Linear Algebra

92 Chapter Two. Vector Spaces2.17 Example The span of this set is all of R 2 .( ) ( )1 1{ , }1 −1To check this we must show that any member of R 2 is a linear combination ofthese two vectors. So we ask: for which vectors (with real components x and y)are there scalars c 1 and c 2 such that this holds?( ) ( ) ( )1 1 xc 1 + c 2 =1 −1 yGauss’s Methodc 1 + c 2 = xc 1 − c 2 = y−ρ 1 +ρ 2−→c 1 + c 2 = x−2c 2 = −x + ywith back substitution gives c 2 = (x − y)/2 and c 1 = (x + y)/2. These twoequations show that for any x and y there are appropriate coefficients c 1 and c 2making the above vector equation true. For instance, for x = 1 and y = 2 thecoefficients c 2 = −1/2 and c 1 = 3/2 will do. That is, we can write any vector inR 2 as a linear combination of the two given vectors.Since spans are subspaces, and we know that a good way to understand asubspace is to parametrize its description, we can try to understand a set’s spanin that way.2.18 Example Consider, in P 2 , the span of the set {3x − x 2 , 2x}. By the definitionof span, it is the set of unrestricted linear combinations of the two{c 1 (3x − x 2 ) + c 2 (2x) ∣ ∣ c1 , c 2 ∈ R}. Clearly polynomials in this span must havea constant term of zero. Is that necessary condition also sufficient?We are asking: for which members a 2 x 2 + a 1 x + a 0 of P 2 are there c 1 andc 2 such that a 2 x 2 + a 1 x + a 0 = c 1 (3x − x 2 ) + c 2 (2x)? Since polynomials areequal if and only if their coefficients are equal, we are looking for conditions ona 2 , a 1 , and a 0 satisfying these.−c 1 = a 23c 1 + 2c 2 = a 10 = a 0Gauss’s Method gives that c 1 = −a 2 , c 2 = (3/2)a 2 +(1/2)a 1 , and 0 = a 0 . Thusthe only condition on polynomials in the span is the condition that we knewof — as long as a 0 = 0, we can give appropriate coefficients c 1 and c 2 to describethe polynomial a 0 + a 1 x + a 2 x 2 as in the span. For instance, for the polynomial0 − 4x + 3x 2 , the coefficients c 1 = −3 and c 2 = 5/2 will do. So the span of thegiven set is {a 1 x + a 2 x 2 ∣ ∣ a1 , a 2 ∈ R}.This shows, incidentally, that the set {x, x 2 } also spans this subspace. A spacecan have more than one spanning set. Two other sets spanning this subspaceare {x, x 2 , −x + 2x 2 } and {x, x + x 2 , x + 2x 2 , . . . }. (Naturally, we usually preferto work with spanning sets that have only a few members.)