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I am wondering if we can find a linear transformation matrix $A$ of size $3\times 3$ over the field of two elements $\mathbb{Z}_2$ i.e. a matrix $A$ of zeros and ones s.t.

$$A \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}, A \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}, A \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}, A \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, A \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, A \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}, A \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, A \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$$

As can be seen above, the vectors I am looking at are all the possible 3D vectors with zeros and ones. $A$ is supposed to be an invertible transformation such that $A\neq A^{-1}$ because the effect of $A$ is a bijection from the space of all the possible 3D vectors with zeros and ones to that space itself.

I found an invertible $A$ using three of the equations above but that $A$ does not work for other vectors and so am wondering if it is possible to find such an $A$?

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A linear transformation can be defined by its values on the basis. So consider the standard basis $e_1,e_2,e_3$ of the $3$ dimensional vector space over the integers modulo $2$. Then your $A$ is defined as a linear map by $Ae_1=(1,1,0),Ae_2=(1,0,1),Ae_3=(0,1,1)$ so we can represent $A$ as a matrix by $[A]_{E,E}=\left[\begin{array}{l}1&1&0\\1&0&1\\0&1&1\end{array}\right]$ We can now easily calculate the determinant of $A$ to be $\det(A)=0+0+0-1-1=-2=0\mod 2$ thus any such matrix is not invertible. Furthermore such a matrix does not satisfy some of your other desired vector mappings, so there is no matrix satisfying all of your desires.

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It's not possible because $A\begin{bmatrix}0\\1\\1\end{bmatrix}\ne A\begin{bmatrix}0\\0\\1\end{bmatrix}+A\begin{bmatrix}0\\1\\0\end{bmatrix}$.

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There is no such $A$, invertible or not. For instance your equations lead to $$ \begin{bmatrix} 1 \\ 0\\ 0\end{bmatrix}=\begin{bmatrix} 1 \\ 1 \\ 1\end{bmatrix}+ \begin{bmatrix} 0 \\ 1 \\ 1\end{bmatrix} =A \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} + A \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} = A \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} $$

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The images of the vector $e_1, e_2,e_3$ of the canonical basis of $\mathbb Z_2^3$ defines entirely $A$.

Then if the image of the other vectors are coherent, $A$ exists. If not, there is no $A$ solution.

Applying this in your particular case, you can see that $A$ doesn’t exist.

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