Linear Algebra

56 Chapter One. Linear SystemsWe can now prove that each matrix is row equivalent to one and only onereduced echelon form matrix. We will find it convenient to break the first halfof the argument off as a preliminary lemma. For one thing, it holds for anyechelon form whatever, not just reduced echelon form.2.6 Lemma If two echelon form matrices are row equivalent then the leadingentries in their first rows lie in the same column. The same is true of all thenonzero rows — the leading entries in their second rows lie in the same column,etc.For the proof we rephrase the result in more technical terms. Define the formof an m×n matrix to be the sequence 〈l 1 , l 2 , . . . , l m 〉 where l i is the columnnumber of the leading entry in row i and l i = ∞ if there is no leading entry inthat row. The lemma says that if two echelon form matrices are row equivalentthen their forms are equal sequences.Proof. Let B and D be echelon form matrices that are row equivalent. Becausethey are row equivalent they must be the same size, say m×n. Let the columnnumber of the leading entry in row i of B be l i and let the column number ofthe leading entry in row j of D be k j . We will show that l 1 = k 1 , that l 2 = k 2 ,etc., by induction.This induction argument relies on the fact that the matrices are row equivalent,because the Linear Combination Lemma and its corollary therefore givethat each row of B is a linear combination of the rows of D and vice versa:β i = s i,1 δ 1 + s i,2 δ 2 + · · · + s i,m δ m and δ j = t j,1 β 1 + t j,2 β 2 + · · · + t j,m β mwhere the s’s and t’s are scalars.The base step of the induction is to verify the lemma for the first rows ofthe matrices, that is, to verify that l 1 = k 1 . If either row is a zero row thenthe entire matrix is a zero matrix since it is in echelon form, and therefore bothmatrices are zero matrices (by Corollary 2.3), and so both l 1 and k 1 are ∞. Forthe case where neither β 1 nor δ 1 is a zero row, consider the i = 1 instance ofthe linear relationship above.(0 · · · b1,l1 · · ·β 1 = s 1,1 δ 1 + s 1,2 δ 2 + · · · + s 1,m δ m) ( )= s1,1 0 · · · d1,k1 · · ·( )+ s 1,2 0 · · · 0 · · · .( )+ s 1,m 0 · · · 0 · · ·First, note that l 1 < k 1 is impossible: in the columns of D to the left of columnk 1 the entries are all zeroes (as d 1,k1 leads the first row) and so if l 1 < k 1 thenthe equation of entries from column l 1 would be b 1,l1 = s 1,1 · 0 + · · · + s 1,m · 0,but b 1,l1 isn’t zero since it leads its row and so this is an impossibility. Next,a symmetric argument shows that k 1 < l 1 also is impossible. Thus the l 1 = k 1base case holds.