Math 2250-4
Tues Oct 8
4.1 - 4.3 Concepts related to linear combinations of vectors, continued: span, linear
independence/dependence, basis, vector space, subspace.
Continue our discussion from Monday. Can you remember this new terminology, which you definitely
want to learn:
A linear combination of the vectors v1 , v2 , ... vn is
The span of v1 , v2 , ... vn is
, Recall that for vectors in =m all linear combination questions can be reduced to matrix (i.e. linear
system) questions because any linear combination like the one on the left is actually just the matrix product
on the right:
a
a
c
1
21
:
a
a
11
m1
a
Cc
2
22
:
a
a
12
a
C...C c
n
0
a
m2
2n
c a
Cc a
C...
c a
Cc a
C ...
1 11
1n
=
mn
1 21
2 12
2 22
1 m1
a
=
:
c a
a
11
21
a
a
:
Cc a
2 m2
C...
a
m1
12
22
:
a
m2
... a
... a
:
... a
1n
2n
:
mn
c
c
1
2
:
c
=Ac .
n
important: The vectors in the linear combination expression on the left are precisely the ordered columns
of the matrix A, and the linear combination coefficients are the ordered entries in the vector c
, Finish Monday's notes; we completed Exercises 1-3 on Monday, and were about to start discussing the
complementary concepts of linear independence and linear dependence, for collections of vectors
v1 , v2 , ... vn . After completing Monday's notes continue the conversation, below.
Exercise 1) Are these four vectors a basis for =3 ?
1
v =
1
0 , v =
2
2
K1
2 ,v =
3
0
K1
6 ,v =
4
4
3
8
66
Exercise 2) For linearly independent vectors v1 , v2 , ... vn , every vector v in their span can be written as
v = d1 v1 C d2 v2 C ... C dn vn uniquely, i.e. for exactly one choice of linear combination coefficients d1 , d2 ,
...dn . This is not true if vectors are dependent. Explain these facts. (This is why independent vectors
v1 , v2 , ... vn are good to have, because there is no redundancy in how we may express vectors in their
span; this is why we call such collections a basis for their span v1 , v2 , ... vn .)
Exercise 3) Use properties of reduced row echelon form matrices to answer the following questions:
3a) Why must more than two vectors in =2 always be linearly dependent?
3b) Why can fewer than two vectors (i.e. one vector) not span =2 ?
(we deduce from 3a,b that every basis of =2 must have exactly two vectors.)
3c) If v1 , v2 are any two vectors in =2 what is the condition on the reduced row echelon form of the 2 # 2
matrix v1 v2 that guarantees they're linearly independent? That guarantees they span =2 ? That
guarantees they're a basis for =2 ?
Exercise 4) Use properties of reduced row echelon form matrices to answer the following questions:
4a) Why must more than 3 vectors in =3 always be linearly dependent?
4b) Why can fewer than 3 vectors never span =3 ?
(So every basis of =3 must have exactly three vectors.)
4c) If you are given 3 vectors v1 , v2 , v3 in =3 , what is the condition on the reduced row echelon form of
the 3 # 3 matrix v1 v2 v3 that guarantees they're linearly independent? That guarantees they span =3 ?
That guarantees they're a basis of =3 ?
Exercise 5) Let v1 , v2 2 =3 be a basis for the plane W which is their span.
5a) Is it possible to pick more than two linearly independent vectors in W?
5b) Could fewer than two vectors span W ?
5c) How many vectors must each possible basis of W contain?
New vocabulary for today:
Definition: A vector space V is a collection of objects (called "vectors"), together with two operations:
addition "+" and scalar multiplication "⋅" so that the following axioms hold:
(α) Whenever f, g 2 V then f C g 2 V . (closure with respect to addition)
(β) Whenever f 2 V and c 2 =, then c$f 2 V. (closure with respect to scalar
multiplication)
As well as:
(a) f C g = g C f (commutative property)
(b) f C g C h = f C g C h (associative property)
(c) d 0 2 V so that f C 0 = f is always true. (additive identity)
(d) c f 2 V dKf 2 V so that f C Kf = 0 (additive inverses)
(e) c$ f C g = c$f C c$g (scalar multiplication distributes over vector addition)
(f) c1 C c2 $f = c1 $f C c2 $f (scalar addition distributes over scalar multiplication)
(g) c1 $ c2 $f = c1 c2 $f (associative property)
(h) 1$f = f, K1 $f =Kf, 0$f = 0 (these last two actually follow from the others).
Remark: If you look at those properties, they are exactly the ones we've been using in order to rewrite and
simplify linear combinations of vectors in =m . If you look at them again, you'll see that they also hold for
the addition and scalar multiplication of functions.
Definition: A subspace W of a vector space V is a (special) subset of V. In order to be called a subspace,
this subset must be closed under addition and scalar multiplication, i.e. properties a, b above hold. As a
result, W is itself a vector space, because it follows from a, b and the fact that W is a part of V that
properties (a)-(h) also hold for all elements of W and all scalars in = .
Exercise 6) The span of any collection of vectors v1 , v2 , ... vn in =m (or actually the span of any
f1 , f2 , ... fn in some vector space V) is closed under addition and scalar multiplication, and so is a subspace
of =m (or V . Show this.
Exercise 7)
7a) W =
7b) W =
7c) W =
Most subsets of a vector space V are actually not subspaces. Show that
x, y T 2 =2 s.t. x2 C y2 = 4 is not a subspace of =2 .
x, y T 2 =2 s.t. y = 3 x C 1 is not a subspace of =2 .
x, y T 2 =2 s.t. y = 3 x is a subspace of =2 . Then find a basis for this subspace.