Linear Algebra

Section I. Definition of Vector Space 97
Since spans are subspaces, and we know that a good way to understand a
subspace is to paramatrize its description, we can try to understand a set’s span
in that way.
2.18 Example Consider, in P2, the span of the set {3x − x 2 , 2x}. By the
definition of span, it is the subspace of unrestricted linear combinations of the
two {c1(3x − x 2 )+c2(2x) � � c1,c2 ∈ R}. Clearly polynomials in this span must
have a constant term of zero. Is that necessary condition also sufficient?
We are asking: for which members a2x 2 + a1x + a0 of P2 are there c1 and c2
such that a2x 2 + a1x + a0 = c1(3x − x 2 )+c2(2x)? Since polynomials are equal
if and only if their coefficients are equal, we are looking for conditions on a2,
a1, anda0 satisfying these.
−c1 = a2
3c1 +2c2 = a1
0=a0
Gauss’ method gives that c1 = −a2, c2 =(3/2)a2 +(1/2)a1, and0=a0. Thus
the only condition on polynomials in the span is the condition that we knew
of — as long as a0 = 0, we can give appropriate coefficients c1 and c2 to describe
the polynomial a0 + a1x + a2x 2 as in the span. For instance, for the polynomial
0 − 4x +3x 2 , the coefficients c1 = −3 andc2 =5/2 will do. So the span of the
given set is {a1x + a2x 2 � � a1,a2 ∈ R}.
This shows, incidentally, that the set {x, x 2 } also spans this subspace. A
space can have more than one spanning set. Two other sets spanning this subspace
are {x, x 2 , −x +2x 2 } and {x, x + x 2 ,x+2x 2 ,...}. (Naturally, we usually
prefer to work with spanning sets that have only a few members.)
2.19 Example These are the subspaces of R 3 that we now know of, the trivial
subspace, the lines through the origin, the planes through the origin, and the
whole space (of course, the picture shows only a few of the infinitely many
subspaces). In the next section we will prove that R 3 has no other type of
subspaces, so in fact this picture shows them all.
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