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LECTURE 12:
RN A N D V E C T O R S PA C E S
Monday, October 3
In the last few lectures we considered various properties of matrices
in order to better understand the coefficient matrices of our linear
systems. In this lecture we’re going to consider shift gears and
consider the solutions of our linear systems.
Suppose we have a linear system with m equations and n variables.
Then we can express this system as
Ax “ b,
where A is the coefficient matrix of the linear system and b is the
constant vector. The solution to this system, if it exists, is the vector
» fi
x1
— x2 ffi
— ffi
x “ — .. ffi
– . fl
xn
where each of the xi are real numbers. That is, x is a n-tuple of real
numbers. Since writing column vectors as above takes up a lot of
space, we often them as follows:
» fi
x1
— x2 ffi
“
‰T
— ffi
— .. ffi “ px1 , x2 , . . . , xn q “ x1 x2 ¨ ¨ ¨ xn ,
– . fl
xn
where the superscript of T denotes the transpose.
Definition 0.1. We denote the set of all n-tuples of real numbers by
Rn . In set builder notation
Rn “ tpx1 , x2 , . . . , xn q : xi P Ru
In the next section we want to study the set Rn , in particular the
algebraic structure of this set, which we call a vector space.
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vector spaces
To motivate the definition of a vector space, let’s consider a set whose
elements are particularly easy to visualize, namely
R2 “ tpx, yq : x, y P Ru.
An element px, yq in the set R2 can be thought of as the arrow from
the origin p0, 0q to the given point px, yq, and it is this perspective
in which these elements are often considered. These arrows were
typically used to represent physical notations like force or velocity,
because they give a direction (whichever way the arrow is pointing)
and a magnitude (the length of the arrow). But their utility doesn’t
end there. For example, if I were to apply a force to an object, let’s
say pushing a chair to the north side of a room, and at the same time
you were to push the chair to the west side of the room, then the
chair will move northwest. The resulting arrow can be obtained by
adding the two arrows giving our respective contributions. How do
we add elements of R2 ? Componentwise, that is, if v1 “ px1 , y1 q and
v2 “ px2 , y2 q are two elements of R2 , then
v1 ` v2 “ px1 ` x2 , y1 ` y2 q.
v1 ` v2
v1
v2
One can show that this vector addition is commutative and associative.
In addition to adding two elements from R2 together, we can also
scale these elements. If a P R and v “ px, yq P R2 , then av “
pax, ayq P R2 . Depending on the value of a, the resulting arrow av
could be longer or shorter than the original vector v.
2
2¨v
v
1
2
¨v
So to summarize, the set R2 can be thought to consist of sets of
arrows that we can add together and scale by multiplying by a
constant. We want to generalize this basic structure to arbitrary sets.
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definition of an abstract vector space
Definition 2.1. A vector space V over R is a set with two operations:
pvector additionq A function that assigns to each v, w P V an
element v ` w P V.
pscalar multiplicationq A function that assigns to each pair
a P R and v P V an element av P V.
such that the following conditions are satisfied:
pcommutativityq
v ` w “ w ` v for all v, w P V
passociativityq
pv ` wq ` u “ v ` pw ` uq for all v, w, u P V,
apbvq “ pabqv for all v P V and a, b P R.
padditive identityq
There exists an element 0 P V, such that v ` 0 “ v for
all v P V.
padditive inverseq
For every v P V, there exists an element w P V such
that v ` w “ 0.
pmultiplicative identityq
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1v “ v for all v P V.
pdistributive propertiesq
apv ` wq “ av ` aw for all v, w P V and a P R,
pa ` bqv “ av ` bv for all v P V and a, b P R.
An element of V is called a vector.
An element of R is called a scalar.
Now, that was a long and abstract definition, so let’s look at some
concrete examples to help make sense of it. We start with our
motivating example.
Example 1. Recall that
Rn “ tpx1 , . . . , xn q : xi P Ru.
If we define vector addition and scalar multiplication by
px1 , . . . , xn q ` py1 , . . . , yn q “ px1 ` y1 , . . . , xn ` yn q
apx1 , . . . , xn q “ pax1 , . . . , axn q,
one can check (and you should!) that Rn is a vector space over the
field R.
Example 2. Denote the set of all polynomials with coefficients in
R by PpRq. That is, every element of the set PpRq is of the form
an xn ` ¨ ¨ ¨ a1 x ` a0
for some n ě 0, where ai P R for all i. If we define vector addition
and scalar multiplication is the natural way, i.e.
pp ` qqpxq “ ppxq ` qpxq
papqpxq “ appxq
then PpRq is a vector space over R.
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Example 3. Let M2 pRq denote the set of 2 ˆ 2 matrices with entries from R. Then M2 pRq is a vector space over R with respect to
vector addition and scalar multiplication given by
ˆ
˙ ˆ
˙
ˆ
˙
a b
e f
a`e b`f
`
“
c d
g h
c`g d`h
ˆ
˙
ˆ
˙
αa αb
a b
α
“
.
c d
αc αd
Let’s look at a couple of examples that aren’t vector spaces.
Example 4. Let’s once again consider the set Rn . Suppose we
define vector addition and scalar multiplication by
pa1 , . . . , an q ` pb1 , . . . , bn q “ pa1 ´ b1 , . . . , an ´ bn q
apa1 , . . . , an q “ paa1 , . . . , aan q.
Is Fn with these operations still a vector space? No, because, for
example, addition is not commutative.
Example 5. Let V “ R. Define addition and scalar multiplication
as follows
u ` v “ uv
av “ av.
Is V a vector space over R? No. Note that in this case, the additive
identity would be 1 since
v`1 “ v¨1 “ v
for all v P V. However, this means that there is no real number v
that can be the additive inverse of 0 since
0`v “ 0¨v “ 0 ‰ 1
for all v. These operations would also fail to satisfy the distributive
properties.
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subspaces
Definition 3.1. Let V be a vector space over R. A subset U Ă V is a
subspace of V if U is a vector space with respect to the addition and
scalar multiplication defined on V.
Example 6. Let n ě 1 be an integer. Recall that the set
Rn “ tpx1 , . . . , xn q : xi P Fu
is a vector space with addition and scalar multiplication given by
px1 , . . . , xn q ` py1 , . . . , yn q “ px1 ` y1 , . . . , xn ` yn q
apx1 , . . . , xn q “ pax1 , . . . , axn q.
Consider the subset
U “ tpx1 , . . . , xn´1 , 0q : xi P Ru Ă Rn .
One can verify that U is vector space over R under the above
operations of addition and scalar multiplication, that is, U is a
subspace of Rn .
Example 7. Recall that the set of polynomials with coefficients in
R, which we denote PpRq, is a vector space over R with addition
and scalar multiplication given by
pp ` qqpxq “ ppxq ` qpxq
papqpxq “ appxq.
Consider the subset
U “ tppxq P PpFq : pp1q “ 0u Ă PpRq.
One can verify that U is vector space over R under the above
operations of addition and scalar multiplication, that is, U is a
subspace of PpRq. We’ll also prove this later.
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Example 8. Suppose we have a homogeneous linear system win
m equations and n variables. Then we can express this system as
Ax “ ~0,
where ~0 here denotes the zero vector. The solutions to this system
form a subspace of Rn
It turns out that verifying a subset of a vector space is a subspace is
easier than verifying that a set is a vector space. The reason for this
is that many of the conditions that a set must satisfied in order to be
a vector space, are automatically satisfied by virtue of the fact that
we are considering a subset of a vector space. In fact, we only need
to check three things!
Theorem 3.1. Let V be a vector space , and let U be a subset of
V. Then U is a subspace of V if and only if the following three
conditions hold:
(i) U is non-empty,
(ii) x ` y P U for all x, y P U,
(iii) ax P U for all x P U and a P R.
In other words, in order for a subset U of a vector space V to be a
subspace, it must be non-empty, closed under addition, and closed
under scalar multiplication.
Proof of Theorem 3.1. The statement of this theorem is an “if and only
if” statement, therefore we have to show two things. First, we have
to show that if U is a subspace of V, then it satisfies conditions (i)-(iii)
(this is the “only if” part). Second, we must also show that if U
satisfies conditions (i)-(iii), then it is a subspace of V (this is the “if”
part). Let’s begin with the “only if” part.
Suppose U is a subspace of V. Then U is a vector space, which
implies it has an additive identity. Therefore, U is non-empty. Furthermore, the fact that U is a vector space implies that it is closed
under addition and scalar multiplication, that is, it satisfies (ii) and
(iii). Next, we consider the “if” part.
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Suppose U satisfies conditions (i)-(iii). Since U is a subset of V, we
know that
• addition is associative and commutative,
• 1v “ v for all v P V,
• addition and scalar multiplication satisfy the distributive properties.
Therefore, all we need to show is that U has an additive identity, and
every v P U has an additive inverse. Let v P U. By (iii) we know that
both 0v and ´1v are elements of U, and we know that 0v “ 0 and
´1v “ ´v. Hence, U has an additive identity and every element of
U has an additive inverse.
When you want to show that a subset U of a vector space V is
non-empty, it is often easiest to show that U contains the additive
identity of V. In fact, we could replace the first condition of the
above theorem with 0 P U, where 0 is the additive identity of V.
Let’s use the above theorem to show that the subsets we considered
in Examples 6 and 8 are subspaces.
Example 9. Recall the subset
U “ tpx1 , . . . , xn´1 , 0q : xi P Fu Ă Rn .
To show that U is a subspace of Rn , we first want to show that U
is non-empty. To do so, we note that
p0, . . . , 0q P U.
Next, we want to show that U is closed under addition. Let
px1 , . . . , xn´1 , 0q, py1 , . . . , yn´1 , 0q P U, then we have
px1 , . . . , xn´1 , 0q ` py1 , . . . , yn´1 , 0q “ px1 ` y1 , . . . , xn´1 ` yn´1 , 0q,
which is an element of U. Hence, U is closed under addition. Finally, we want to show that U is closed under scalar multiplication.
Let px1 , . . . , xn´1 , 0q P U and a P R. Then,
apx1 , . . . , xn´1 , 0q “ pax1 , . . . , axn´1 , 0q,
which is an element of U. Hence, U is closed under scalar multiplication. Since U is non-empty, closed under addition, and scalar
multiplication, it is a subspace of V.
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Example 10. Recall the subset
U “ tppxq P PpRq : pp1q “ 0u Ă PpRq.
To show that U is a subspace of V we first want to show that it is
non-empty. Since the zero polynomial 0 P U, we know that this is
the case.
Next, we want to show that U is closed under addition. Let
ppxq, qpxq P U. We want to show that pp ` qqpxq “ ppxq ` qpxq is
an element of U. We have
pp ` qqp1q “ pp1q ` qp1q “ 0
which means pp ` qqpxq P U. Hence, U is closed under addition.
Finally, we want to show that U is closed under scalar multiplication. Let a P R and ppxq P U. We want to show that
papqpxq “ appxq is an element of U. We have
papqp1q “ app1q “ a0 “ 0,
which implies papqpxq P U. Hence, U is closed under scalar
multiplication. Since U is non-empty, closed under addition, and
scalar multiplication, it is a subspace of V.
Example 11. Suppose we have a homogeneous linear system
Ax “ ~0 with n-variables. Let U denote the set of solutions to
this system. Then U Ă Rn . To show that U is a subspace of Rn
we first want to show that it is non-empty. Since the zero vector
~0 P U, we know that this is the case.
Next, we want to show that U is closed under addition. Let
x1 , x2 P U. We want to show that x1 ` x2 is also a solution. By the
distributive property of matrix multiplication, we have
Apx1 ` x2 q “ Ax1 ` Ax2 “ ~0 ` ~0 “ ~0,
which means x1 ` x2 P U. Hence, U is closed under addition.
Finally, we want to show that U is closed under scalar multiplication. Let a P R and let x P U. We want to show that ax is also a
solution. We have
Apaxq “ apAxq “ a0 “ 0,
which implies ax P U. Hence, U is closed under scalar multiplication. Since U is non-empty, closed under addition, and scalar
multiplication, it is a subspace of Rn .
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