06.09.2021 Views

Linear Algebra, 2020a

Linear Algebra, 2020a

Linear Algebra, 2020a

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

Section I. Definition of Vector Space 101<br />

is a linear combination of elements of S and so is an element of [S] (possibly<br />

some of the ⃗s i ’s from ⃗v equal some of the ⃗s j ’s from ⃗w but that does not matter).<br />

QED<br />

The converse of the lemma holds: any subspace is the span of some set,<br />

because a subspace is obviously the span of itself, the set of all of its members.<br />

Thus a subset of a vector space is a subspace if and only if it is a span. This<br />

fits the intuition that a good way to think of a vector space is as a collection in<br />

which linear combinations are sensible.<br />

Taken together, Lemma 2.9 and Lemma 2.15 show that the span of a subset<br />

S of a vector space is the smallest subspace containing all of the members of S.<br />

2.16 Example In any vector space V, for any vector ⃗v ∈ V, the set {r · ⃗v | r ∈ R}<br />

is a subspace of V. For instance, for any vector ⃗v ∈ R 3 the line through the<br />

origin containing that vector {k⃗v | k ∈ R} is a subspace of R 3 . This is true even<br />

if ⃗v is the zero vector, in which case it is the degenerate line, the trivial subspace.<br />

2.17 Example The span of this set is all of R 2 .<br />

( ) ( )<br />

1 1<br />

{ , }<br />

1 −1<br />

We know that the span is some subspace of R 2 . To check that it is all of R 2 we<br />

must show that any member of R 2 is a linear combination of these two vectors.<br />

So we ask: for which vectors with real components x and y are there scalars c 1<br />

and c 2 such that this holds?<br />

( ) ( ) ( )<br />

1 1 x<br />

c 1 + c 2 =<br />

(∗)<br />

1 −1 y<br />

Gauss’s Method<br />

c 1 + c 2 = x<br />

c 1 − c 2 = y<br />

−ρ 1 +ρ 2<br />

−→<br />

c 1 + c 2 = x<br />

−2c 2 =−x + y<br />

with back substitution gives c 2 =(x − y)/2 and c 1 =(x + y)/2. This shows<br />

that for any x, y there are appropriate coefficients c 1 ,c 2 making (∗) true — we<br />

can write any element of R 2 as a linear combination of the two given ones. For<br />

instance, for x = 1 and y = 2 the coefficients c 2 =−1/2 and c 1 = 3/2 will do.<br />

Since spans are subspaces, and we know that a good way to understand a<br />

subspace is to parametrize its description, we can try to understand a set’s span<br />

in that way.<br />

2.18 Example Consider, in the vector space of quadratic polynomials P 2 , the<br />

span of the set S = {3x − x 2 ,2x}. By the definition of span, it is the set of

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!