# Finding a linear transformation matrix for a set of vectors

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 $$A$$?

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