Lecture 12: Some remarks about Echelon form Definition: An m n matrix (aij)1im , 1jn is in Echelon form if the following Summary: Lecture 12: Some remarks about Echelon form Definition: An m × n matrix (aij)1im , 1jn is in Echelon form if the following conditions are satisfied: (1) If there is 1 i m with ai1, . . . , ain = 0 then for all k i and 1 j n we have akj = 0. (2) Assume that 1 i m such that (ai1, . . . , ain) = (0, . . . , 0) and (ai-1,1, . . . , ai-1,n) = (0, . . . , 0) Then min{j | aij = 0} > min{j | ai-i,j = 0}. Theorem: For any m×n matrix A = (aij)1im , 1jn there is a natural number k and elementary matrices E1, . . . Ek such that E1 · · · EkA is in Echelon form. Proof: Since multiplication from the left by elementary matrices is equivalent to performing row operations it suffices to describe the necessary row operations. If the matrix A has all zero entries then it is already in Echelon form and there is nothing to prove. Hence we may assume that there is an entry aij = 0. Let 1 j0 n be the index of the first column containing a non-zero entry. More precisely, j0 = min{1 j n | 1 i m : aij = 0} by definition of j0 there exists 1 i0 m such that ai0j0 = 0. If i0 = 1 then exchange the first row with row #i0 such that a1j0 = 0. If already i0 = 1 then this Collections: Mathematics