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3 The Vector Space Rn
Core Sections
3.2 Vector space Properties of Rn
3.3 Examples of Subspaces
3.4 Bases for Subspaces
3.5 Dimension
3.6 Orthogonal Bases for Subspaces
3.1 Introduction
In mathematics and the physical sciences, the term
vector is applied to a wide variety of objects. Perhaps
the most familiar application of the term is to
quantities, such as force and velocity, that have both
magnitude and direction. Such vectors can be
represented in two space or in three space as directed
line segments or arrows. As we will see in chapter
5,the term vector may also be used to describe objects
such as matrices , polynomials, and continuous realvalued functions.
In this section we demonstrate that Rn, the set of
n-dimensional vectors, provides a natural bridge
between the intuitive and natural concept of a
geometric vector and that of an abstract vector in a
general vector space.
3.2 VECTOR SPACE PROPERTIES OF Rn
 x1 
x 
R n  { X : X   2 , x1 , x2 ,, xn real numbers }.

 
 xn 
The Definition of Subspaces of Rn
A subset W of Rn is a subspace of Rn if and only if the
following conditions are met:
(s1)* The zero vector, θ, is in W.
(s2)X+Y is in W whenever X and Y are in W.
(s3)aX is in W whenever X is in W and a is any scalar.
Example 1: Let W be the subset of R3 defined by
 x1 
W  {x : x   x2 , x1  x2  x3 , x2 and x 3 any real numbers }.
 x3 
Verify that W is a subspace of R3 and give a geometric
interpretation of W.
Solution:
Verifying that W is a subspace of Rn
Step 1. An algebraic specification for the subset W is
given, and this specification serves as a test for
determining whether a vector is in Rn or is not in W.
Step 2.Test the zero vector, θ, of Rn to see whether it
satisfies the algebraic specification required to be in W.
(This sows that W is nonempty.)
Step 3.Choose two arbitrary vectors X and Y from W.
Thus X and Y are in Rn, and both vectors satisfy the
algebraic specification of W.
Step 4. Test the sum X+Y to see whether it meets the
specification of W.
Step 5. For an arbitrary scalar, a, test the scalar multiple
aX to see whether it meets the specification of W.
Example 2: Let W be the subset of R3 defined by
 x1 
W  {x : x   x2 , x 2  2x 1 , x 3  3x 1 , x 1 any real number }.
 x3 
Verify that W is a subspace of R3 and give a
geometric interpretation of W.
Example 3: Let W be the subset of R3 defined by
 x1 


W  {x : x   x2 , x 1 and x 2 any real numbers }.
 1 
Show that W is not a subspace of R3.
Example 4:Let W be the subset of R2 defined by
 x1 
W  {x : x   , x 1 and x 2 any integers }.
 x2 
Demonstrate that W is not a subspace of R2.
Example 5:Let W be the subset of R2 defined by
 x1 
W  { X : X   , where either x 1  0 or x 2  0 }.
 x2 
Demonstrate that W is not a subspace of R2.
Exercise P175 18
32
3.3 EXAMPLES OF SUBSPACES
In this section we introduce several important and
particularly useful examples of subspaces of Rn.
The span of a subset
Theorem 3:If v1, …vr are vectors in Rn, then the set W
consisting of all linear combinations of v1, …,vr is a
subspace of Rn.
If S={v1, …,vr} is a subset of Rn, then the subspace W
consisting of all linear combinations of v1, …,vr is called
the subspace spanned by S and will be denoted by Sp(S)
or Sp{v1, …,vr}.
For example:
(1)For a single vector v in Rn, Sp{v} is the subspace
Sp{v}={av:a is any real number} .
(2)If u and v are noncollinear geometric vectors, then
Sp{u,v}={au+bv:a,b any real numbers}
(3) If u, v, w are vectors in R3,and are not on the same
space,then
Sp{u,v,w}={au+bv+cw : a,b,c any real numbers}
Example 1: Let u and v be the three-dimensional vectors
 2
u  1
0
0 
and v  1
2
Determine W=Sp{u,v} and give a geometric
interpretation of W.
The null space of a matrix
We now introduce two subspaces that have particular
relevance to the linear system of equations Ax=b,
where A is an (m×n) matrix. The first of these
subspaces is called the null space of A (or the kernel of
A) and consists of all solutions of Ax=θ.
Definition 1:Let A be an (m × n) matrix. The null
space of A [denoted N(A)] is the set of vectors in Rn
defined by
N(A)={x:Ax= θ, x in Rn}.
Theorem 4:If A is an (m × n) matrix, then N(A) is a
subspace of Rn.
where A is the (3 × 4)
3 1

5 4 .
4  1
Solution: N(A)
is determined by solving the
homogeneous system
Ax= θ.
This is accomplished by reducing the augmented matrix
[A| θ] to echelon form. It is easy to verify that [A| θ] is
row equivalent to
Example 2:Describe N(A),
matrix
1 1

A  2 1
1 2
1 0 2 3 0
0 1 1  2 0.


0 0 0 0 0
Solving the corresponding reduced system yields
x1=-2x3-3x4
x2=-x3+2x4
 2 x3  3 x 4 
  2
 3
  x  2x 
  1
2
4 
x 3
 x3    x 4  ,


1
0
x3


 
 
x4
0
1


Where x3 and x4 are arbitrary; that is,
  2
 3
  1
2
N ( A)  {x : x  x3    x 4  , x3 and x 4 any real numbers }.
1
0
 
 
0
1
Example 5:Let S={v1,v2,v3,v4} be a subset of R3,
where
1
2 
1
2
v1  2, v 2  3 , , v 3   4 , and v 4   5 .
1
5 
 5
- 1
Show that there exists a set T={w1,w2} consisting of
two vectors in R3 such that Sp(S)=Sp(T).
Solution: let
1 2 1 2 
A  2 3 4 5 .
1 5  5  1
Set row operation to A and reduce A to the following matrix:
1
A  2
1
1
 0
0
2
3
5
2
1
0
2
1
2
1 2
4
5 .  0  1 2
1 
0 3  6  3
 5  1
1
2
4
1 0 5
 2  1  0 1  2  1
0 0 0
0
0 
0 
1
So, Sp(S)={av1+bv2:a,b any real number}
Because Sp(T)=Sp(S), then Sp(T)={av1+bv2:a,b any real
number}
For example, we set
1
2 1
w1  3v1  2v2  32  23  0
1
5 7
1 2  0 
w2  2v1  v 2  22  3   1
1 5  3 
1 2 1 
The solution on P184
2 3 5 
1
2
1
2


T

,
A

1 4  5
A  2 3 4
5 .


1 5  5  1
 2 5  1
And the row vectors of AT are precisely the vectors v1T,v2T,v3T,
and v4T. It is straightforward to see that AT reduces to the matrix
1
0
BT  
0

0
0 7

1  3
.
0 0

0 0
So, by Theorem 6, AT and BT have the same row space. Thus A
and B have the same column space where
1 0 0 0
B  0 1 0 0.
7  3 0 0
In particular, Sp(S)=Sp(T), where T={w1,w2},
1
0
w1  0 and w 2  - 1.
7
 3 
3.4 BASES FOR SUBSPACES
Two of the most fundamental concepts of geometry
are those of dimension and the use of coordinates to
locate a point in space. In this section and the next,
we extend these notions to an arbitrary subspace of Rn
by introducing the idea of a basis for a subspace.
An example from R2 will serve to illustrate the transition
from geometry to algebra. We have already seen that each
vector v in R2,
a 
v   ,
b 
can be interpreted geometrically as the point with
coordinates a and b. Recall that in R2 the vectors e1 and e2
are defined by
1
0 
e1    and e 2   .
0
1
Clearly the vector v in (1) can be expressed uniquely as a
linear combination of e1 and e2:
v=ae1+be2
(2)
As we will see later, the set {e1,e2} is an example of a
basis for R2 (indeed, it is called the natural basis for
R2). In Eq.(2), the vector v is determined by the
coefficients a and b (see Fig.3.12). Thus the geometric
concept of characterizing a point by its coordinates
can be interpreted algebraically as determining a
vector by its coefficients when the vector is expressed
as a linear combination of “basis” vectors.
Spanning sets
Let W be a subspace of Rn, and let S be a subset of
W. The discussion above suggests that the first
requirement for S to be a basis for W is that each
vector in W be expressible as a linear combination of
the vectors in S. This leads to the following
definition.
Definition 3:Let W be a subspace of Rn and let
S={w1,…,wm} be a subset of W. we say that S is a
spanning set for W, or simply that S spans W, if every
vector w in W can be expressed as a linear
combination of vectors in S;
w=a1w1+…+amwm.
A restatement of Definition 3 in the notation of the
previous section is that S is a spanning set of W
provided that Sp(S)=W. It is evident that the set
S={e1,e2,e3}, consisting of the unit vectors in R3, is a
spanning set for R3. Specifically, if v is in R3,
a 


v  b  ,
 c 
Then v=ae1+be2+ce3. The next two examples consider
other subset of R3.
Example 1: In R3, let S={u1,u2,u3}, where
1
- 2
1
u1   1, u 2   3 , and u 3  2.
 0 
 1 
4
Determine whether S is a spanning set for R3.
 1  2 1 a
[ A | v]   1 3 2 b ,
The augmented matrix
 0
1 4 c 
Solution:
this matrix is row
equivalent to
1 0 0 10a  9b  7c 
0 1 0 4a  4b  3c .


0 0 1
 a  b  c 
Example 2: Let S={v1,v2,v3} be the subset of R3
defined by
1
 - 1
 2
v1  2, v 2   0 , and v3  7.
3
- 7
0
Does S span R3?
Solution:
1  1 2 a 


[ A | v]  2 0 7 b ,
3  7 0 c 
and the matrix [A|v] is row equivalent to
b/2
1 0 7 / 2

0 1 3 / 2  a  (1 / 2)b .


0 0 0  7a  2b  c 
So,
a 


Sp ( S )  {v : v  b , where - 7a  2b  c  0}.
 c 
For example, the vector
1
w  1
1
is in R3 but is not in Sp(S); that is, w cannot be
expressed as a linear combination of v1, v2,and v3.
The next example illustrates a procedure for constructing
a spanning set for the null space, N(A), of a matrix A.
Example 3:Let A be the (3×4) matrix
1 1 3 1 
A  2 1 5 4 .
1 2 4  1
Exhibit a spanning set for N(A), the null space of A.
Solution: The first step toward obtaining a spanning set
for N(A) is to obtain an algebraic specification for N(A)
by solving the homogeneous system Ax=θ.
  2 x 3  3 x4 
  x  2x 
3
4 
N ( A)  { x : x  
, x 3 and x 4 any real numbers }.


x3


x4


  2 x 3  3 x4    2 x 3    3 x4 
  2
  3
  x  2x    x   2x 
  1
 2 
3
4 
4 
3 
x


 x 3    x4   .(8)

  x3   0 
 1 
 0 
x3

 

 
 
 
x4
 0 
 1 

  0   x4 
Let u1 and u2 be the vectors
  3
  2
 2 
  1
u1    and u2    .
 0 
 1 
 
 
 1 
 0 
Therefore, N(A)=Sp{u1,u2}
Minimal spanning sets
If W is a subspace of Rn, W≠{θ}, then spanning sets for W
abound. For example a vector v in a spanning set can always
be replaced by av, where a is any nonzero scalar. It is easy to
demonstrate, however, that not all spanning sets are equally
describe. For example, define u in R2 by
1
u   .
1
The set S={e1,e2,u} is a spanning set for R2. indeed, for an
arbitrary vector v in R2,
a 
v   ,
b 
V=(a-c)e1+(b-c)e2+cu, where c is any real number
whatsoever. But the subset {e1,e2} already spans R2, so
the vector u is unnecessary .
Recall that a set {v1,…,vm} of vectors in Rn is linearly
independent if the vector equation
x1v1+…+xmvm=θ
(9)
has only the trivial solution x1=…=xm=0; if Eq.(9) has
a nontrivial solution, then the set is linearly dependent.
The set S={e1,e2,u} is linearly dependent because
e1+e2-u=θ.
Our next example illustrates that a linearly dependent
set is not an efficient spanning set; that is, fewer
vectors will span the same space.
Example 4: Let S={v1,v2,v3} be the subset of R3, where
1
 2
 3
v1  1 , v 2   3 , and v 3  5 .
1
1
1
Show that S is a linearly dependent set, and exhibit a subset
T of S such that T contains only two vectors but
Sp(T)=Sp(S).
Solution: The vector equation
x1v1+x2v2+x3v3=θ
(10)
is equivalent to the (3 × 3) homogeneous system of
equations with augmented matrix
1 2 3 0
A  1 3 5 0
1 1 1 0
Matrix is row equivalent to
1 0  1 0
B  0 1 2 0
0 0 0 0
So
v3=-1v1+2v2
Sp(T )  {av1  bv2 : a, b, any real number}  Sp(S )
The lesson to be drawn from example 4 is that a linearly
dependent spanning set contains redundant information.
That is, if S={w1,…,wr} is a linearly dependent
spanning set for a subspace W, then at least one vector
from S is a linear combination of the other r-1 vectors
and can be discarded from S to produce a smaller
spanning set.
On the other hand, if B={v1,…,vm} is a linearly
independent spanning set for W, then no vector in B is a
linear combination of the other m-1 vectors in B.
Hence if a vector is removed from B, this smaller
set cannot be a spanning set for W (in particular, the
vector removed from B is in W but cannot be
expressed as a linear combination of the vectors
retained). In this sense a linearly independent spanning
set is a minimal spanning set and hence represents the
most efficient way of characterizing the subspace. This
idea leads to the following definition.
Definition 4:Let W be a nonzero subspace of Rn. A
basis for W is a linearly independent spanning set for
W.
Uniqueness of representation
Remark Let B={v1,v2, …,vp} be a basis for W, where W
is a subspace of Rn. If x is in W, then x can be
represented uniquely in terms of the basis B. That is,
there are unique scalars a1,a2, …,ap such that
x=a1v1+a2v2+…+apvp.
As we see later, these scalars are called the coordinates x
with respect to the basis.
Example of bases
It is easy to show that the unit vectors
1
0
0
e1  0, e2  1, e3  0,
0
0
1
is a basis for R3
In general, the n-dimensional vectors e1,e2,…,en form a
basis for Rn, frequently called the natural basis.
And the vectors
1
1
1
v1  0, v 2  1, v3  1,
0
0
1
Provide another basis for R3.
Example 6:Let W be the subspace of R4 spanned by the
set S={v1,v2,v3,v4,v5}, where
1
1 
1
1
 2
1
 2
4
0
5 
v1   , v 2   , v3   , v 4   , v5   ,
2
1 
 1
4
0 
 
 
 
 
 
 1
1 
5
 1
 2
Find a subset of S that is a basis for W.
Solution:
1
1
A
2

 1
1 1 1
2 4 0
1 1 4
1 5 1
2 1
5 0

0  0
 
2  0
So {v1,v2,v4}is a basis for W.
0 -2
1 3
0 0
0 0
0 1
0 2 
1  1

0 0
The procedure demonstrated in the preceding example
can be outlined as follows:
1.A spanning set S{v1,…,vm} for a subspace W is given.
2.Solve the vector equation
x1v1+…+xmvm=θ
(20)
3.If Eq.(20) has only the trivial solution x1=…=xm=0,
then S is a linearly independent set and hence is a basis
for W.
4.If Eq.(20) has nontrivial solutions, then there are
unconstrained variables. For each xj that is designated as
an unconstrained variable, delete the vector vj from the
set S. The remaining vectors constitute a basis for W.
Theorem 7:If the nonzero matrix A is row equivalent
to the matrix B in echelon form, then the nonzero rows
of B form a basis for the row space of A.
3.5 DIMENSION
In this section we translate the geometric concept of
dimension into algebraic terms. Clearly R2 and R3 have
dimension 2 and 3, respectively, since these vector
spaces are simply algebraic interpretations of two-space
and three-space. It would be natural to extrapolate from
these two cases and declare that Rn has dimension n for
each positive integer n; indeed, we have earlier referred
to elements of Rn as n-dimensional vectors. But if W is a
subspace of Rn, how is the dimension of W to be
determined? An examination of the subspace ,W, of R3
defined by
 x 2  2 x3 


W  {x : x   x2 , x2 and x 3 any real numbers }
 x3 
Suggests a possibility. Geometrically, W is the plane
with equation x=y-2z, so naturally the dimension of W is
2. The techniques of the previous section show that W
has a basis [v1,v2] consisting of the two vectors
1
- 2
v1  1 and v 2   0 .
0
 1 
Thus in this case the dimension of W is equal to the
number of vectors in a basis for W.
The definition of dimension
More generally, for any subspace W of Rn, we wish
to define the dimension of W to be the number of
vectors in a basis for W.
We have seen, however, that a subspace W may have many
different bases. In fact, Exercise 30 of section 3.4 shows
that any set of three linearly independent vectors in R3 is a
basis for R3. Therefore, for the concept of dimension to
make sense, we must show that all bases for a given
subspace W contain the same number of vectors. This fact
will be an easy consequence of the following theorem.
Theorem 8: Let W be a subspace of Rn, and let
B={w1,w2,…,wp} be a spanning set for W containing p
vectors. Then an set of p+1 or more vectors in W is
linearly dependent.
As an immediate corollary of Theorem 8, we can show
that all bases for a subspace contain the same number of
vectors.
Corollary: Let W be a subspace of Rn, and let
B={w1,w2,…wp} be a basis for W containing p vectors.
Then every basis for W contains p vectors.
Given that every basis for a subspace contains the
same number of vectors, we can make the following
definition without any possibility of ambiguity.
Definition 5 : Let W be a subspace of Rn. If W has a
basis B={w1,w2,…,wp} of p vectors, then we say that W
is a subspace of dimension p, and we write dim(W)=p.
In exercise 30, the reader is asked to show that every
nonzero subspace of Rn does have a basis. Thus a value
for dimension can be assigned to any subspace of Rn,
where for completeness we define dim(W)=0 if W is the
zero subspace.
Since R3 has a basis {e1,e2,e3} containing three vectors,
we see that dim(R)=3. In general, Rn has a basis
{e1,e2,…,en} that contains n vectors; so dim(Rn)=n.
Thus the definition of dimension– the number of vectors
in a basis– agrees with the usual terminology; R3 is
three—dimensional, and in general, Rn is n-dimensional.
Example 1: Let W be the subspace of R3 defined by
 x1 
W  {x : x   x2 , x1  2 x3 , x2  x3 , x3 arbitrary }.
 x3 
Exhibit a basis for W and determine dim(W).
Solution: A vector x in W can be written in the form
 2 x3 
  2
x   x3   x3  1 .
 x3 
 1 
Therefore, the set {u} is a basis for W, where
 2
u   1 .
 1 
Example 2 Let W be the subspace of R3,
W=span{u1,u2,u3,u4},where
1
 2
 3
2
u1  1, u 2  4, u 3  5, u 4   5 
2
0
2
 2
Find three different bases for W and give the dimension of W.
Properties of a p-Dimensional subspace
An important feature of dimension is that a p-dimensional
subspace W has many of the same properties as Rp. For
example, Theorem 11 of section 1.7 shows that any set of
p+1 or more vectors in Rp is linearly dependent. The
following theorem shows that this same property and others
hold in W when dim(W)=p.
Theorem 9:Let W be a subspace of Rn with dim(W)=p.
1.Any set of p+1 or more vectors in W is linearly
dependent.
2.Any set of fewer than p vectors in W does not span W.
3.Any set of p linearly independent vectors in W is a
basis for W.
4.Any set of p vectors that spans W is a basis for W.
Example 3: Let W be the subspace of R3 given in
Example 2, and let {v1,v2,v3} be the subset of W defined
by
1
1
 2
v1   1, v 2  2,
v 3  1.
 6 
0
6
Determine which of the subsets {v1} {v2} {v1,v2} {v1,v3}
{v2,v3},and {v1,v2,v3} is a basis for W.
The Rank of matrix
In this subsection we use the concept of dimension
to characterize nonsingular matrices and to determine
precisely when a system of linear equation Ax=b is
consistent. For an (m×n) matrix A, the dimension of
the null space is called the nullity of A, and the
dimension of the range of A is called the rank of A.
Example 4: Find the rank, nullity, and dimension of
the row space for the matrix A, where
1 1 1 2
A   1 0 2  3.
 2 4 8 5 
Solution: To find the dimension of the row space of A,
observe that A is row equivalent to the matrix
1 0  2 0
B  0 1 3 0,
0 0 0 1
and B is in echelon form. Since the nonzero rows of B
form a basis for the row space of A, the row space of A
has dimension 3.
 2 x3 
 3 x 
3
N ( A)  {x : x  
, x3 any real number }.
 x3 


 0 
It now follows that the nullity of A is 1 because the vector
2
 3
v 
1
 
0
forms a basis for N(A).
A is row equivalent to matrix C, where
1 0 0 0


C  0 1 0 0  ,
0 0 1 0
form a basis for R(A). Thus the rank of A is 3
Note in the previous example that the row space of A is
a subspace of R4, whereas the column space (or range)
of A is a subspace of R3. Thus they are entirely different
subspaces; even so, the dimensions are the same, and the
next theorem states that this is always the case.
Theorem 10: If A is an (m×n) matrix, then the rank
of A is equal to the rank of AT.
Remark: If A is an (m × n) matrix, then
n=rank(A)+nullity(A).
The following theorem uses the concept of the rank of
a matrix to establish necessary and sufficient
conditions for a system of equations, Ax=b, to be
consistent.
Theorem 11: An (m × n) system of linear
equations , Ax=b, is consistent if and only if
rank(A)=rank([A|b]).
Theorem 12: An (n × n) matrix A is nonsingular if
and only if the rank of A is n.
3.6 ORTHOGONAL BASES FOR SUBSPACES
We have seen that a basis provides a very efficient
way to characterize a subspace. Also, given a subspace
w, we know that there are many different ways to
construct a basis for w. In this section we focus on a
particular type of basis called an orthogonal basis.
Orthogonal Bases
The idea of orthogonality is a generalization of the
vector geometry concept of perpendicularity. If u and v
are two vectors in R2 or R3, then we know that u and v
are perpendicular if uTv=0 . For example, consider the
vectors u and v given by
1
6
u   , and
v 
  2
3
Clearly uTv=0 , and these two vectors are
perpendicular when viewed as directed line
segments in the plane.
In general , for vectors in Rn, we use the
term orthogonal rather than the term
perpendicular. Specially, if u and v are vectors
in Rn, we say that u and v are orthogonal if
uTv=0
We will also find the concept of an
orthogonal set of vectors to be useful.
Definition 6: Let S ={u1 u2 …up,} be a set of vectors in
Rn, The set S is said to be an orthogonal set if each pair
of distinct vectors form S is orthogonal; that is
uiT u j  0when
i j
Example 1 verify that S is an orthogonal set of
vectors , where
 1   1   1  
    


 0   1    2 
S 
,
,

1   1   1  
   


  2  0   0  
Theorem 13 : let S ={u1 u2 …up,} be a set of nonzero
vectors in Rn,. If S is an orthogonal set of vectors , then
S is a linearly independent set of vectors.
Proof:
Definition 7: Let W be a subspace of Rn, and let
B={u1, u2 up} be a basis for W. If B is an orthogonal set
of vectors, then B is called an orthogonal basis for W.
ui  1 for1  i  p
Furthermore, if
Then B is said to be an orthonormal basis for W
The word orthonormal suggests both orthogonal and
normalized. Thus an orthonormal basis is an
orthogonal basis consisting of vectors having length 1,
where a vector of length 1 is a unit vector or a
normalized vector. Observe that the unit vectors e1
form an orthonormal basis for Rn.
Example 2 Verify that the set B={v1 v2 v3}, is an
orthogonal basis for R3,where
1
v1  2,
1
3
v2   1
 1
1
v3   4
 7 
Corollary: Let W be a subspace of Rn, where
dim(W)=p, If S is an orthogonal set of p nonzero vectors
and is also a subset of W, then S is an orthogonal basis
for W.
Orthonormal Bases
If B={u1, u2…,up} is an orthogonal set, then C={a1u1,
a2u2,apup} is also an orthogonal set for any scalars a1
a2,ap. If B contains only nonzero vectors and if we
define the scalars a i by
ai 
1
uiT ui
Then C is an orthonormal set. That is , we can convert
an orthogonal set of nonzero vectors into an orthonormal
set by dividing each vector by its length.
Example3: Recall that the set B in Example 2 is an
orthogonal basis for R3. Modify B so that it is an
orthonormal basis.
1
v1  2,
1
Solution: v1  6 ,
1

1
w1 
 2
6
1

w3 
3
v2   1
 1
v2  11
6

6 ,
6

 1 66 


1
  4
66 
66
 7

66


w2 
1
v3   4
 7 
v3  66
 3 11 


1
  1 11
11
 1 11


Determining Coordinates
Suppose that W is a p-dimensional subspace of Rn, and
B={w1 w2 ….wp} is a basis for W. if v is any vector in W,
then v can be written uniquely in the form
v=a1w1+a2w2+…+ apwp
(3)
The scalars a1,a2,…,ap, in Eq.(3) are called the
coordinates of v with respect to the basis B
As we will see, it is fairly easy to determine the
coordinates of a vector with respect to an orthogonal
basis. To appreciate the savings in computation, consider
how coordinates are found when the basis is not
orthogonal.
Example4: Express the vector v in terms of the
orthogonal basis B={w1 w2 w3},where
 12 
v   3,
 6 
1
3
1 
w1  2, w2   1, and , w3   4
1
 1
 7 
In general , let W be a subspace of Rn, and let B={w1
w2 …wp} be an orthogonal basis for W. If v is any vector
in W, then v can be expressed uniquely in the form
v=a1w1+a2w2+…+apwp;
(5a)
Where
wiT v
ai  T , 1  i  p.
(5b)
wi wi
Constructing an Orthogonal Basis
The next theorem gives a procedure that can be used to
generate an orthogonal basis from any given basis. This
procedure, called the Gram-Schmidt process, is quite
practical from a computational standpoint
Theorem 14 Gram-Schmidt
Let W be a p-dimensional subspace of Rn, and let {w1
w2… wp} be any basis for W. Then the set of vectors
{u1 u2…up} is an orthogonal basis for W, where
u1  w1
u2
T
1
T
1
u w2
 w2 
u1
u u1
u1T w3
u 2T w3
u3  w3  T
u1  T
u2
u1 u1
u2 u2
And where, in general
i 1
T
k i
T
k k
u w
ui  wi  
uk ,
k 1 u u
2i p
(6)
Example 3 : let W be the subspace of R3 defined by
W=Sp{w1,w2},where
1
 0 




w1  1, and , w2   2 
2
 4
Use the Gram-Schmidt process to construct an
orthogonal basis for W
Example 6 : Use the Gram-Schmidt orthogonalization
process to generate an orthogonal basis for
W=Sp{w1w2 w3},where
0 
1 
w1   ,
 2
 
1 
0 
1
w2   ,
 3
 
1
Exercise P224 15, 19
1
1
w2   ,
1
 
0 