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, N of a polynomial R such that R(0) = 0, from which we deduce that M 3 = ((R(N ) ?1 ) T , R(N )) and M 3 ? Stab(T )

, Given the form of the elements of Stab(I ? ) ? Stab(N ), its cardinality is equal to the number of polynomials R of degree at most ? ? 1 such that R(0) = 0

A. Appendix, Stabilizer of the matrix product We denote by T p,q,r the vector space given by the product of matrices p × q by q × r, which is isomorphic to M p,r ? I q (we do not use the canonical basis for this representation), For the group action M ·

P. , Q. ?-;-for-p-?-gl-p, R. , and Q. , Theorem 17. For the group action M · (X, Y ) ? X T M Y , the subgroup stabilizing the vector space T p,q,r can be described as the group given by the pairs

X. T-t-p,q,r-y-=-t-p,q,r.-for-any-i-?-{0, .. .. {0, and .. , q ? 1}, we denote by M i,j the matrix X T · (e i,j ) · Y , where e i,j is the canonical basis of M p,r. Denoting by X i,h the q × q blocks of X and Y ?,j the q × q blocks of Y , we have M i,j = (X i,h Y j,? ) h,? for any i and j. Consequently, since X T · (e i,j ) · Y ? T p,q,r , we have ?i

, Let (i, h) such that X i,h is not null and j any integer in {0,. .. , q ? 1}. We have the inclusion X i

, Thus, for any (i, h) such that X i,h is not null, we have shown that X i,h is invertible. We have the same property for the blocks of Y. Combining the fact that the blocks of X and Y that are not null are invertible and Equation (A.1), we can conclude that the stabilizer of T p,q,r is generated by matrices, Y is invertible, we even have the equality

B. Appendix, Using the Hamming weight for the matrix product We describe in this section a trick allowing one to speed-up the execution of our approach for the matrix product. However, this part is technical and can be skipped on a first read

C. =-{span,

, We define the following sets