \in {1, \ldots n}$ which I would do like this

n = 4;
matC = Array[Subscript[c, #1, #2] &, {n, n}]

Now I want to define each $c_{i,j}=\{0,1\}$ but I have no idea how to effectively delay the assignment! I have tried using the Notation package but I just can't put my mind around it. I would go about it like this:

Do[Subscript[c,i,j] := {0,1}, {i,1,n}, {j,1,n}]

Finally I would like to calculate $det(tI_n - C) = 0$ and return all possible $C$ which I would do as follows:

Solve[Det[t*IdentityMatrix[n] - matC] == 0, matC]

Any help would be much appreciated!

I forgot to mention that I would actually like to find those $C$ where $det(tI_n - C) = a_nt^n - a_{n-1}t^{n-1} - \ldots - a_1t^1 - a_0 = a_nt^n - \sum_{i=0}^{n-1} a_it^i = 0$ for every $a_i = 1$.

As an example if $n = 3$ then

$C = \left\{ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}, \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix} \right\} $

Alright, I'm very new to Mathematica and I have yet to learn a lot so please bear with me (sorry if this is a duplicate!):

I want to create an $n\times n$ matrix $C$ with elements $c_{i,j}$ with $i,j \in {1, \ldots n}$ which I would do like this

n = 4;
matC = Array[Subscript[c, #1, #2] &, {n, n}]

Now I want to define each $c_{i,j}=\{0,1\}$ but I have no idea how to effectively delay the assignment! I have tried using the Notation package but I just can't put my mind around it. I would go about it like this:

Do[Subscript[c,i,j] := {0,1}, {i,1,n}, {j,1,n}]

Finally I would like to calculate $det(tI_n - C) = 0$ and return all possible $C$ which I would do as follows:

Solve[Det[t*IdentityMatrix[n] - matC] == 0, matC]

Any help would be much appreciated!

I forgot to mention that I would actually like to find those $C$ where $det(tI_n - C) = a_nt^n - a_{n-1}t^{n-1} - \ldots - a_1t^1 - a_0 = a_nt^n - \sum_{i=0}^{n-1} a_it^i = 0$ for every $a_i = 1$.

As an example if $n = 3$ then

$C = \left\{ \begin{pmatrix} 1 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 1 & 0 & 1 \end{pmatrix} \right\} $