Linear Algebra

54 Chapter One. Linear SystemsBy the inductive hypothesis the coefficients c 1 , . . . c i are all 0 so the equationreduces to 0 = c i+1 r i+1,li+1 + · · · + c m r m,li+1 . As in the base case, because thematrix is in echelon form r i+2,li+1 = · · · = r m,li+1 = 0 and r i+1,li+1 ≠ 0. Thusc i+1 = 0.QED2.6 Theorem Each matrix is row equivalent to a unique reduced echelon formmatrix.Proof [Yuster] Fix a number of rows m. We will proceed by induction on thenumber of columns n.The base case is that the matrix has n = 1 column. If this is the zero matrixthen its unique echelon form is the zero matrix. If instead it has any nonzeroentries then when the matrix is brought to reduced echelon form it must haveat least one nonzero entry, so it has a 1 in the first row. Either way, its reducedechelon form is unique.For the inductive step we assume that n > 1 and that all m row matriceswith fewer than n columns have a unique reduced echelon form. Consider anm×n matrix A and suppose that B and C are two reduced echelon form matricesderived from A. We will show that these two must be equal.Let  be the matrix consisting of the first n − 1 columns of A. Observethat any sequence of row operations that bring A to reduced echelon form willalso bring  to reduced echelon form. By the inductive hypothesis this reducedechelon form of  is unique, so if B and C differ then the difference must occurin their n-th columns.We finish the inductive step, and the argument, by showing that the twocannot differ only in that column. Consider a homogeneous system of equationsfor which A is the matrix of coefficients.a 1,1 x 1 + a 1,2 x 2 + · · · + a 1,n x n = 0a 2,1 x 1 + a 2,2 x 2 + · · · + a 2,n x n = 0.a m,1 x 1 + a m,2 x 2 + · · · + a m,n x n = 0(∗)By Theorem One.I.1.5 the set of solutions to that system is the same as the setof solutions to B’s systemand to C’s.b 1,1 x 1 + b 1,2 x 2 + · · · + b 1,n x n = 0b 2,1 x 1 + b 2,2 x 2 + · · · + b 2,n x n = 0.b m,1 x 1 + b m,2 x 2 + · · · + b m,n x n = 0c 1,1 x 1 + c 1,2 x 2 + · · · + c 1,n x n = 0c 2,1 x 1 + c 2,2 x 2 + · · · + c 2,n x n = 0.c m,1 x 1 + c m,2 x 2 + · · · + c m,n x n = 0(∗∗)(∗∗∗)