Linear Algebra

Section III. Reduced Echelon Form 51(c) Expand the proof of that lemma to make explicit exactly where it uses thei ≠ j condition on combining.III.2The Linear Combination LemmaWe will close this section and this chapter by proving that every matrix is rowequivalent to one and only one reduced echelon form matrix. The ideas thatappear here will reappear, and be further developed, in the next chapter.The crucial observation concerns how row operations act to transform onematrix into another: they combine the rows linearly.2.1 Example In this reduction( )2 1 01 3 5−→2 1 00 5/2 5( )−(1/2)ρ 1 +ρ 2((1/2)ρ 1−→(2/5)ρ 2)1 1/2 00 1 2( )1 0 −1−→0 1 2−(1/2)ρ 2 +ρ 1denoting those matrices A → D → G → B and writing the rows of A as α 1 andα 2 , etc., we have this.( )()α 1 −(1/2)ρ 1 +ρ 2 δ 1 = α 1−→α 2 δ 2 = −(1/2)α 1 + α 2((1/2)ρ 1−→(2/5)ρ 2(−(1/2)ρ 2 +ρ 1−→)γ 1 = (1/2)α 1γ 2 = −(1/5)α 1 + (2/5)α 2)β 1 = (3/5)α 1 − (1/5)α 2β 2 = −(1/5)α 1 + (2/5)α 22.2 Example This also holds if there is a row swap. With this A, D, G, and B( ) ( ) ( ) ( )0 2 ρ 1 ↔ρ 2 1 1 (1/2)ρ 2 1 1 −ρ 2 +ρ 1 1 0−→ −→ −→1 1 0 20 10 1we get these linear relationships.() ( ) (⃗α 1 ρ 1 ↔ρ 2 ⃗ δ 1 = ⃗α 2 (1/2)ρ 2−→ −→⃗α 2⃗δ 2 = ⃗α 1)⃗γ 1 = ⃗α 2⃗γ 2 = (1/2)⃗α 1( )−ρ 2 +ρ 1 β ⃗ 1 = (−1/2)⃗α 1 + 1 · ⃗α 2−→⃗β 2 = (1/2)⃗α 1In summary, Gauss’s Method systemmatically finds a suitable sequence oflinear combinations of the rows.