text: The matrix  of  is the -submatrix of  which is obtained by removing the last  rows of zeros in the submatrix of the columns 1 to  and  to  of  (that is removing  columns in each block and the  last rows of zeros). The principal subresultant coefficient  is the determinant of the  first rows of .
  Let  be the  matrix defined as follows. First we add  columns of zeros to the right of the  identity matrix. Then we border the bottom of the resulting matrix by a row consisting in  zeros followed by :
  With this notation, the -th subresultant polynomial is the determinant of the matrix product . Its coefficient of degree  is the determinant of the square submatrix of Ti consisting in its  first rows and the -th row.
  Sketch of the proof
  It is not obvious that, as defined, the subresultants have the desired properties. Nevertheless, the proof is rather simple if the properties of linear algebra and those of polynomials are put together.
  As defined, the columns of the matrix  are the vectors of the coefficients of some polynomials belonging to the image of . The definition of the -th subresultant polynomial  shows that the vector of its coefficients is a linear combination of these column vectors, and thus that Si belongs to the image of
  If the degree of the GCD is greater than , then Bézout's identity shows that every non zero polynomial in the image of  has a degree larger than . This implies that .