Linear Algebra

54 Chapter One. Linear SystemsIn the base step, that zero reduction operations suffice, the two matricesare equal and each row of B is obviously a combination of A’s rows: βi ⃗ =0 · ⃗α 1 + · · · + 1 · ⃗α i + · · · + 0 · ⃗α m .For the inductive step, assume the inductive hypothesis: with t ≥ 0, if amatrix can be derived from A in t or fewer operations then its rows are linearcombinations of the A’s rows. Consider a B that takes t+1 operations. Becausethere are more than zero operations, there must be a next-to-last matrix G sothat A −→ · · · −→ G −→ B. This G is only t operations away from A and so theinductive hypothesis applies to it, that is, each row of G is a linear combinationof the rows of A.If the last operation, the one from G to B, is a row swap then the rowsof B are just the rows of G reordered and thus each row of B is also a linearcombination of the rows of A. The other two possibilities for this last operation,that it multiplies a row by a scalar and that it adds a multiple of one row toanother, both result in the rows of B being linear combinations of the rows ofG. But therefore, by the Linear Combination Lemma, each row of B is a linearcombination of the rows of A.With that, we have both the base step and the inductive step, and so theproposition follows.QED2.4 Example In the reduction( ) ( )0 2 ρ 1↔ρ 2 1 1−→1 1 0 2( ) ( )(1/2)ρ 2 1 1 −ρ 2+ρ 1 1 0−→ −→0 1 0 1call the matrices A, D, G, and B. The methods of the proof show that thereare three sets of linear relationships.δ 1 = 0 · α 1 + 1 · α 2 γ 1 = 0 · α 1 + 1 · α 2 β 1 = (−1/2)α 1 + 1 · α 2δ 2 = 1 · α 1 + 0 · α 2 γ 2 = (1/2)α 1 + 0 · α 2 β 2 = (1/2)α 1 + 0 · α 2The prior result gives us the insight that Gauss’ method works by takinglinear combinations of the rows. But to what end; why do we go to echelonform as a particularly simple, or basic, version of a linear system? The answer,of course, is that echelon form is suitable for back substitution, because we haveisolated the variables. For instance, in this matrix⎛⎞2 3 7 8 0 0R = ⎜0 0 1 5 1 1⎟⎝0 0 0 3 3 0⎠0 0 0 0 2 1x 1 has been removed from x 5 ’s equation. That is, Gauss’ method has made x 5 ’srow independent of x 1 ’s row.Independence of a collection of row vectors, or of any kind of vectors, willbe precisely defined and explored in the next chapter. But a first take on it isthat we can show that, say, the third row above is not comprised of the other