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I am working on a problem involving nilpotent matrices over $\mathbb{F}_2$ and I was able to reduce it to proving that the system \begin{equation} \begin{cases} A^2+ BC+ BCA+ ABC+A = I_4 \\ AB+ABD+BCB = 0 \\ CA+DCA+CBC = 0 \\ DCB+CBD = I_4 \\ A^3+BCA+ABC+BDC=0 \\ A^2B+BCB+ABD+BD^2=0 \\ CA^2+DCA+CBC+D^2C=0 \\ CAB+DCB+CBD+D^3=0 \end{cases}\,, \end{equation} has no solution, where $A, B, C, D$ are $4 \times 4$ matrices over $\mathbb{F}_2.$ The first four equations were obtained by plugging-in matrices to some polynomial in $\mathbb{F}_2$ while the other four came from the condition that $$\left[ \begin{array}{c|c} A & B \\ \hline C & D \end{array} \right]^3 = 0.$$

Any advice on how to prove this? Any help would be appreciated.

The original problem is if $$M=\begin{bmatrix} N & 0 \\ 0 & N \end{bmatrix}, \mbox{ where } N=\begin{bmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1\end{bmatrix}$$ then $M$ cannot be written as a sum $X+Y$ such that $X^3=0$ and $Y^2=Y.$ Here $M, X, Y$ are matrices over $\mathbb{F}_2$.

Write $X=[x_{i,j}], Y=[y_{i,j}]$. For each $(i,j)$, $i,j \in \{1, \ldots, 8\}$, $x_{i,j}$ and $y_{i,j}$ satisfies the following equations: \begin{equation} \begin{cases} \displaystyle \sum_{k=1}\sum_{l=1} x_{i,k}x_{k,l}x_{l,j}=0\\ \displaystyle \sum_{k=1} y_{i,k}y_{k,j}=y_{i,j}\\ x_{i,j}+y_{i,j}=m_{i,j} \\ x_{i,j}^2= x_{i,j} \\ y_{i,j}^2= y_{i,j} \end{cases}\,, \end{equation}\begin{equation} \begin{cases} \displaystyle \sum_{k=1}^8\sum_{l=1}^8 x_{i,k}x_{k,l}x_{l,j}=0\\ \displaystyle \sum_{k=1}^8 y_{i,k}y_{k,j}=y_{i,j}\\ x_{i,j}+y_{i,j}=m_{i,j} \\ x_{i,j}^2= x_{i,j} \\ y_{i,j}^2= y_{i,j} \end{cases}\,, \end{equation}

So we want to show that this system has no solution via Groebner basis and implement it using Maple. I have zero background in Maple so I need help with the code. So what I've researched is we start with

with(Groebner);

and define matrix $M$:

M:=[[0,0,0,1,0,0,0,0], [1,0,0,0,0,0,0,0], [0,1,0,0,0,0,0,0], [0,0,1,1,0,0,0,0], [0,0,0,0,0,0,0,1], [0,0,0,0,1,0,0,0], [0,0,0,0,0,1,0,0], [0,0,0,0,0,0,1,1]];

Define the following polynomials for all $(i,j)$:

f_{i,j}= \sum_{k=1}\sum_^8\sum_{l=1}^8 x_{i,k}x_{k,l}x_{l,j};
g_{i,j}= \sum_{k=1}^8 y_{i,k}y_{k,j}-y_{i,j};
h_{i,j}=x_{i,j}+y_{i,j}-M[i,j];
q_{i,j}=x_{i,j}^2-x_{i,j};
s_{i,j}=y_{i,j}^2-y_{i,j};

and we let for all $(i,j)$

B:=[f_{i,j},g_{i,j},h_{i,j},q_{i,j},s_{i,j}];

and call

G:=Groebner[Basis](B, plex(x_{i,j},y_{i,j}), characteristic=2);

We expect that the output is

G:=[1]

Is this correct? Also, how do I typeset the summations in Maple, and how do I typeset the polynomials so that I don't have to manually type for all 64 pairs $(i,j)$? Thank you so much.

I am working on a problem involving nilpotent matrices over $\mathbb{F}_2$ and I was able to reduce it to proving that the system \begin{equation} \begin{cases} A^2+ BC+ BCA+ ABC+A = I_4 \\ AB+ABD+BCB = 0 \\ CA+DCA+CBC = 0 \\ DCB+CBD = I_4 \\ A^3+BCA+ABC+BDC=0 \\ A^2B+BCB+ABD+BD^2=0 \\ CA^2+DCA+CBC+D^2C=0 \\ CAB+DCB+CBD+D^3=0 \end{cases}\,, \end{equation} has no solution, where $A, B, C, D$ are $4 \times 4$ matrices over $\mathbb{F}_2.$ The first four equations were obtained by plugging-in matrices to some polynomial in $\mathbb{F}_2$ while the other four came from the condition that $$\left[ \begin{array}{c|c} A & B \\ \hline C & D \end{array} \right]^3 = 0.$$

Any advice on how to prove this? Any help would be appreciated.

The original problem is if $$M=\begin{bmatrix} N & 0 \\ 0 & N \end{bmatrix}, \mbox{ where } N=\begin{bmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1\end{bmatrix}$$ then $M$ cannot be written as a sum $X+Y$ such that $X^3=0$ and $Y^2=Y.$ Here $M, X, Y$ are matrices over $\mathbb{F}_2$.

Write $X=[x_{i,j}], Y=[y_{i,j}]$. For each $(i,j)$, $i,j \in \{1, \ldots, 8\}$, $x_{i,j}$ and $y_{i,j}$ satisfies the following equations: \begin{equation} \begin{cases} \displaystyle \sum_{k=1}\sum_{l=1} x_{i,k}x_{k,l}x_{l,j}=0\\ \displaystyle \sum_{k=1} y_{i,k}y_{k,j}=y_{i,j}\\ x_{i,j}+y_{i,j}=m_{i,j} \\ x_{i,j}^2= x_{i,j} \\ y_{i,j}^2= y_{i,j} \end{cases}\,, \end{equation}

So we want to show that this system has no solution via Groebner basis and implement it using Maple. I have zero background in Maple so I need help with the code. So what I've researched is we start with

with(Groebner);

and define matrix $M$:

M:=[[0,0,0,1,0,0,0,0], [1,0,0,0,0,0,0,0], [0,1,0,0,0,0,0,0], [0,0,1,1,0,0,0,0], [0,0,0,0,0,0,0,1], [0,0,0,0,1,0,0,0], [0,0,0,0,0,1,0,0], [0,0,0,0,0,0,1,1]];

Define the following polynomials for all $(i,j)$:

f_{i,j}= \sum_{k=1}\sum_{l=1} x_{i,k}x_{k,l}x_{l,j};
g_{i,j}= \sum_{k=1} y_{i,k}y_{k,j}-y_{i,j};
h_{i,j}=x_{i,j}+y_{i,j}-M[i,j];
q_{i,j}=x_{i,j}^2-x_{i,j};
s_{i,j}=y_{i,j}^2-y_{i,j};

and we let for all $(i,j)$

B:=[f_{i,j},g_{i,j},h_{i,j},q_{i,j},s_{i,j}];

and call

G:=Groebner[Basis](B, plex(x_{i,j},y_{i,j}), characteristic=2);

We expect that the output is

G:=[1]

Is this correct? Also, how do I typeset the summations in Maple, and how do I typeset the polynomials so that I don't have to manually type for all 64 pairs $(i,j)$? Thank you so much.

I am working on a problem involving nilpotent matrices over $\mathbb{F}_2$ and I was able to reduce it to proving that the system \begin{equation} \begin{cases} A^2+ BC+ BCA+ ABC+A = I_4 \\ AB+ABD+BCB = 0 \\ CA+DCA+CBC = 0 \\ DCB+CBD = I_4 \\ A^3+BCA+ABC+BDC=0 \\ A^2B+BCB+ABD+BD^2=0 \\ CA^2+DCA+CBC+D^2C=0 \\ CAB+DCB+CBD+D^3=0 \end{cases}\,, \end{equation} has no solution, where $A, B, C, D$ are $4 \times 4$ matrices over $\mathbb{F}_2.$ The first four equations were obtained by plugging-in matrices to some polynomial in $\mathbb{F}_2$ while the other four came from the condition that $$\left[ \begin{array}{c|c} A & B \\ \hline C & D \end{array} \right]^3 = 0.$$

Any advice on how to prove this? Any help would be appreciated.

The original problem is if $$M=\begin{bmatrix} N & 0 \\ 0 & N \end{bmatrix}, \mbox{ where } N=\begin{bmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1\end{bmatrix}$$ then $M$ cannot be written as a sum $X+Y$ such that $X^3=0$ and $Y^2=Y.$ Here $M, X, Y$ are matrices over $\mathbb{F}_2$.

Write $X=[x_{i,j}], Y=[y_{i,j}]$. For each $(i,j)$, $i,j \in \{1, \ldots, 8\}$, $x_{i,j}$ and $y_{i,j}$ satisfies the following equations: \begin{equation} \begin{cases} \displaystyle \sum_{k=1}^8\sum_{l=1}^8 x_{i,k}x_{k,l}x_{l,j}=0\\ \displaystyle \sum_{k=1}^8 y_{i,k}y_{k,j}=y_{i,j}\\ x_{i,j}+y_{i,j}=m_{i,j} \\ x_{i,j}^2= x_{i,j} \\ y_{i,j}^2= y_{i,j} \end{cases}\,, \end{equation}

So we want to show that this system has no solution via Groebner basis and implement it using Maple. I have zero background in Maple so I need help with the code. So what I've researched is we start with

with(Groebner);

and define matrix $M$:

M:=[[0,0,0,1,0,0,0,0], [1,0,0,0,0,0,0,0], [0,1,0,0,0,0,0,0], [0,0,1,1,0,0,0,0], [0,0,0,0,0,0,0,1], [0,0,0,0,1,0,0,0], [0,0,0,0,0,1,0,0], [0,0,0,0,0,0,1,1]];

Define the following polynomials for all $(i,j)$:

f_{i,j}= \sum_{k=1}^8\sum_{l=1}^8 x_{i,k}x_{k,l}x_{l,j};
g_{i,j}= \sum_{k=1}^8 y_{i,k}y_{k,j}-y_{i,j};
h_{i,j}=x_{i,j}+y_{i,j}-M[i,j];
q_{i,j}=x_{i,j}^2-x_{i,j};
s_{i,j}=y_{i,j}^2-y_{i,j};

and we let for all $(i,j)$

B:=[f_{i,j},g_{i,j},h_{i,j},q_{i,j},s_{i,j}];

and call

G:=Groebner[Basis](B, plex(x_{i,j},y_{i,j}), characteristic=2);

We expect that the output is

G:=[1]

Is this correct? Also, how do I typeset the summations in Maple, and how do I typeset the polynomials so that I don't have to manually type for all 64 pairs $(i,j)$? Thank you so much.

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I am working on a problem involving nilpotent matrices over $\mathbb{F}_2$ and I was able to reduce it to proving that the system \begin{equation} \begin{cases} A^2+ BC+ BCA+ ABC+A = I_4 \\ AB+ABD+BCB = 0 \\ CA+DCA+CBC = 0 \\ DCB+CBD = I_4 \\ A^3+BCA+ABC+BDC=0 \\ A^2B+BCB+ABD+BD^2=0 \\ CA^2+DCA+CBC+D^2C=0 \\ CAB+DCB+CBD+D^3=0 \end{cases}\,, \end{equation} has no solution, where $A, B, C, D$ are $4 \times 4$ matrices over $\mathbb{F}_2.$ The first four equations were obtained by plugging-in matrices to some polynomial in $\mathbb{F}_2$ while the other four came from the condition that $$\left[ \begin{array}{c|c} A & B \\ \hline C & D \end{array} \right]^3 = 0.$$

Any advice on how to prove this? Any help would be appreciated.

The original problem is if $$M=\begin{bmatrix} N & 0 \\ 0 & N \end{bmatrix}, \mbox{ where } N=\begin{bmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1\end{bmatrix}$$ then $M$ cannot be written as a sum $X+Y$ such that $X^3=0$ and $Y^2=Y.$ Here $M, X, Y$ are matrices over $\mathbb{F}_2$.

Write $X=[x_{i,j}], Y=[y_{i,j}]$. For each $(i,j)$, $i,j \in \{1, \ldots, 8\}$, $x_{i,j}$ and $y_{i,j}$ satisfies the following equations: \begin{equation} \begin{cases} \displaystyle \sum_{k=1}\sum_{l=1} x_{i,k}x_{k,l}x_{l,j}=0\\ \displaystyle \sum_{k=1} y_{i,k}y_{k,j}=y_{i,j}\\ x_{i,j}+y_{i,j}=m_{i,j} \\ x_{i,j}^2= x_{i,j} \\ y_{i,j}^2= y_{i,j} \end{cases}\,, \end{equation}

So we want to show that this system has no solution via Groebner basis and implement it using Maple. I have zero background in Maple so I need help with the code. So what I've researched is we start with

with(Groebner);

and define matrix $M$:

M:=[[0,0,0,1,0,0,0,0], [1,0,0,0,0,0,0,0], [0,1,0,0,0,0,0,0], [0,0,1,1,0,0,0,0], [0,0,0,0,0,0,0,1], [0,0,0,0,1,0,0,0], [0,0,0,0,0,1,0,0], [0,0,0,0,0,0,1,1]];

Define the following polynomials for all $(i,j)$:

f_{i,j}= \sum_{k=1}\sum_{l=1} x_{i,k}x_{k,l}x_{l,j};
g_{i,j}= \sum_{k=1} y_{i,k}y_{k,j}-y_{i,j};
h_{i,j}=x_{i,j}+y_{i,j}-M[i,j];
q_{i,j}=x_{i,j}^2-x_{i,j};
s_{i,j}=y_{i,j}^2-y_{i,j};

and we let for all $(i,j)$

B:=[f_{i,j},g_{i,j},h_{i,j},q_{i,j},s_{i,j}];

and call

G:=Groebner[Basis](B, plex(x_{i,j},y_{i,j}), characteristic=2);

We expect that the output is

G:=[1]

Is this correct? Also, how do I typeset the summations in Maple, and how do I typeset the polynomials so that I don't have to manually type for all 64 pairs $(i,j)$? Thank you so much.

I am working on a problem involving nilpotent matrices over $\mathbb{F}_2$ and I was able to reduce it to proving that the system \begin{equation} \begin{cases} A^2+ BC+ BCA+ ABC+A = I_4 \\ AB+ABD+BCB = 0 \\ CA+DCA+CBC = 0 \\ DCB+CBD = I_4 \\ A^3+BCA+ABC+BDC=0 \\ A^2B+BCB+ABD+BD^2=0 \\ CA^2+DCA+CBC+D^2C=0 \\ CAB+DCB+CBD+D^3=0 \end{cases}\,, \end{equation} has no solution, where $A, B, C, D$ are $4 \times 4$ matrices over $\mathbb{F}_2.$ The first four equations were obtained by plugging-in matrices to some polynomial in $\mathbb{F}_2$ while the other four came from the condition that $$\left[ \begin{array}{c|c} A & B \\ \hline C & D \end{array} \right]^3 = 0.$$

Any advice on how to prove this? Any help would be appreciated.

The original problem is if $$M=\begin{bmatrix} N & 0 \\ 0 & N \end{bmatrix}, \mbox{ where } N=\begin{bmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1\end{bmatrix}$$ then $M$ cannot be written as a sum $X+Y$ such that $X^3=0$ and $Y^2=Y.$ Here $M, X, Y$ are matrices over $\mathbb{F}_2$.

I am working on a problem involving nilpotent matrices over $\mathbb{F}_2$ and I was able to reduce it to proving that the system \begin{equation} \begin{cases} A^2+ BC+ BCA+ ABC+A = I_4 \\ AB+ABD+BCB = 0 \\ CA+DCA+CBC = 0 \\ DCB+CBD = I_4 \\ A^3+BCA+ABC+BDC=0 \\ A^2B+BCB+ABD+BD^2=0 \\ CA^2+DCA+CBC+D^2C=0 \\ CAB+DCB+CBD+D^3=0 \end{cases}\,, \end{equation} has no solution, where $A, B, C, D$ are $4 \times 4$ matrices over $\mathbb{F}_2.$ The first four equations were obtained by plugging-in matrices to some polynomial in $\mathbb{F}_2$ while the other four came from the condition that $$\left[ \begin{array}{c|c} A & B \\ \hline C & D \end{array} \right]^3 = 0.$$

Any advice on how to prove this? Any help would be appreciated.

The original problem is if $$M=\begin{bmatrix} N & 0 \\ 0 & N \end{bmatrix}, \mbox{ where } N=\begin{bmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1\end{bmatrix}$$ then $M$ cannot be written as a sum $X+Y$ such that $X^3=0$ and $Y^2=Y.$ Here $M, X, Y$ are matrices over $\mathbb{F}_2$.

Write $X=[x_{i,j}], Y=[y_{i,j}]$. For each $(i,j)$, $i,j \in \{1, \ldots, 8\}$, $x_{i,j}$ and $y_{i,j}$ satisfies the following equations: \begin{equation} \begin{cases} \displaystyle \sum_{k=1}\sum_{l=1} x_{i,k}x_{k,l}x_{l,j}=0\\ \displaystyle \sum_{k=1} y_{i,k}y_{k,j}=y_{i,j}\\ x_{i,j}+y_{i,j}=m_{i,j} \\ x_{i,j}^2= x_{i,j} \\ y_{i,j}^2= y_{i,j} \end{cases}\,, \end{equation}

So we want to show that this system has no solution via Groebner basis and implement it using Maple. I have zero background in Maple so I need help with the code. So what I've researched is we start with

with(Groebner);

and define matrix $M$:

M:=[[0,0,0,1,0,0,0,0], [1,0,0,0,0,0,0,0], [0,1,0,0,0,0,0,0], [0,0,1,1,0,0,0,0], [0,0,0,0,0,0,0,1], [0,0,0,0,1,0,0,0], [0,0,0,0,0,1,0,0], [0,0,0,0,0,0,1,1]];

Define the following polynomials for all $(i,j)$:

f_{i,j}= \sum_{k=1}\sum_{l=1} x_{i,k}x_{k,l}x_{l,j};
g_{i,j}= \sum_{k=1} y_{i,k}y_{k,j}-y_{i,j};
h_{i,j}=x_{i,j}+y_{i,j}-M[i,j];
q_{i,j}=x_{i,j}^2-x_{i,j};
s_{i,j}=y_{i,j}^2-y_{i,j};

and we let for all $(i,j)$

B:=[f_{i,j},g_{i,j},h_{i,j},q_{i,j},s_{i,j}];

and call

G:=Groebner[Basis](B, plex(x_{i,j},y_{i,j}), characteristic=2);

We expect that the output is

G:=[1]

Is this correct? Also, how do I typeset the summations in Maple, and how do I typeset the polynomials so that I don't have to manually type for all 64 pairs $(i,j)$? Thank you so much.

added 326 characters in body
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I am working on a problem involving nilpotent matrices over $\mathbb{F}_2$ and I was able to reduce it to proving that the system \begin{equation} \begin{cases} A^2+ BC+ BCA+ ABC+A = I_4 \\ AB+ABD+BCB = 0 \\ CA+DCA+CBC = 0 \\ DCB+CBD = I_4 \\ A^3+BCA+ABC+BDC=0 \\ A^2B+BCB+ABD+BD^2=0 \\ CA^2+DCA+CBC+D^2C=0 \\ CAB+DCB+CBD+D^3=0 \end{cases}\,, \end{equation} has no solution, where $A, B, C, D$ are $4 \times 4$ matrices over $\mathbb{F}_2.$ The first four equations were obtained by plugging-in matrices to some polynomial in $\mathbb{F}_2$ while the other four came from the condition that $$\left[ \begin{array}{c|c} A & B \\ \hline C & D \end{array} \right]^3 = 0.$$

Any advice on how to prove this? Any help would be appreciated.

The original problem is if $$M=\begin{bmatrix} N & 0 \\ 0 & N \end{bmatrix}, \mbox{ where } N=\begin{bmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1\end{bmatrix}$$ then $M$ cannot be written as a sum $X+Y$ such that $X^3=0$ and $Y^2=Y.$ Here $M, X, Y$ are matrices over $\mathbb{F}_2$.

I am working on a problem involving nilpotent matrices over $\mathbb{F}_2$ and I was able to reduce it to proving that the system \begin{equation} \begin{cases} A^2+ BC+ BCA+ ABC+A = I_4 \\ AB+ABD+BCB = 0 \\ CA+DCA+CBC = 0 \\ DCB+CBD = I_4 \\ A^3+BCA+ABC+BDC=0 \\ A^2B+BCB+ABD+BD^2=0 \\ CA^2+DCA+CBC+D^2C=0 \\ CAB+DCB+CBD+D^3=0 \end{cases}\,, \end{equation} has no solution, where $A, B, C, D$ are $4 \times 4$ matrices over $\mathbb{F}_2.$ The first four equations were obtained by plugging-in matrices to some polynomial in $\mathbb{F}_2$ while the other four came from the condition that $$\left[ \begin{array}{c|c} A & B \\ \hline C & D \end{array} \right]^3 = 0.$$

Any advice on how to prove this? Any help would be appreciated.

I am working on a problem involving nilpotent matrices over $\mathbb{F}_2$ and I was able to reduce it to proving that the system \begin{equation} \begin{cases} A^2+ BC+ BCA+ ABC+A = I_4 \\ AB+ABD+BCB = 0 \\ CA+DCA+CBC = 0 \\ DCB+CBD = I_4 \\ A^3+BCA+ABC+BDC=0 \\ A^2B+BCB+ABD+BD^2=0 \\ CA^2+DCA+CBC+D^2C=0 \\ CAB+DCB+CBD+D^3=0 \end{cases}\,, \end{equation} has no solution, where $A, B, C, D$ are $4 \times 4$ matrices over $\mathbb{F}_2.$ The first four equations were obtained by plugging-in matrices to some polynomial in $\mathbb{F}_2$ while the other four came from the condition that $$\left[ \begin{array}{c|c} A & B \\ \hline C & D \end{array} \right]^3 = 0.$$

Any advice on how to prove this? Any help would be appreciated.

The original problem is if $$M=\begin{bmatrix} N & 0 \\ 0 & N \end{bmatrix}, \mbox{ where } N=\begin{bmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1\end{bmatrix}$$ then $M$ cannot be written as a sum $X+Y$ such that $X^3=0$ and $Y^2=Y.$ Here $M, X, Y$ are matrices over $\mathbb{F}_2$.

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