Notes for Chapter 4
Intro
If W is a subset of V which also happens to be a vector space, then we say W is a subspace of V. Sometimes
we call W a subspace to emphasize it lies within a bigger space and sometimes we call it just a vector space.
Not all subsets of a vector space form another vector space. Since the set must be closed under scalar
multiplication, it must contain all multiples of every vector in the set. Some subsets are too “small”.
Ex 1: The subset S={ (1,2), (2,4) } of 2 is not a vector space.
This is because even though 2(1,2) is in the set 3(1,2) is not. (An easier reason is that (0,0) is not in the
set.) Can you see that a set S={(0,0)} is a vector space called the trivial space, but that all other vector
spaces of 2 must contain infinite number of vectors? (Think about closure under scalar multiplication
using every possible integer scalar.)
Even if a subset is infinite it may not be a vector space.
Ex 2: The subset S={(x,y) such that 2x+3y=4} of
vector.
2
is not a subspace because it doesn’t include the zero
Some subsets that are infinite do form vector spaces.
Ex 3: The subset S  { p  P1 | p(0)  0} forms a vector space in P2 .
Since S is a subset of a known vector space and we are using standard operations we need only prove:
1. S contains the zero polynomial z(x)=0 since z( x)  0  z(0)  0 .
2. If p and q are in S then p(0)  0 and q(0)  0 . Notice  p  cq  (0)  p(0)  cq(0)  0  c0  0 so
the polynomial p  cq satisfies the condition required for vectors in the subset S so p  cq is in S
which makes S closed.
Thus S is a vector space.
Spans and Spanning
Some subsets of V form vector spaces others don’t, but from every subset of V we can “create” a vector
subspace Span(S) of V. Unless otherwise stated S  {v1, v2 ,...., vk } is a set of vectors in a vector space V.
k
Defn: Span(S)=set of all vectors w that can be written as w   ci vi where the ci are real numbers
i 1
The subset S of V is said to span the subspace W (of V) if and only if S is a subset of W and every vector in
W is a linear combination of vectors in S. In other words:
Defn: A subset S of a vector space W is said to span W iff for every w in W there exists scalars ci such that
k
w   ci vi iff Span(S)=W.
i 1
Combining these two definitions we find S spans Span(S); hence, the confusing terminology. Unfortunately
the confusion only increases when one realizes that there can be another subset T with Span(T)=Span(S).
Our goal is to find the “smallest” subset T of Span(S) such that Span(T)=Span(S). This set T will be called a
basis.
Defn 1: A basis of a vector space W is a subset S of W such that W=Span(S) and if T is any subset of W
with fewer elements than S then Span(T) is not W. This definition is hard to work with.
Defn 2: A basis of a vector space W is a subset S of W such that W=Span(S) and the vectors in S are
k
linearly independent. (linearly independent means that if
c v
i 1
i i
 0 you can conclude that c1  c2  ...ck  0 )
This is the most common approach. Notice it requires two things first span and then independent. Our goal
is always to do both simultaneously.
The choice of a basis for a vector space is not unique, but the quantity of vectors in the set is unique. This
quantity is called the dimension of the vector space. You are required to know the dimension of the standard
vector spaces we work with in this class as well as the standard basis vectors. You can find this on a chart on
my homepage. This is useful for determining that a set is not a basis for a standard space. If the set has too
few or too many vectors it cannot be a basis. See 5, 8, 12, 13, 15 in 4.5
Lemma: If W is a subspace of a vector space V then
1. If W<V (ie. There is some vector in V that is not in W) then dim(W)<dim(V).
2. If W=V then dimW=dimV
Lemma (Rephrased) Given two vector spaces W and V with W contained in V. W=V iff dimW=dimV.
This lemma is useful because if dimV=n and if you are given a set of n vectors in V which are independent
then they must form a basis for V. If the set is dependent by definition it can’t be a basis. You will use this
in #9,11,14,16,21-34 in 4.5. In all of these problems a set S is given to you. It spans some vector space
W=Span(S). The question here becomes are these vectors independent making dimW=n thus W=V.
Example 4a: Determine if the set S={(-1,2,5), (3,0,3), (5,1,8) } is a basis for 3 .
Notice that the dim 3 =3 so we need 3 vectors in a basis which is what we have so we have to keep
working. We have to determine if they are independent. If
a(-1,2,5)+b(3,0,3)+c(5,1,8)=(0,0,0)
then
(-a+3b+5c,2a+c,5a+3b+8c)=(0,0,0).
 1 3 5 0 
This forms a system of three equations three unknowns:  2 0 1 0  Notice the vectors became the
 5 3 8 0


 1 0 1/ 2 0 
columns! The rref of this is  0 1 11/ 6 0  . Since the system has infinite solutions, the original set is
0 0
0
0 

dependent and not a basis.
Example 4b: (slightly more involved but basically the same)
Find a basis for W=Span{ (-1,2,5), (3,0,3), (5,1,8) }
You are already given the spanning set S={ (-1,2,5), (3,0,3), (5,1,8) } so all you need to do is determine if
this set is independent if it is your done it forms a basis. If it is not, then you need to discover the
“problem”. Above we discovered that the set was not independent since we had infinite solutions to our
equation. Notice that the reason it has infinite solutions is because of the third column. If we were to
“throw” out the third vector (5,1,8) then the system would have 3 equations two unknowns
1 0 1 / 2 0


 0 1 11 / 6 0  and would have a unique solution. Thus the vectors (-1,25) and (3,0,3) are
0 0
0
0 

independent in W. They also span W since if we finish solving the above we get a  (1/ 2)c  0 and
b  (11/ 6)c  0 . This implies a  (1/ 2)c and b  (11/ 6)c . Choosing c=1 Our original equation
becomes
(-1/2)(-1,2,5)+(-11/6)(3,0,3)+(5,1,8)=(0,0,0). Rearranging we get
(5,1,8)= (1/2)(-1,2,5)+(11/6)(3,0,3)
Now if w is in W then w= x(-1,2,5)+y(3,0,3)+z(5,1,8) if we substitute the above equation we get that w
is a linear combination of just (-1,2,5) and (3,0,3). Thus every vector in W is a linear combination of (1,2,5) and (3,0,3). By definition then Span{ (-1,2,5) , (3,0,3)}=W. This set { (-1,2,5) , (3,0,3) } is a basis
for W which therefore has dimension two.
One comment here is needed: we could just as well have thrown out column 2 because if we imagine it gone
 1 0 1/ 2 0 


we would have  0 1 11/ 6 0  . Hopefully you can see that this system will also have a unique solution.
0 0
0
0 

This is the idea behind problem 49 in 4.6.
To find a basis for a space W given a spanning set of W.
This method always works to create an independent subset of the original set. It can be used in 4.6
problems 9-12. To restate what we did:
Step 1: Start with a spanning set
Step 2: Write the vectors as columns in a matrix with a zero last column if you like
Step 3: Write the matrix in rref.
Step 4: Locate the columns which contain the pivots (the leading 1’s) The original vectors
corresponding to these columns will be an independent set and they will span the original space. The
dimension of the space is the number of columns with a leading 1.
Another method for finding a basis for W is found on page 211. This uses theory discussed in the textbook
that I didn’t discuss here. I will not test on rowspace. You can skip all these questions in the book.
Sometimes you are given a space W whose spanning set is not explicitly stated. #47-52 in 4.5 and 13-30 in
4.6. You need to find a spanning set.
Example 5: Find a basis for S  { p  P1 | p(0)  0} from example 3 above.
We have to create the spanning set and prove it is independent. The normal approach here is to get your
hands dirty and go into components. Rewrite your condition somehow and substitute it back into your
original vector form. Our vector has the form p  a  bx  cx 2 with the condition that p(0)  0 . Then
since p( x)  b  2cx , your condition becomes b  0 . So you really have p  a  cx 2 . Thus we have
S  {a  cx 2 | a, c real}  {a(1)  c( x 2 )| a, c real}  Span{1, x 2} . Since we could write S as a span it is a
vector space, without doing the stuff we did in example 3. Our spanning set is 1, x 2  . This set is
independent because we can use the fact that since there are only two we need only check that both are
not zero and that they are not multiples of each other. Or if you have more than two vectors use the
1 0 0


2
2
2
above method 1  1(1)  0 x  0 x , x  0(1)  0 x  1x so the matrix of interest would be  0 0 0 
0 1 0


which clearly has a unique solution so they are independent.
A basis for S would be 1, x 2  .
This last approach is important because it actually shows that the set is vector space and finds a basis all in a
few easy steps. When you do it it will look a lot more compact because all the words I used to explain it to
you won’t be needed. See example 7 for how short it could actually look.
Nullspace and Range
Example 6: Find a basis for the set of all solutions to the system Ax  0 where
4 0 2
5 0
1


0
0 1 3
1 0

A
.
1
4 2 8
7 1


 3 12 1 3 14 0 
You must first find a spanning set. So we are looking for vectors x  ( x1, x2, ..., x6 )T that satisfy Ax  0 .
We know how to do this.
1 4 0 2 5 0


0 0 1 3 1 0

rref(A)=
0 0 0 0 0 1


0 0 0 0 0 0
Thus we need x1  4 x2  2 x4  5x5  0 , x3  3x4  x5  0 , and x6  0 . There are three columns with
leading ones namely column 1, 3 and 6. Isolate these coordinates in the above equations yielding:
x1  4 x2  2 x4  5x5 , x3  3x4  x5 , and x6  0 . Substitute these into our original
x  ( x1, x2, ..., x6 )T  (4 x2  2 x4  5x5 , x2 , 3x4  x5 , x4 , x5 ,0)T . This can be rewritten as
x  (4 x2  2 x4  5 x5 , x2 , 3x4  x5 , x4 , x5 ,0)T  x2 (4,1,0,0,0,0)T  x4 ( 2,0, 3,1,0,0)T  x5 ( 5,0, 1,0,1,0)T
Thus a spanning set is (4,1,0,0,0,0)T , (2,0, 3,1,0,0)T , (5,0, 1,0,1,0)T . You still need to see if these
are independent. You can use the idea from example 3 or easier yet you can easily see that these are
independent because if x2 (4,1,0,0,0,0)T  x4 (2,0, 3,1,0,0)T  x5 (5,0, 1,0,1,0)T  (0,0,0,0,0,0)T then
from component 2 we find x2  0 , from component 4 we find x4  0 and from component 5 we find
x5  0 .
Thus {(4,1,0,0,0,0)T , (2,0, 3,1,0,0)T , (5,0, 1,0,1,0)T } is a basis for the set of all solutions to the
system Ax  0 . So this space has dimension 3. Notice that the dimension was determined because
x2, x4 , and x5 could be any real values. So the dimension of this space corresponds to the number of
columns that don’t have a leading one.
This problem (see homework 13-24 in 4.6) is so common it has been given a special name.
Defn: Let A be an m by n matrix. The set of all solutions to the system Ax  0 is called the Null Space of
A. Abbreviated NS(A). It is also called Kernel of A abbreviated Ker(A).
Lemma: NS(A) is a subspace of
n
.
Defn: The dimension of the subspace NS(A) is called the nullity of A.
To find the nullity of A and a basis for the nullspace of A.
Step 1: Write A in rref.
Step 2: The nullity of A will be the number of columns in rref(A) that don’t have a leading 1.
Step 3: To actually find a basis treat rref(A) as the coefficient matrix of a homogeneous system. Locate the
leading 1’s in this matrix. Isolate the variables corresponding to the columns that have a leading 1.
Substitute these equations into x  ( x1, x2, ..., xn )T . Be careful you should still have n entries here separated by
commas.
Step 4: Write this parametrically. These vectors will be a basis for NS(A)
There is more information here than we have discussed. Consider the subset S  {v1, v2 ,...., vn } of
a matrix A whose columns are these vectors. Thus A will be be a m by n matrix.
m
. Form
Defn: Range(A)=Span(S). This will also be called the column space of A. Notice Range(A) is a subspace
of m .
Defn: The rank of A is the dimension of Range(A). In other words it is the dimension of the space spanned
by the columns of A which lie in m . Rank(A) is at most m. If Rank(A)=m then Range(A)= m .
To find a more compact description of Range(A) we need to find a basis for Span(S), but we already know
how to do this from Example 4b!
To find Rank(A) and a basis for Range(A)
Step 1: Write A in rref
Step 2: Rank A = number of columns with a leading 1. Since there are n columns rank(A) is at most n.
Step 3: Range(A) = span of the original column vectors that correspond to the columns in rref(A) with a
leading 1
This idea will be used in 1-8 in 4.6.
4 0 2
5 0
1


0
0 1 3
1 0

A

Example 7: Find the rank of A and a basis for the Range(A) where
.
1
4 2 8
7 1


 3 12 1 3 14 0 
1 4 0 2 5 0


0 0 1 3 1 0

We had rref(A)=
. Thus there are three leading ones (column 1, 3, and 6). So
0 0 0 0 0 1


0 0 0 0 0 0
Rank(A)=3; furthermore, a basis for Range(A)= {(1,0,1, 3)T ,(0,1,2,1)T ,(0,0,1,0)T } These are the original
columns 1, 3 and 6.
Misc: Problems in section 4.6 #25-34
There are three questions that are asked that are essentially the same: Determine if b is in the column space
of A. Determine if Ax=b is consistent. Determine if Ax=b has a solution. To answer any of these questions
just do them the old way of row reducing the augmented matrix to determine if there are solutions are not.
Let me explain why these questions are the same. Consider the subset S  {v1 , v 2 ,...., vn } of m where the
vectors are the columns of A. Thus A will be be an m by n matrix. Recall that multiplication on the left by a
column corresponds to adding up multiples of the columns, thus x  ( x1, x2, ..., xn )T is a solution to Ax=b iff
n
x v
i
i
 b . If this happens it can be interpreted that b is in the span of the columns hence it relates to
1
section 4.6.
In example 9 pg 221 and in problem 45 pg 223, your text considers the relationship between rank of the
augmented matrix and rank of the coefficient matrix. Notice that you can’t solve a system if you end up with
an augmented matrix with a row of n zeros and the last entry nonzero. Notice those zeros don’t contribute to
the rank of the coefficient matrix, but that nonzero does contribute to the rank of the augmented matrix. You
need to read and think about this. Arguments can also be made using ideas similar to ones used in Example
4b above.