Section 4.1
Vectors in ℝn
ℝn
Vectors
Vector addition
Scalar multiplication.
Def.
Let v1, v2, . . . vn be vectors in ℝn. A linear combination of these vectors is an
expression c1v1 + c2v2 + . . . + cnvn where c1, c2, . . . . , cn are some scalars.
Ex.
(a) If possible, write <6, 7> as a linear combination of <2, 3> and <1, 1>.
Ex.
(b) If possible, write <10, 4> as a linear combination of <5, 2> and <3, 1>.
Ex.
(c) If possible, write <7, 10> as a linear combination of <1, 2> and <3, 6>.
Ex. Verify your answers geometrically for the last example.
(a) <6, 7> written as a linear combination of <2, 3> and <1, 1>.
Ex. Verify your answers geometrically for the last example.
(b) <10, 4> written as a linear combination of <5, 2> and <3, 1>.
Ex. Verify your answers geometrically for the last example.
(c) <7, 10> written as a linear combination of <1, 2> and <3, 6>.
Properties of vectors in ℝn
Let u, v, and w be vectors in ℝn and let c and d be scalars.
1. u + v is in ℝn
2. u + v = v + u
3. (u + v) + w = u + (v + w)
4. There is a vector 0 such that u + 0 = u for all u in ℝn
5. For all u in ℝn there is a vector –u such that u + (–u) = 0
6. cu is in ℝn
7. c(u + v) = cu + cv
8. (c + d)u = cu + du
9. c(du) = (cd)u
10. 1(u) = u
Section 4.2
Vector Spaces
Def. Let V be a set on which two operations (called vector addition and
scalar multiplication) are defined. V is called a vector space if V satisfies the
following properties.
For any u, v, and w be vectors in V and any scalars c and d:
1. u + v is in V
2. u + v = v + u
3. (u + v) + w = u + (v + w)
4. There is a vector 0 such that u + 0 = u for all u in V
5. For all u in V there is a vector –u such that u + (–u) = 0
6. cu is in V
7. c(u + v) = cu + cv
8. (c + d)u = cu + du
9. c(du) = (cd)u
10. 1(u) = u
Examples of various vector spaces (each with the standard operations of addition
and scalar multiplication):
1. ℝn – the set of all n–tuples.
Examples of various vector spaces (each with the standard operations of addition
and scalar multiplication):
2. C(–∞,∞) – the set of all continuous functions defined on the real line.
Examples of various vector spaces (each with the standard operations of addition
and scalar multiplication):
3. C[a, b] – the set of all continuous functions defined on a closed interval [a, b].
Examples of various vector spaces (each with the standard operations of addition
and scalar multiplication):
4. P – the set of all polynomials.
Examples of various vector spaces (each with the standard operations of addition
and scalar multiplication):
5. Pn – the set of all polynomials of degree ≤ n.
Examples of various vector spaces (each with the standard operations of addition
and scalar multiplication):
6. Mm,n – the set of all mxn matrices.
Ex. Show that the following are not examples of vector spaces.
(a) The set of all integers (with the standard operations).
Ex. Show that the following are not examples of vector spaces.
(b) The set of all polynomials with degree 2 (with the standard operations).
Ex. Show that the following are not examples of vector spaces.
(c) The set of all invertible 2x2 matrices (with the standard operations).
Ex. Show that the following are not examples of vector spaces.
(d) The set of all matrices (with the standard operations).
Ex. Let V = ℝ2 with the standard operation of addition and the following
nonstandard definition of scalar multiplication: c (x, y) = (cx, 0). Determine
whether or not V is a vector space.
Section 4.3
Subspaces of Vector Spaces
Def. Let V be a vector space and let W be a subset of V. If W is itself a
vector space then W is said to be a subspace of V.
Note: Every vector space contains at least two subspaces, the vector space
itself and the set consisting only of the zero element. These are called the
trivial subspaces.
Ex. Determine whether or not the following are subspaces.
(a) Let V = ℝ3 with the standard operations.
Define W to be { (x,y,0) : x and y are real numbers}.
Ex. Determine whether or not the following are subspaces.
(b) Let V = ℝ3 with the standard operations.
Define W to be { (x,y,1) : x and y are real numbers}.
Theorem: Test for a Subspace
Let V be a vector space and suppose W is a subset of V. W is a subspace of V
if and only if the following hold:
1. If u and v are in W then u + v is in W.
2. If u is in W then cu is in W.
Ex. Let a, b, and c be constants. Show that the set of all solutions to the
differential equation ay″ + by′ + cy = 0 is a subspace (a subspace of the
vector space of all functions).
Note: W cannot be a subspace of V unless W contains the zero element.
Ex. Which of the following are subspaces?
(a) V = ℝ2 . W = the solution set of 2x + 3y = 5.
Ex. Which of the following are subspaces?
(b) V = ℝ2 . W = the solution set of -5x + y = 0.
Ex. Which of the following are subspaces?
(c) V = ℝ3 . W = the solution set of 3x + 8y – 4z = 7.
Ex. Which of the following are subspaces?
(d) V = ℝ3 . W = the solution set of -2x + 12y – z = 0.
All nontrivial subspaces of ℝ2 are lines which pass through the origin.
All nontrivial subspaces of ℝ3 are planes which pass through the origin or
lines which pass through the origin.
Ex. Is W = {(x, x+ z, z) : x and z are real numbers} a subspace of ℝ3?
Ex. Show that the solution set to x2 + y2 = 1 is not a subspace of ℝ2.
Ex.
(a) Let V = ℝ3 with the standard operations. Define W to be {(x,y,0) : x and
y are real numbers}. What can W be identified with?
Ex.
(b) Let V = ℝ3 with the standard operations. Define W to be {(t, t, 3t) : t is
a real number}. What can W be identified with?
Notation:
P0, P1, P2, P3, . . . . , P
Section 4.4
Spanning Sets and Linear Independence
Recall:
Let v1, v2, . . . vn be vectors in ℝn. A linear combination of these vectors is an
expression c1v1 + c2v2 + . . . + cnvn where c1, c2, . . . . , cn are some scalars.
Ex.
(a) If possible, write f (x) = x2 + x + 1 as a linear combination of
g1(x) = x2 + 2x + 3, g2(x) = x + 2, and g3(x) = –x2 + 1.
Ex.
2
0 8
(b) If possible, write 
 as a linear combination of 
2
1
 1
4

4
and
,
1

1
3

2
2

 1
0

3
,
Def.
Let S = {v1, v2, . . . vn} be a subset of a vector space V. The span of S is the
set of all linear combinations of elements of S.
span(S) = {c1v1 + c2v2 + . . . + cnvn : c1, c2, . . . . cn are scalars}
Def. Let S = {v1, v2, . . . vn} be a subset of a vector space V. If every vector of
V can be written as a linear combination of elements of S then we say S spans V.
Put another way, S spans V if span(S) = V.
Ex. Show that {(1, 0, 0), (0, 1, 0), (0, 0, 1)} spans ℝ3.
Ex. Does {(1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 1)} span ℝ3?
Ex. Does {(1, 0, 0), (0, 1, 0)} span ℝ3?
Ex. Determine whether or not {(1, 2, 3), (4, 5, 6), (7, 8, 9)} spans ℝ3. If
it does, then write a general vector (x, y, z) as a linear combination of these
vectors. If it does not, then find a vector in ℝ3 which it does not span.
Ex.
Determine whether or not {(1, 2, 3), (4, 5, 6), (–1, 1, 3)} spans
ℝ3. If it does, then write a general vector (x, y, z) as a linear combination of
these vectors. If it does not, then find a vector in ℝ3 which it does not span.
Ex.
(a) ℝ2
(b) M23
(c) P2
Give a set which spans . . .
We shall be interested in spanning a vector space with as few vectors as
possible.
Def. Suppose v1, v2, . . . vn are vectors in a vector space V. v1, v2, . . . vn are
said to be linearly independent if the only linear combination of these vectors
which produces the zero element is the trivial one.
(i.e. if the only solution to the equation c1v1 + c2v2 + . . . + cnvn = 0
happens to be the trivial solution c1 = c2 = . . . = cn = 0)
Ex. Use the definition of linear independence to determine whether the
following are linearly independent or linearly dependent.
(a) (1,0), (0,1)
Ex. Use the definition of linear independence to determine whether the
following are linearly independent or linearly dependent.
(b) (1,2,3), (4,5,6), (7,8,9)
Ex. Use the definition of linear independence to determine whether the
following are linearly independent or linearly dependent.
(c) (1,0), (0,1), (2,3)
Ex. Use the definition of linear independence to determine whether the
following are linearly independent or linearly dependent.
(d) (1,2), (3,6)
Ex. Use the definition of linear independence to determine whether the
following are linearly independent or linearly dependent.
(e) (1,2,3), (4,5,6), (-1,1,3)
Theorem: v1, v2, . . . vn are linearly dependent if and only if one of the
vectors can be written as a linear combination of the others.
Theorem: Let v1, v2, . . . vn be vectors in a vector space V. Then
span({v1, v2, . . . vn}) is a subspace of V.
Section 4.5
Basis and Dimension
We seek a set of vectors in a vector space which will span the entire vector
space, but has as few vectors as possible to do so.
Def.
{v1, v2, . . . vn} is a basis for a vector space V if:
1. {v1, v2, . . . vn} spans V.
2. {v1, v2, . . . vn} is a linearly independent set.
Standard bases:
ℝ2
Standard bases:
ℝ3
Standard bases:
ℝn
Standard bases:
M22
Standard bases:
P2
Nonstandard bases.
Ex. Show that {(1,2,3), (0,1,2), (-2,0,1)} is a basis for ℝ3.
Compare the following three sets to see why we impose these two restrictions
in our definition of a basis:
1. span( {(1, 0, 0), (0, 1, 0), (0, 0, 1)} )
Compare the following three sets to see why we impose these two restrictions
in our definition of a basis:
1. span( {(1, 0, 0), (0, 1, 0), (0, 0, 1)} )
2. span( {(1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 1)} )
Compare the following three sets to see why we impose these two restrictions
in our definition of a basis:
1. span( {(1, 0, 0), (0, 1, 0), (0, 0, 1)} )
2. span( {(1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 1)} )
3. span( {(1, 0, 0), (0, 1, 0)} )
Theorem: S = {v1, v2, . . . , vn} is a basis for V if and only if every element
of V can be written as a linear combination of v1, v2, . . . , and vn in one and
only one way.
Theorem: If a vector space V has a basis with n vectors, then every basis
for V has n vectors.
Def. The dimension of a vector space V, denoted dim(V), is the number of
vectors in a basis for V.
ℝn has dimension __________
Pn has dimension __________
Mmn has dimension _________
Ex. Quickly show that {(1, 2, 6), (-4, 1, 4), (24, 0, 1), (1, 6, 1)} is not a
basis for ℝ3.
Note: If you know V has dimension n then to ascertain whether or not
S = {v1, v2, . . . , vn} is a basis, you only need to check one of the two defining
properties of a basis (i.e. check whether S is linearly independent or check
whether S spans V, but you do not need to check both.)
Recall: The solution set to the differential equation ay″ + by′ + cy = 0 is a
vector space. Notice what an impact these last theorems have on us trying to
find a way to describe the solutions to a differential equation like this.
Ex. Find all the solutions to y″ – 3y′ + 2y = 0.
Ex.
Find the dimension of the following vector spaces:
(a) V = {(d, c-d, d) : c and d are real numbers}
Ex.
Find the dimension of the following vector spaces:
(b) V = {(2b, b, 0) : b is a real number}
Ex.
Find the dimension of the following vector spaces:
a
(c) V =  
b

b
:
a
,
b
,
and
c
are
real
num
bers


c

Section 4.6
Rank of a Matrix and
Systems of Linear Equations
Def.
Given a matrix A, . . . .
(a) the row space of A is the span of the row vectors of A.
(b) the column space of A is the span of the column vectors of A.
Ex.
1
Let A  
4
2
5
3

6
(a) List a few vectors in the row space of A.
(b) List a few vectors in the column space of A.
Ex.
1
Let A  
4
2
5
3

6
(c) Find a basis for the row space of A. Then write a general vector in the row
space, (x, y, z) as a linear combination of these
Ex.
1
Let A  
4
2
5
3

6
(d) Find a basis for the column space of A. Then write a general vector in the
column space, (x, y) as a linear combination of these
Ex.
1
Let A  
4
2
5
3

6
(e) Solve AX = B using B = [11 29]T by using your answer from part (d).
Ex.
1
Let A  
4
2
5
3

6
(f) The dimension of the row space of A is _______. The dimension of the
column space of A is ______.
Theorem: Let A be a matrix. The dimension of the row space of A equals
the dimension of the column space of A.
Def. The rank of a matrix A, denoted rank(A), is the dimension of the row
space of A (which is also the dimension of the column space of A).
Ex.
1
Give the rank of 
4
2
5
3

6
Ex. Give the rank of
1

1

 0

1
3
2
2
0
5
2
1
1
6

1

7

1
Theorem: Let A be a given matrix and A* be a matrix produced after applying
elementary row operations on A. Then the row space of A is equal to the row
space of A*.
(Note: This does not say that the column space of A is equal to the column space of A*.)
Ex. Give the rank of
1

1

 0

1
3
2
2
0
5
2
1
1
6

1

7

1
Def. Given a matrix A, the nullspace of A is defined to be all of the column
matrices X so that AX = O. The dimension of the nullspace of A is called the
nullity of A.
Ex. Verify that the nullspace of a matrix A is indeed a subspace.
Ex. Let A =
1

3

 1
2
2
6
5
2
0
1

4

3 
14 
(a) Verify that 
is in the nullspace of A.

4

 2

 2




Ex. Let A =
1

3

 1
2
2
6
5
2
0
1

4

3 
(b) Find the nullspace of A. (Give your answer as a span of basis vectors.)
Ex. Let A =
1

3

 1
2
2
6
5
2
0
1

4

3 
(c) Determine the linear combination of basis vectors that generates
14 


4


 2 


2


Given a matrix A, we know that the solutions to AX = O form a vector space.
The solutions to AX = B do not form a solution space (if B ≠ O), but we can
use the solutions to AX = O to help us give a general solution to AX = B.
Theorem: Suppose we are given an mxn matrix A and an nx1 column
matrix B. Further suppose that Xh is the general solution to AX = O, and Xp is
one particular solution to AX = B. The general solution to AX = B is then
given by Xp + Xh.
Ex. Use this theorem to describe all the solutions to the following system:
x1
– 2x3 + x4 = 5
3x1 + x2 – 5x3
=8
x1 + 2x2
– 5x4 = –9
Recall that the solutions to the differential equation ay″ + by′ + cy = 0 form a
vector space. The solutions to the differential equation ay″ + by′ + cy = f (x)
do not form a vector space (if f (x) ≠ 0). Compare this with the info in the
paragraph at the top of this page. We can use the previous theorem to
describe the solution set to ay″ + by′ + cy = f (x).
Ex. Find all the solutions to y″ – 3y′ + 2y = 14e–x.