Finding non-singular transformation mapping one tensor to other in $(Bbb F_2)^{otimes 3}$












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Let $u, v in Vdoteq mathbb{F}_2^{2 times 2 times 2}= mathbb{F}_2 otimes mathbb{F}_2 otimes mathbb{F}_2$ be given by



$$u = e_1 otimes e_1 otimes e_1 + e_2 otimes e_2 otimes e_1 + e_1 otimes e_2 otimes e_2$$



and



$$v = (e_1+e_2) otimes (e_1+e_2) otimes (e_1+e_2) + e_1 otimes e_2 otimes (e_1+e_2) + e_2 otimes e_1 otimes e_2,$$



where $e_1, e_2$ are the usual standard basis vectors and $otimes$ is the usual tensor product.



How can I construct an element $A in {rm GL}(V)$ such that $Au = v$? What is the explicit element $A$ that transforms $u$ to $v$? Thank you.






linear-algebra linear-transformations tensor-products multilinear-algebra tensor-rank







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    1












    $begingroup$


    Let $u, v in Vdoteq mathbb{F}_2^{2 times 2 times 2}= mathbb{F}_2 otimes mathbb{F}_2 otimes mathbb{F}_2$ be given by



    $$u = e_1 otimes e_1 otimes e_1 + e_2 otimes e_2 otimes e_1 + e_1 otimes e_2 otimes e_2$$



    and



    $$v = (e_1+e_2) otimes (e_1+e_2) otimes (e_1+e_2) + e_1 otimes e_2 otimes (e_1+e_2) + e_2 otimes e_1 otimes e_2,$$



    where $e_1, e_2$ are the usual standard basis vectors and $otimes$ is the usual tensor product.



    How can I construct an element $A in {rm GL}(V)$ such that $Au = v$? What is the explicit element $A$ that transforms $u$ to $v$? Thank you.






    linear-algebra linear-transformations tensor-products multilinear-algebra tensor-rank







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Let $u, v in Vdoteq mathbb{F}_2^{2 times 2 times 2}= mathbb{F}_2 otimes mathbb{F}_2 otimes mathbb{F}_2$ be given by



      $$u = e_1 otimes e_1 otimes e_1 + e_2 otimes e_2 otimes e_1 + e_1 otimes e_2 otimes e_2$$



      and



      $$v = (e_1+e_2) otimes (e_1+e_2) otimes (e_1+e_2) + e_1 otimes e_2 otimes (e_1+e_2) + e_2 otimes e_1 otimes e_2,$$



      where $e_1, e_2$ are the usual standard basis vectors and $otimes$ is the usual tensor product.



      How can I construct an element $A in {rm GL}(V)$ such that $Au = v$? What is the explicit element $A$ that transforms $u$ to $v$? Thank you.






      linear-algebra linear-transformations tensor-products multilinear-algebra tensor-rank







      share|cite|improve this question











      $endgroup$




      Let $u, v in Vdoteq mathbb{F}_2^{2 times 2 times 2}= mathbb{F}_2 otimes mathbb{F}_2 otimes mathbb{F}_2$ be given by



      $$u = e_1 otimes e_1 otimes e_1 + e_2 otimes e_2 otimes e_1 + e_1 otimes e_2 otimes e_2$$



      and



      $$v = (e_1+e_2) otimes (e_1+e_2) otimes (e_1+e_2) + e_1 otimes e_2 otimes (e_1+e_2) + e_2 otimes e_1 otimes e_2,$$



      where $e_1, e_2$ are the usual standard basis vectors and $otimes$ is the usual tensor product.



      How can I construct an element $A in {rm GL}(V)$ such that $Au = v$? What is the explicit element $A$ that transforms $u$ to $v$? Thank you.






      linear-algebra linear-transformations tensor-products multilinear-algebra tensor-rank




      linear-algebra linear-transformations tensor-products multilinear-algebra tensor-rank






      share|cite|improve this question















      share|cite|improve this question













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      edited Jan 26 at 22:03









      Ivo Terek

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      46.5k954143










      asked Jan 26 at 20:03









      SteveSteve

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      133






















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          First of all, simplify $v$ using multilinearity of $otimes$ together with the relations $e_1+e_1=e_2+e_2=0$ (since you are working over the field $Bbb F_2$). In what follows, terms with the same colors will cancel: $$begin{align} v &= e_1otimes e_1otimes e_1+e_1otimes e_1otimes e_2 + color{blue}{e_1otimes e_2otimes e_1} + e_2otimes e_1otimes e_1 \ &quad+color{green}{e_1otimes e_2otimes e_2} + color{red}{e_2otimes e_1otimes e_2}+e_2otimes e_2otimes e_1 +e_2otimes e_2otimes e_2 \ &quad + color{blue}{e_1otimes e_2otimes e_1}+color{darkgreen}{e_1otimes e_2otimes e_2} + color{red}{e_2otimes e_1otimes e_2} end{align}$$Further grouping terms, we get $$v=e_1otimes e_1otimes e_1+e_1otimes e_1otimes e_2 + e_2otimes e_2otimes e_1+e_2otimes e_2otimes e_2 + e_2otimes e_1otimes e_1$$Now, if $e_{ijk} = e_iotimes e_jotimes e_k$, consider the ordered basis of $(Bbb F_2)^{otimes 3}$ given by $$mathcal{B} = {e_{111}, e_{112}, e_{121}, e_{122}, e_{211}, e_{212}, e_{221}, e_{222}}$$Then $$[u]_{mathcal{B}} = begin{pmatrix} 1 \ 0 \ 0 \ 1 \
          0 \ 0 \1 \ 0end{pmatrix}quad mbox{and}quad [v]_{mathcal{B}} = begin{pmatrix} 1 \1 \ 0 \ 0 \1 \ 0 \ 1 \ 1end{pmatrix}.$$          Given any $8times 8$ matrix $A in {rm Mat}(8, Bbb F_2)$ which sends the first column vector into the second one, there is an unique $F in {rm Lin}((Bbb F_2)^{otimes 3})$ such that $[F]_{mathcal{B}} = A$. The map you're looking for won't be unique because the matrix $A$ will also not be. Then you only have to pick a matrix which is invertible over $Bbb F_2$. Here's one concrete matrix doing this $$A = begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 end{pmatrix}$$Identifying blocks of orders $2$, $2$ and $3$ and multiplying their determinants, it is easy to see that $det A = 1$, so the map $F$ corresponding to $A$ is actually in ${rm GL}((Bbb F_2)^{otimes 3})$. Here's how I came up with this matrix. I searched in lower dimension for non-singular matrices taking $$begin{pmatrix} 1 \ 0 end{pmatrix}mapsto begin{pmatrix} 1 \ 1 end{pmatrix}, quad begin{pmatrix} 0 \ 1 \ 0end{pmatrix} mapsto begin{pmatrix} 0 \ 0 \ 1end{pmatrix}quadmbox{and}quad begin{pmatrix} 0 \ 1 \ 0end{pmatrix} mapsto begin{pmatrix} 0 \ 1 \ 1end{pmatrix}.$$The matrix $A$ above is the direct sum of said smaller matrices. Concretely, the map $F$ is the unique linear endomorphism of $(Bbb F_2)^{otimes 3}$ satisfying $$begin{align} F(e_1otimes e_1otimes e_1) &= e_1otimes e_1 otimes(e_1+e_2) \ F(e_1otimes e_1otimes e_2) &= e_1otimes e_1otimes e_2 \ F(e_1otimes e_2otimes e_1) &= e_1otimes e_2otimes e_1 \ F(e_1otimes e_2otimes e_2) &= e_2otimes e_1 otimes e_1 \ F(e_2otimes e_1otimes e_1) &= e_1otimes e_2otimes e_2 \ F(e_2otimes e_1otimes e_2) &= e_2otimes e_1otimes e_2 \ F(e_2otimes e_2otimes e_1) &= e_2otimes e_2otimes (e_1+e_2) \ F(e_2otimes e_2otimes e_2) &= e_2otimes e_2otimes e_2.end{align}$$






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            $begingroup$

            First of all, simplify $v$ using multilinearity of $otimes$ together with the relations $e_1+e_1=e_2+e_2=0$ (since you are working over the field $Bbb F_2$). In what follows, terms with the same colors will cancel: $$begin{align} v &= e_1otimes e_1otimes e_1+e_1otimes e_1otimes e_2 + color{blue}{e_1otimes e_2otimes e_1} + e_2otimes e_1otimes e_1 \ &quad+color{green}{e_1otimes e_2otimes e_2} + color{red}{e_2otimes e_1otimes e_2}+e_2otimes e_2otimes e_1 +e_2otimes e_2otimes e_2 \ &quad + color{blue}{e_1otimes e_2otimes e_1}+color{darkgreen}{e_1otimes e_2otimes e_2} + color{red}{e_2otimes e_1otimes e_2} end{align}$$Further grouping terms, we get $$v=e_1otimes e_1otimes e_1+e_1otimes e_1otimes e_2 + e_2otimes e_2otimes e_1+e_2otimes e_2otimes e_2 + e_2otimes e_1otimes e_1$$Now, if $e_{ijk} = e_iotimes e_jotimes e_k$, consider the ordered basis of $(Bbb F_2)^{otimes 3}$ given by $$mathcal{B} = {e_{111}, e_{112}, e_{121}, e_{122}, e_{211}, e_{212}, e_{221}, e_{222}}$$Then $$[u]_{mathcal{B}} = begin{pmatrix} 1 \ 0 \ 0 \ 1 \
            0 \ 0 \1 \ 0end{pmatrix}quad mbox{and}quad [v]_{mathcal{B}} = begin{pmatrix} 1 \1 \ 0 \ 0 \1 \ 0 \ 1 \ 1end{pmatrix}.$$            Given any $8times 8$ matrix $A in {rm Mat}(8, Bbb F_2)$ which sends the first column vector into the second one, there is an unique $F in {rm Lin}((Bbb F_2)^{otimes 3})$ such that $[F]_{mathcal{B}} = A$. The map you're looking for won't be unique because the matrix $A$ will also not be. Then you only have to pick a matrix which is invertible over $Bbb F_2$. Here's one concrete matrix doing this $$A = begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 end{pmatrix}$$Identifying blocks of orders $2$, $2$ and $3$ and multiplying their determinants, it is easy to see that $det A = 1$, so the map $F$ corresponding to $A$ is actually in ${rm GL}((Bbb F_2)^{otimes 3})$. Here's how I came up with this matrix. I searched in lower dimension for non-singular matrices taking $$begin{pmatrix} 1 \ 0 end{pmatrix}mapsto begin{pmatrix} 1 \ 1 end{pmatrix}, quad begin{pmatrix} 0 \ 1 \ 0end{pmatrix} mapsto begin{pmatrix} 0 \ 0 \ 1end{pmatrix}quadmbox{and}quad begin{pmatrix} 0 \ 1 \ 0end{pmatrix} mapsto begin{pmatrix} 0 \ 1 \ 1end{pmatrix}.$$The matrix $A$ above is the direct sum of said smaller matrices. Concretely, the map $F$ is the unique linear endomorphism of $(Bbb F_2)^{otimes 3}$ satisfying $$begin{align} F(e_1otimes e_1otimes e_1) &= e_1otimes e_1 otimes(e_1+e_2) \ F(e_1otimes e_1otimes e_2) &= e_1otimes e_1otimes e_2 \ F(e_1otimes e_2otimes e_1) &= e_1otimes e_2otimes e_1 \ F(e_1otimes e_2otimes e_2) &= e_2otimes e_1 otimes e_1 \ F(e_2otimes e_1otimes e_1) &= e_1otimes e_2otimes e_2 \ F(e_2otimes e_1otimes e_2) &= e_2otimes e_1otimes e_2 \ F(e_2otimes e_2otimes e_1) &= e_2otimes e_2otimes (e_1+e_2) \ F(e_2otimes e_2otimes e_2) &= e_2otimes e_2otimes e_2.end{align}$$






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              $begingroup$

              First of all, simplify $v$ using multilinearity of $otimes$ together with the relations $e_1+e_1=e_2+e_2=0$ (since you are working over the field $Bbb F_2$). In what follows, terms with the same colors will cancel: $$begin{align} v &= e_1otimes e_1otimes e_1+e_1otimes e_1otimes e_2 + color{blue}{e_1otimes e_2otimes e_1} + e_2otimes e_1otimes e_1 \ &quad+color{green}{e_1otimes e_2otimes e_2} + color{red}{e_2otimes e_1otimes e_2}+e_2otimes e_2otimes e_1 +e_2otimes e_2otimes e_2 \ &quad + color{blue}{e_1otimes e_2otimes e_1}+color{darkgreen}{e_1otimes e_2otimes e_2} + color{red}{e_2otimes e_1otimes e_2} end{align}$$Further grouping terms, we get $$v=e_1otimes e_1otimes e_1+e_1otimes e_1otimes e_2 + e_2otimes e_2otimes e_1+e_2otimes e_2otimes e_2 + e_2otimes e_1otimes e_1$$Now, if $e_{ijk} = e_iotimes e_jotimes e_k$, consider the ordered basis of $(Bbb F_2)^{otimes 3}$ given by $$mathcal{B} = {e_{111}, e_{112}, e_{121}, e_{122}, e_{211}, e_{212}, e_{221}, e_{222}}$$Then $$[u]_{mathcal{B}} = begin{pmatrix} 1 \ 0 \ 0 \ 1 \
              0 \ 0 \1 \ 0end{pmatrix}quad mbox{and}quad [v]_{mathcal{B}} = begin{pmatrix} 1 \1 \ 0 \ 0 \1 \ 0 \ 1 \ 1end{pmatrix}.$$              Given any $8times 8$ matrix $A in {rm Mat}(8, Bbb F_2)$ which sends the first column vector into the second one, there is an unique $F in {rm Lin}((Bbb F_2)^{otimes 3})$ such that $[F]_{mathcal{B}} = A$. The map you're looking for won't be unique because the matrix $A$ will also not be. Then you only have to pick a matrix which is invertible over $Bbb F_2$. Here's one concrete matrix doing this $$A = begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 end{pmatrix}$$Identifying blocks of orders $2$, $2$ and $3$ and multiplying their determinants, it is easy to see that $det A = 1$, so the map $F$ corresponding to $A$ is actually in ${rm GL}((Bbb F_2)^{otimes 3})$. Here's how I came up with this matrix. I searched in lower dimension for non-singular matrices taking $$begin{pmatrix} 1 \ 0 end{pmatrix}mapsto begin{pmatrix} 1 \ 1 end{pmatrix}, quad begin{pmatrix} 0 \ 1 \ 0end{pmatrix} mapsto begin{pmatrix} 0 \ 0 \ 1end{pmatrix}quadmbox{and}quad begin{pmatrix} 0 \ 1 \ 0end{pmatrix} mapsto begin{pmatrix} 0 \ 1 \ 1end{pmatrix}.$$The matrix $A$ above is the direct sum of said smaller matrices. Concretely, the map $F$ is the unique linear endomorphism of $(Bbb F_2)^{otimes 3}$ satisfying $$begin{align} F(e_1otimes e_1otimes e_1) &= e_1otimes e_1 otimes(e_1+e_2) \ F(e_1otimes e_1otimes e_2) &= e_1otimes e_1otimes e_2 \ F(e_1otimes e_2otimes e_1) &= e_1otimes e_2otimes e_1 \ F(e_1otimes e_2otimes e_2) &= e_2otimes e_1 otimes e_1 \ F(e_2otimes e_1otimes e_1) &= e_1otimes e_2otimes e_2 \ F(e_2otimes e_1otimes e_2) &= e_2otimes e_1otimes e_2 \ F(e_2otimes e_2otimes e_1) &= e_2otimes e_2otimes (e_1+e_2) \ F(e_2otimes e_2otimes e_2) &= e_2otimes e_2otimes e_2.end{align}$$






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                $begingroup$

                First of all, simplify $v$ using multilinearity of $otimes$ together with the relations $e_1+e_1=e_2+e_2=0$ (since you are working over the field $Bbb F_2$). In what follows, terms with the same colors will cancel: $$begin{align} v &= e_1otimes e_1otimes e_1+e_1otimes e_1otimes e_2 + color{blue}{e_1otimes e_2otimes e_1} + e_2otimes e_1otimes e_1 \ &quad+color{green}{e_1otimes e_2otimes e_2} + color{red}{e_2otimes e_1otimes e_2}+e_2otimes e_2otimes e_1 +e_2otimes e_2otimes e_2 \ &quad + color{blue}{e_1otimes e_2otimes e_1}+color{darkgreen}{e_1otimes e_2otimes e_2} + color{red}{e_2otimes e_1otimes e_2} end{align}$$Further grouping terms, we get $$v=e_1otimes e_1otimes e_1+e_1otimes e_1otimes e_2 + e_2otimes e_2otimes e_1+e_2otimes e_2otimes e_2 + e_2otimes e_1otimes e_1$$Now, if $e_{ijk} = e_iotimes e_jotimes e_k$, consider the ordered basis of $(Bbb F_2)^{otimes 3}$ given by $$mathcal{B} = {e_{111}, e_{112}, e_{121}, e_{122}, e_{211}, e_{212}, e_{221}, e_{222}}$$Then $$[u]_{mathcal{B}} = begin{pmatrix} 1 \ 0 \ 0 \ 1 \
                0 \ 0 \1 \ 0end{pmatrix}quad mbox{and}quad [v]_{mathcal{B}} = begin{pmatrix} 1 \1 \ 0 \ 0 \1 \ 0 \ 1 \ 1end{pmatrix}.$$                Given any $8times 8$ matrix $A in {rm Mat}(8, Bbb F_2)$ which sends the first column vector into the second one, there is an unique $F in {rm Lin}((Bbb F_2)^{otimes 3})$ such that $[F]_{mathcal{B}} = A$. The map you're looking for won't be unique because the matrix $A$ will also not be. Then you only have to pick a matrix which is invertible over $Bbb F_2$. Here's one concrete matrix doing this $$A = begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 end{pmatrix}$$Identifying blocks of orders $2$, $2$ and $3$ and multiplying their determinants, it is easy to see that $det A = 1$, so the map $F$ corresponding to $A$ is actually in ${rm GL}((Bbb F_2)^{otimes 3})$. Here's how I came up with this matrix. I searched in lower dimension for non-singular matrices taking $$begin{pmatrix} 1 \ 0 end{pmatrix}mapsto begin{pmatrix} 1 \ 1 end{pmatrix}, quad begin{pmatrix} 0 \ 1 \ 0end{pmatrix} mapsto begin{pmatrix} 0 \ 0 \ 1end{pmatrix}quadmbox{and}quad begin{pmatrix} 0 \ 1 \ 0end{pmatrix} mapsto begin{pmatrix} 0 \ 1 \ 1end{pmatrix}.$$The matrix $A$ above is the direct sum of said smaller matrices. Concretely, the map $F$ is the unique linear endomorphism of $(Bbb F_2)^{otimes 3}$ satisfying $$begin{align} F(e_1otimes e_1otimes e_1) &= e_1otimes e_1 otimes(e_1+e_2) \ F(e_1otimes e_1otimes e_2) &= e_1otimes e_1otimes e_2 \ F(e_1otimes e_2otimes e_1) &= e_1otimes e_2otimes e_1 \ F(e_1otimes e_2otimes e_2) &= e_2otimes e_1 otimes e_1 \ F(e_2otimes e_1otimes e_1) &= e_1otimes e_2otimes e_2 \ F(e_2otimes e_1otimes e_2) &= e_2otimes e_1otimes e_2 \ F(e_2otimes e_2otimes e_1) &= e_2otimes e_2otimes (e_1+e_2) \ F(e_2otimes e_2otimes e_2) &= e_2otimes e_2otimes e_2.end{align}$$






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                $endgroup$



                First of all, simplify $v$ using multilinearity of $otimes$ together with the relations $e_1+e_1=e_2+e_2=0$ (since you are working over the field $Bbb F_2$). In what follows, terms with the same colors will cancel: $$begin{align} v &= e_1otimes e_1otimes e_1+e_1otimes e_1otimes e_2 + color{blue}{e_1otimes e_2otimes e_1} + e_2otimes e_1otimes e_1 \ &quad+color{green}{e_1otimes e_2otimes e_2} + color{red}{e_2otimes e_1otimes e_2}+e_2otimes e_2otimes e_1 +e_2otimes e_2otimes e_2 \ &quad + color{blue}{e_1otimes e_2otimes e_1}+color{darkgreen}{e_1otimes e_2otimes e_2} + color{red}{e_2otimes e_1otimes e_2} end{align}$$Further grouping terms, we get $$v=e_1otimes e_1otimes e_1+e_1otimes e_1otimes e_2 + e_2otimes e_2otimes e_1+e_2otimes e_2otimes e_2 + e_2otimes e_1otimes e_1$$Now, if $e_{ijk} = e_iotimes e_jotimes e_k$, consider the ordered basis of $(Bbb F_2)^{otimes 3}$ given by $$mathcal{B} = {e_{111}, e_{112}, e_{121}, e_{122}, e_{211}, e_{212}, e_{221}, e_{222}}$$Then $$[u]_{mathcal{B}} = begin{pmatrix} 1 \ 0 \ 0 \ 1 \
                0 \ 0 \1 \ 0end{pmatrix}quad mbox{and}quad [v]_{mathcal{B}} = begin{pmatrix} 1 \1 \ 0 \ 0 \1 \ 0 \ 1 \ 1end{pmatrix}.$$                Given any $8times 8$ matrix $A in {rm Mat}(8, Bbb F_2)$ which sends the first column vector into the second one, there is an unique $F in {rm Lin}((Bbb F_2)^{otimes 3})$ such that $[F]_{mathcal{B}} = A$. The map you're looking for won't be unique because the matrix $A$ will also not be. Then you only have to pick a matrix which is invertible over $Bbb F_2$. Here's one concrete matrix doing this $$A = begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 end{pmatrix}$$Identifying blocks of orders $2$, $2$ and $3$ and multiplying their determinants, it is easy to see that $det A = 1$, so the map $F$ corresponding to $A$ is actually in ${rm GL}((Bbb F_2)^{otimes 3})$. Here's how I came up with this matrix. I searched in lower dimension for non-singular matrices taking $$begin{pmatrix} 1 \ 0 end{pmatrix}mapsto begin{pmatrix} 1 \ 1 end{pmatrix}, quad begin{pmatrix} 0 \ 1 \ 0end{pmatrix} mapsto begin{pmatrix} 0 \ 0 \ 1end{pmatrix}quadmbox{and}quad begin{pmatrix} 0 \ 1 \ 0end{pmatrix} mapsto begin{pmatrix} 0 \ 1 \ 1end{pmatrix}.$$The matrix $A$ above is the direct sum of said smaller matrices. Concretely, the map $F$ is the unique linear endomorphism of $(Bbb F_2)^{otimes 3}$ satisfying $$begin{align} F(e_1otimes e_1otimes e_1) &= e_1otimes e_1 otimes(e_1+e_2) \ F(e_1otimes e_1otimes e_2) &= e_1otimes e_1otimes e_2 \ F(e_1otimes e_2otimes e_1) &= e_1otimes e_2otimes e_1 \ F(e_1otimes e_2otimes e_2) &= e_2otimes e_1 otimes e_1 \ F(e_2otimes e_1otimes e_1) &= e_1otimes e_2otimes e_2 \ F(e_2otimes e_1otimes e_2) &= e_2otimes e_1otimes e_2 \ F(e_2otimes e_2otimes e_1) &= e_2otimes e_2otimes (e_1+e_2) \ F(e_2otimes e_2otimes e_2) &= e_2otimes e_2otimes e_2.end{align}$$







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                edited Jan 27 at 1:07

























                answered Jan 26 at 21:53









                Ivo TerekIvo Terek

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