MAC 2103
Module 10
lnner Product Spaces I
1
Learning Objectives
Upon completing this module, you should be able to:
1.
Define and find the inner product, norm and distance in a
given inner product space.
2.
Find the cosine of the angle between two vectors and
determine if the vectors are orthogonal in an inner product
space.
3.
Find the orthogonal complement of a subspace of an
inner product space.
4.
Find a basis for the orthogonal complement of a subspace
of ℜⁿ spanned by a set of row vectors.
5.
Identify the four fundamental matrix spaces of an m x n
matrix A, and know the column rank of A, the row rank of
A, and the rank of A.
6.
Know the equivalent statements of an n x n matrix A.
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2
General Vector Spaces II
The major topics in this module:
Inner Product Spaces, Inner Products,
Norm, Distance, Fundamental Matrix Spaces,
Rank, and Orthogonality
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3
Definition of an Inner Product and Orthogonal Vectors
A. Inner Product: Let , :V  V  °
be any function from V V into that satisfies the
following conditions for any u, v, z in V:
r r
r r
a) u, v  v, u,
r r r
r r
r r
b) u  v, z  u, z  v, z,
r r
r r
c) ku, v  ku, v, r r
r r
d) v, v  0, and v, v  0 iff v  0.
r r
Then u, v defines an inner product for all u, v in V.
r r
u and v in V are Orthogonal Vectors iff u, v  0.
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4
Definition of a Norm of a Vector

B. Norm: Let  :V  °  [0,)
be any function from V into  that satisfies the
following conditions for any u, v in V:
r
r
r r
a) u  0, and u  0 iff u  0,
r
r
b) su  s u , and
r r
r
r
c) u  v  u  v
Triangle inequality.
Then u defines a norm for all u in V.
The special norm that is induced by an inner product is
r r
u  u, u.
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5
Definition of the Distance Between Two Vectors

C. Distance: Let d(,) :V  V  °  [0,)
be any function from V V into  that satisfies the
following conditions for any u, v, w in V:
r r
r r
r r
a) d(u, v)  0, and d(u, v)  0 iff u  v,
r r
r r
b) d(u, v)  d(v, u), and
r r
r r
r r
c) d(u, v)  d(u, w)  d(w, v)
Triangle inequality.
r r
Then d(u, v) defines the distance for all u, v in V.
The special distance that is induced by a norm is
r r
r r
r r r r
d(u, v)  u  v  u  v, u  v.
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6
How to Find an Inner Product, Norm, and Distance for
Vectors in ℜⁿ?
For u and v in ℜⁿ, we define
r r
r r
u, v  u  v  u1v1  u2 v2  ...  un vn ,
1
r r 2
r r
u  u, u  (u  u)  u12  u22  ...  un2 ,
r r
r r
and d(u, v)  u  v
r r r r 12
r r r r
 u  v, u  v  (u  v)  (u  v)
 (u1  v1 )2  (u2  v2 )2  ...  (un  vn )2 .
Note: These are our usual norm and distance in ℜⁿ. ℜⁿ is an
inner product space with the dot product as its inner product.
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7
How to Find an Inner Product, Norm, and Distance for
Vectors in ℜⁿ? (Cont.)
Example 1: Let u = (1,2,-1) and v = (2,0,1) in ℜ³. Find the
inner product, norms, and distance.
r r
r r
u, v  u  v  u1v1  u2 v2  u3v3  (1)(2)  (2)(0)  (1)(1)  1,
r
r r 12
u  u, u  u12  u22  u32  12  2 2  (1)2  6,
r
r r 12
v  v, v  v12  v22  v32  2 2  0 2  12  5, and
r r
r r
r r r r 12
d(u, v)  u  v  u  v, u  v
 (1  2)2  (2  0)2  (1  1)2  9  3.
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8
How to Find an Inner Product, Norm, and Distance for
Vectors (Matrices) in M22?
In M22, the vectors are 2 x 2 matrices. Thus, given two matrices
U and V in M22, we define
U,V  tr(U T V)  tr(V TU)  V,U  u1v1  u2v2  u3v3  u4 v4 .
Recall: tr(A) =
sum of the
entries on the
main diagonal of
A.
 u1 u2 
 v1 v2 
 ,V  
.
U
 u3 u4 
 v3 v4 
U  U,U  2  u12  u22  u32  u42 ,
1
and
d(U,V )  U  V  U  V,U  V  2  U  V,U  V 
1
 (u1  v1 )2  (u2  v2 )2  (u3  v3 )2  (u4  v4 )2 .
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9
How to Find an Inner Product, Norm, and Distance for
Vectors (Matrices) in M22? (Cont.)
Example 2: Let
 u1 u2   2 1 
 v1 v2   0 9 


U
 ,V  

u
u
3
1
v
v
4
6
 3 4  
 3 4  


be matrices in the vector space M22. Find the inner product,
norms and distance for U and V in M22.
U,V   tr(U T V )  tr(V TU)  u1v1  u2 v2  u3v3  u4 v4
 (2)(0)  (3)(4)  (1)(9)  (1)(6)  9.
Recall: tr(A) = sum
of the entries on the
main diagonal of A.
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10
How to Find an Inner Product, Norm, and Distance for
Vectors (Matrices) in M22? (Cont.)
U  U,U  2  u12  u22  u32  u 42  2 2  (1)2  32  12  15,
1
V  V,V  2  u12  u22  u 32  u 42  0 2  9 2  4 2  6 2  133,
1
d(U,V )  U  V  U  V,U  V  2  U  V,U  V 
1
 (u1  v1 )2  (u2  v2 )2  (u 3  v3 )2  (u4  v4 )2
 (2  0)2  ((1)  9)2  (3  4)2  (1  6)2
 4  100  1  25  130.
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11
and
How to Find an Inner Product, Norm, and Distance for
Vectors (Polynomials) in P2?
Let P2 be the set of all polynomials
r of degree less than or equal
to two. For two polynomials p, q P2 ,
p(x)  a0 x 0  a1 x1  a2 x 2 ,
q(x)  b0 x 0  b1 x1  b2 x 2
we define
r r
 p, q  a0b0  a1b1  a2b2 ,
r
r r 12
p   p, p  a 20  a12  a22 , and
r r
r r
r r r r 12
d( p, q)  p  q   p  q, p  q
 (a0  b0 )2  (a1  b1 )12  (a3  b3 )2 .
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12
How to Find an Inner Product, Norm, and Distance for
Vectors (Polynomials) in P2? (Cont.)
Example 3: Find the inner product, norms, and distance for p
and q in P2, where
p(x)  8  3x  5x 2 and q(x)  1  2x  7x 2 .
r r
 p, q  a0b0  a1b1  a2b2  (8)(1)  (3)(2)  (5)(7)  8  6  35  49,
r
r r 12
p   p, p  a02  a12  a22  8 2  (3)2  5 2  98  7 2,
r
r r 1
q  q, q 2  b02  b12  b22  12  (2)2  7 2  54  3 6, and
r r
r r
r r r r 1
d( p, q)  p  q   p  q, p  q 2  (a0  b0 )2  (a1  b1 )12  (a3  b3 )2
 (8  1)2  ((3)  (2))2  (5  7)2  49  1  4  54  3 6.
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13
How to Find an Inner Product, Norm, and Distance for
Vectors (Functions) in C[a,b]?
Let C[a,b] be the set of all continuous functions on [a,b].
For all f and g in C[a,b], we define
r r
 f , g 
b
 f (x)g(x)dx,
a
1
2
r
r r 1 

f   f , f  2    [ f (x)]2 dx  
a

b
b

f 2 (x)dx ,
a
1
2
r r
r r
r r r r 1 

and d( f , g)  f  g   f  g, f  g 2    [ f (x)  g(x)]2 dx  .
a

b
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14
How to Find an Inner Product, Norm, and Distance for
Vectors (Functions) in C[a,b]? (Cont.)
Example 4: Let f = f(x) = 1 and g = g(x) = -x. Find the inner
product, norms, and distance on C[2,4].
r r
 f , g 
4
4

f (x)g(x)dx   (1)(x)dx  
2
2
1
2
r
r r 1 4

f   f , f  2    [ f (x)]2 dx  
2

1

r
r r 1 
g   g, g 2    [g(x)]2 dx  
2

4
2
2 4
x
2
 (8)  (2)  6,
2
4

4
 dx  x 2  2,
f (x)dx 
4
2
2
2
4
4
2
g
 (x)dx 
2
(x)
dx 

2
2
r r
r r
r r r r 1 

2
2
and d( f , g)  f  g   f  g, f  g    [ f (x)  g(x)] dx 
2

4
4

4
2
(1

x)
dx 

2
2
(1

2x

x
)dx

(x

x


2
2
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3
x3
3
1
4
2
2
2
4
x
2
) 7 .
3 2
3
15
14
,
3
The Cauchy-Schwarz Inequality
r r
r r
 f , g  f g
Cauchy-Schwarz Inequality:
for all f, g in an inner product space V, with the norm induced by
the inner product. We evaluate the Cauchy-Schwarz inequality
for the previous examples.
r r
r r
1. u, v  1  6 5  u v
from Example 1.
2 U,V   9  15 133  U V
r r
r r
3.  p, q  49  (7 2 )(3 6 )  p q
r r
r r
14
4.  f , g  6  ( 2 )(2
) f g
3
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from Example 2.
from Example 3.
from Example 4.
16
Orthogonality for Vectors (Functions) in C[a,b]
Example 5: Let f = f(x) = 1 and g = g(x) = ex. Find the inner
product on C[0,2].
2
2
r r
 f , g   f (x)g(x)dx   (1)(ex )dx  e2  e0  e2  1  0
0
0
Since the inner product is not zero, f and g are not orthogonal
functions in C[0,2].
Example 6: Let f = f(x) = 5 and g = g(x) = cos(x). Find the
inner product on C[0,π].



r r
 f , g   f (x)g(x)dx   5 cos(x)dx  5  cos(x)dx
0
0
0
 5[sin( )  sin(0)]  5[0  0]  0
Since the inner product is zero, f and g are orthogonal functions
in C[0,π].
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17
Orthogonality for Vectors (Functions) in C[a,b] (Cont.)
Example 7: Let f = f(x) = 4x and g = g(x) = x2. Find the inner
product on C[-1,1].
r r
 f , g 
1
1
 f (x)g(x)dx   (4x)(x
1
1
1
2
 
)dx  4  x dx  x
3
1
4
1
1
 11  0
Since the inner product is zero, f and g are orthogonal functions
in C[-1,1].
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18
Inner Product and Orthogonality for
Vectors (Functions) in C[a,b] (Cont.)
Example 8: Let f = f(x) = 1 and g = g(x) = sin(2x). Find the
inner product on C[-π,π].
r r
 f , g 






1
f (x)g(x)dx   (1)(sin(2x))dx   sin(2x)dx   cos(2x)
2



1
1
 1
  1

   cos(2 )     cos(2 )    cos(2 )  cos(2 )  0,
2
2
 2
  2

cos(2x) is even.
Since the inner product is zero, f and g are orthogonal functions
in C[-π,π].
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19
The Generalized Theorem of Pythogoras
Generalized Theorem of Pythogoras: If f and g are orthogonal
r 2 r 2
r 2
vectors in an inner product space V, then f  g  f  g .
r r
Proof: Let f and g be orthogonal in V, then  f , g  0.
r r 2
r r r r
r r r
r r r
Thus, f  g   f  g, f  g   f , f  g   g, f  g
r r
r r
r r
r r
  f , f    f , g   g, f    g, g
r 2
r r
r 2
 f  2 f , g  g
r 2 r 2
 f  g .
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20
The Generalized Theorem of Pythogoras (Cont.)
r
r 2
f g  f
2
r
 g
The Generalized Theorem of Pythogoras
We evaluate the theorem for the previous examples. From
example 6, we have


0
0
2
r
f  g   [5  cos(x)]2 dx   (25  (2)(5)cos(x)  cos 2 (x))dx
2



0
0
0
  25 dx  2  (5)cos(x)dx   cos 2 (x)dx
r
 f
Rev.F09
2
r
r 2
 25, cos(x)  g  f
2
r
r 2
 2(0)  g  f
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2
r 2
 g .
21
The Generalized Theorem of Pythogoras (Cont.)
From Example 7, we have
1
1
1
1
r
f  g   [4x  x 2 ]2 dx   (16x 2  (2)(4x)x 2  x 4 )dx
2
1
1
1
1
1
1
  16x 2 dx  2  4x 3 dx   x 4 dx
r
 f
2
r
r 2
 24x, x   g  f
2
2
r
r 2
 2(0)  g  f
2
r 2
 g .
Notice that we have used orthogonality to obtain our results
without directly computing the integrals.
Next, we will compute the integrals directly to obtain the result.
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22
The Generalized Theorem of Pythogoras (Cont.)
2
f
1
r r
 f, f 

1
1
1
1
f 2 (x)dx   [ f (x)]2 dx   (4x)(4x)dx  16  x 2 dx  (16)
1
1
1
1
1
1
1
1
x5
r 2
r r
2
2
2
2
4
g   g, g   g (x)dx   [g(x)] dx   (x )(x )dx   x dx 
5
1
1
1
1

1
3 1
x
3
2
5
1
1
r r 2
r r r r
2
f  g   f  g, f  g   [ f (x)  g(x)] dx   (4x  x 2 )(4x  x 2 )dx
1
1
 16x
8x
x 
  (16x  8x  x )dx  

 
4
5
 3
1
1
3
2
Thus,
Rev.F09
3
4
5
4
r
r 2
f g  f
2
1
r
166 32 2


  f
15
3 5
1
2
r 2
 g .
r 2
 g .
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23
1

32
3
How to Find the Cosine of the Angle Between Two
Vectors in an Inner Product Space?
Example 9: Let ℜ⁵ have the Euclidean inner product.
Find the cosine of the angle between
u = (2,1,0,-1,5) and v = (1,-1,-2,3,1).
r r
u, v
u1v1  u2 v2  u3v3  .u4 v4  u5 v5
cos( )  r r 
u v
u12  u22  u32  u42  u52 v12  v22  v32  v42  v52


(2)(1)  (1)(1)  (0)(2)  (1)(3)  (5)(1)
2 2  12  0  (1)2  5 2 12  (1)2  (2)2  (3)2  12
3
.
4 31
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24
How to Find the Cosine of the Angle Between Two
Vectors in an Inner Product Space? (Cont.)
Example 10: Let P2 have the previous inner product.
Find the cosine of the angle between p and q.
p  p(x)  a0 x 0  a1 x1  a2 x 2  2  6x  3x 2 ,
r
q  q(x)  b0 x 0  b1 x1  b2 x 2  2  x  4x 2
r r
 p, q
cos( )  r r 
p q

a02  a12  a22 b02  b12  b22
(2)(2)  (6)(1)  (3)(4)
2 2  6 2  32 (2)2  (1)2  (4)2
2

.
7 21
Rev.F09
a0b0  a1b1  a2b2
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Quick Review on a Basis for a Vector Space and the
Dimension of a Vector Space
r
r
Let v1 , v2 ,..., vn be nonzero vectors in V. Then for
r r
r , we have that
S  {v1, v2 ,..., vn }
r
r
W = span(S) = {all linear combinations of v1 , v2 ,..., vn } is a
subspace of V.
If S contains some linearly dependent vectors, then the
dim(W) < n.
If S is a linearly independent set of vectors, then S is a basis
for the vector space W. W is a proper subspace of V, if
dim(W) < dim(V). If dim(W) = dim(V) = n, then S is a basis
for V.
A non-trivial vector space V always have two subspaces:
the trivial vector space, {0} and V.
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Quick Review on a Basis for a Vector Space and the
Dimension of a Vector Space
So, we need two conditions for the set S to be a basis of a
vector space, it must be a linearly independent set and it
must span the vector space. The number of basis vectors
in S is the dimension of the vector space.
For example: dim(ℜ²)=2, dim(ℜ³)=3, dim(ℜ⁵)=5,
dim (P2)=3, dim(M22)=4, and dim(C[a,b]) is infinite.
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Rank, Row Rank, and Column Rank of a Matrix
The dimensions of the row space of A, column space of A, and
nullspace of A are also called the row rank of A, column rank
of A, and nullity of A, respectively.
The rank of A = the row rank of A = the column rank of A, and
is the number of linearly independent columns in the matrix
A.
The nullity of A = dim(null(A)) and is the number of free
variables in the solution space.
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The Orthogonal Complement of a Subspace W

The orthogonal complement of W, W is the set of all vectors
in a finite dimensional inner product space V that are
orthogonal to every vector in W.
Both W and W  are subspaces of V where

dim(W) + dim( W ) = dim(V).
 
The only vector common to W and W  is 0, and W  (W ) .
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29
The Four Fundamental Matrix Spaces
Let A be an m x n matrix, then:
1.The null(A) and the row(A) are orthogonal complements in
ℜⁿ with respect to the Euclidean inner product.
2.The null(AT) and the col(A) are orthogonal complements in
ℜᵐ with respect to the Euclidean inner product.
Note that row(A) = col(AT) and col(A) = row(AT).
The four fundamental matrix spaces are:
1. row(A) = col(AT),
2. col(A) = row(AT),
3. null(A), and
4. null(AT).
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How to Find a Basis for the Orthogonal Complement of
the Subspace of ℜⁿ Spanned by the Vectors?
Example 11: Find a basis for the orthogonal complement of
the subspace of ℜⁿ spanned by the vectors v1, v2, and v3.
Let V = span({v1, v2, v3}). Then, V is a subspace of of ℜⁿ if v1,
v2, v3 ∈ ℜⁿ. Let
r
r
v1  (1,1, 3), v2  (5,4,4), v3  (7,6,2).
From the last module, we know that the null(A) is the
orthogonal complement to the row(A) = col(AT), and that the
null(A) is the solution space of the homogeneous system, Ax =
0. Let v1, v2, v3 be the row vectors of the matrix A. Then,
row(A) is the orthogonal complement to null(A) which is the
solution space to Ax = 0. We shall use Gauss Elimination to
reduce A to a row echelon form to find the basis for null(A).
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How to Find a Basis for the Orthogonal Complement of
the Subspace of ℜⁿ Spanned by the Vectors? (Cont.)
r1
r2
r3
 1

 5
 7
r1
5r1  r2
7r1  r3
r1
r2
r2  r3
1
4
6
 1

 0
 0
 1

 0
 0
3
4
2
1
1
1
1
1
0

A


3
19
19
3
19
0








Since the null(A) is the
solution space of the
homogeneous system, Ax =
0, the general solution of the
system is:
x3  s,
x2  19x3  0, x2  19s  0, x2  19s.
x1  x2  3x3  0, x1  19s  3s  0, x1  16s.
Note that column 1 and column 2, with the red leading 1’s, are
linearly independent. The corresponding columns of A form a basis
for the col(A). Then, dim(col(A)) = 2.
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How to Find a Basis for the Orthogonal Complement of
the Subspace of ℜⁿ Spanned by the Vectors? (Cont.)
The nonzero rows with the leading 1’s are linearly
independent. These rows form a basis for row(A). Then,
dim(row(A)) = 2.
Note that the row space and column space are both two
dimensional, so rank(A) = 2. The rank(A) is the common
dimension of the row space of A and the column space of A.
So, rank(A) = rowrank(A) = colrank(A) = dim(col(A))
=dim(row(A)) is always true for any m x n matrix A.
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How to Find a Basis for the Orthogonal Complement of
the Subspace of ℜⁿ Spanned by the Vectors? (Cont.)
Thus, the solution vector is:
 x1  

 16 
16s

 



 x2    19s   s  19 
 x   s 
 1 


3


 16 
and the vector
w1   19 


 1 
{w1} forms a basis for the null(A). In other words, {w1} =
{(16,19,1)} forms a basis for the orthogonal complement of
row(A).
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How to Find a Basis for the Orthogonal Complement of
the Subspace of ℜⁿ Spanned by the Vectors? (Cont.)
Since B = {w1} is a basis for null(A), span(B) = null(A), and
dim(null(A)) = 1 is the nullity of A. In this example, rank(A) is 2
and the nullity of A is 1, and the number of columns of A is 3.
For any m x n matrix A, rank(A) + nullity(A) is n, which is the
number of columns of A.
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35
Let’s Review Some Equivalent Statements
If A is an n x n matrix, and if TA: ℜⁿ→ ℜⁿ is multiplication by
A, than the following are equivalent. (See Theorem
6.2.7)
•
•
•
•
•
•
•
•
•
Rev.F09
A is invertible.
Ax = 0 has only the trivial solution
The reduced row-echelon form of A is In.
A is expressible as a product of elementary matrices.
Ax = b is consistent for every n x 1 matrix b.
Ax = b has exactly one solution for every n x 1 matrix b.
det(A) ≠ 0.
The range of TA is ℜⁿ.
TA is one-to-one.
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Let’s Review Some Equivalent Statements (Cont.)
•
•
•
•
•
•
•
•
•
•
Rev.F09
The column vectors of A are linearly independent.
The row vectors of A are linearly independent.
The column vectors of A span ℜⁿ.
The row vectors of A span ℜⁿ.
The column vectors of A form a basis for ℜⁿ.
The row vectors of A form a basis for ℜⁿ.
A has rank n = row rank n = column rank n =
dim(col(A)).
A has nullity 0 = dim(null(A)).
The orthogonal complement of the nullspace of A,
null(A), is ℜⁿ.
The orthogonal complement of the row space of A,
row(A), is {0}.
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What have we learned?
We have learned to:
1.
Define and find the inner product, norm and distance in a given
inner product space.
2.
Find the cosine of the angle between two vectors and determine if
the vectors are orthogonal in an inner product space.
3.
Find the orthogonal complement of a subspace of an inner
product space.
4.
Find a basis for the orthogonal complement of a subspace of ℜⁿ
spanned by a set of row vectors.
5.
Identify the four fundamental matrix spaces of an m x n matrix A,
and know the column rank of A, the row rank of A, and the rank of
A.
6.
Know the equivalent statements of an n x n matrix A.
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38
Credit
Some of these slides have been adapted/modified in part/whole from the
following textbook:
• Anton, Howard: Elementary Linear Algebra with Applications, 9th Edition
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39