 
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 (ai1,1, . . . , ai1,n) = (0, . . . , 0)
Then
min{j  aij = 0} > min{j  aii,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 nonzero 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
