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Preface
The work presented in this book is a result of teaching my students for close to two decades.
Much of the work can be found in other excellent books written by imminent authors. The
approach used in the book is student friendly and can be read by anyone familiar with early years
of undergraduate study. The material presented is sufficient for an introductory single semester
course in linear Algebra. A number of theorems have just been stated to allow ease of reading.
A Wafula
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TABLE OF CONTENTS
Chapter I
Vector spaces Page 4
Chapter II
Vector subspaces Page16
Chapter III
Bases of vector spaces Page 20
Chapter IV
Linear systems of equations Page31
Chapter V
Linear transformations Page 40
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CHAPTER I
VECTOR SPACES
We discuss a basic structure in linear Algebra that forms a foundation of solutions to many
problems in diverse fields namely vector spaces.
Definition;
A vector space V is a non-empty set of elements called vectors on which is defined a concept of
equality of vectors and two operation called vector addition and scalar multiplication with the
following properties.
i.
If a, b V then a  b  V for all a, b V . (Closure of vector addition).
ii.
k, a  V for all the aV and kR.(Closure of scalar multiplication).
iii.
 a  b   c  a   b  c  for all vectors a, b, c  V .(Associative law).
iv.
v.
There exists a vector called zero and written 0 such that 0  a  a for all aV.
For every vector a  V there exists a vector called minus a and denoted by -a such
that a+(-a)=0
vi.
a  b  b  a for all vectors a and b in V
vii.
k  a  b   ka  kb for all a and b in V and k in R
ix.
 p  q  a  pa  qa for all vectors a in V and scalars p and q in R
 pq  a  p  qa  for all scalars p and q in R and vector a in V
x.
1a  a for all a in V
viii.
Examples of vector space
1. The space R2 of all column vectors x=(𝑥𝑥1 ).
2
Note:
i.
4
1
𝑥 + 𝑦 = (𝑥𝑥1 ) + (𝑦𝑦1 ) = (𝑥𝑥1 +𝑦
)
+𝑦
2
2
2
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1
(𝑘𝑥) = 𝑘 (𝑥𝑥1 ) = (𝑘𝑥
) 𝑅 2
𝑘𝑥
ii.
2
(00)
iii.
0=
iv.
1
For 𝑥 = (𝑥𝑥1 ) 𝑡ℎ𝑒𝑛 − 𝑥 = (−𝑥
)
−𝑥
v.
∈𝑅
2
2
2
2
(𝑥 + 𝑦) + 𝑧 =
(𝑥𝑥1 ) +
2
(𝑦𝑦1 )
2
+𝑦1 +𝑧1
1
+ (𝑧𝑧1 ) = (𝑥𝑥1 +𝑦
) + (𝑧𝑧1 ) = (𝑥𝑥1+𝑦
)
+𝑦
+𝑦
2
2
2
2
2
2
3
One can verify that the remaining condition for a vector space are satisfied.
𝑥1
2. Consider R the space of all column vectors of the type 𝑥 = (𝑥2 ) where addition in R n
𝑥3
is performed according to corresponding entries, namely
n
𝑥1
𝑦1
𝑥1 + 𝑦1
𝑦2
𝑥2
𝑥 +𝑦
𝑥 + 𝑦 = ( ⋮ ) + ( ⋮ ) = ( 2 ⋮ 2)
𝑥𝑛
𝑦𝑛
𝑥𝑛 + 𝑦𝑛
𝑘𝑥1
𝑘𝑥
𝑘𝑥 = ( 2 ) for any value 𝑘 in R
⋮
𝑘𝑥𝑛
0
The zero in R is the vector with zero entries i.e. 0 = (0).
⋮
0
n
One can verify that R n is a vector space. This is called the Euclidean vector space
3. Let M(2) be the space of all 2x2 matrices where vector addition is performed by
𝑥11
𝑥 + 𝑦 = [𝑥
21
𝑥12
𝑦11
𝑥22 ] + [𝑦21
The zero vector is 0 = [
𝑦21
𝑥11 + 𝑦11, 𝑥12+ 𝑦12
𝑦22 ]=[𝑥21 + 𝑦21, 𝑥22 + 𝑦22 ] for all 𝑖𝑛 𝑅 .
0 0
]
0 0
One can verify that M(2) is a vector space.
The space p n consisting of all polynomial in x of degree n or less is a vector space where addition
and scalar multiplication are defined as follows
a x
n
5
n
 a n 1 a1x  a 0   b n x n  b n 1x n 1  b1x  b 0   a n  b n  x n   a n 1  b n 1  x n 1  a1  b1  x  a 0  b 0
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And
k  a n x n +a n 1x n a1x  a   ka n x n  ka n 1x n ka1x  ka 0
The zero in p n is the zero polynomial. p n is a vector space
4. Let C  0,1 be the space of all continuous real valued functions from the interval [0,1]
Addition and scalar multiplication are defined as follows
 f  g  x   f  x   g  x 
.
 kf  x   kf  x  for all k  R and x  0,1
That the above functions are continuous is well established in calculus.
The zero in C  0,1 is the constant zero function. C  0,1 is an example of an infinite dimensional
vector space.
In the Euclidean space we have a concept of geometry. A vital tool in this vector space that can be
used to investigate its geometry is called the dot product.
Definition
If x, y  R n the dot product (scalar product) of x and y is defined by:
x.y  x1y1  x 2 y2 x n yn
Example
1
2
(𝑖) (−1) . ( 4 ) = (1𝑥2) + (−1𝑥4) + (2𝑥10) = 2 − 4 + 20 = 18
2
10
2
(ii) (−1
). (−1
) = (−1𝑥2) + [2𝑥(- − 1)] = −2 + (−2) = −4.
2
0
1
(𝑖𝑖𝑖) (0) . (0) = (0𝑥1) + (0𝑥0) + (1𝑥0) = 0 + 0 + 0 = 0.
1
0
Definition
The vector x and y are said to be orthogonal (perpendicular) if x. y=0
properties of the dot product.
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x. x  0 and x. x  0 if and only if x  0
1.
2. x. y  y.x for all 𝑥 and 𝑦 in V.
3. ( x. y ).z  x.( y.z ) for all 𝑥, 𝑦 and z in V.
x. y  z   x. y  x. z for all x, y, z  V
4.
5. 𝒌𝒙. 𝒚 = 𝒙. 𝒌𝒚 = 𝒌(𝒙. 𝒚) for all scalars k and vectors x and y
Definition
In R n we define the magnitude of a vector (the norm) to be
x = √𝑥. 𝑥
Properties of the norm
From the definition of the norm we identify the properties of the norm as follows
(i)
x ≥ 0 and x = 0 if and only if 𝑥 = 0
(ii)
kx = kx for all x in R .
(iii)
x + y ≤ x + y for all vectors 𝑥, 𝑦 in R
n
n
Example
1
2
1. Let 𝑥 = (−1) 𝑦 = ( 3 )
0
−1
Evaluate 2x + 3y.
1
2
2
6
8
2x + 3y = 2 (−1) + 3 ( 3 ) = (−2) + ( 9 ) = ( 7 ).
0
−1
0
−3
−3
8
8
We first compute ( 7 ) . ( 7 ) = 64 + 49 + 9 = 122.
−3
−3
So 2x + 3y = √122 = 11.045.
2. Given that x and y are orthogonal vectors, simplify the expression
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 x  2y.3x  2y 
Solution
 x  2y.3x  2y  3x. x  y   2y. x  2y   3x 2  6xy  2yx  4y2
 3x 2  4yx  4y2 =3x.x  4y.y= x  y
2
2
(𝑥. 𝑦 = 0 . Since the vectors are orthogonal)
3. Given that x+2y2=4 and 2x-3y2=9 find x and y if x and y are
Orthogonal
4. Prove the parallelogram law that
x y  x y  2 x 2 y
2
5.
2
2
2
.
Prove that the diagonals of a Rhombus intersect at right angles.
6. Show in a right-angled triangle whose longest side is 𝑐 it must be that
c  a  b
2
2
2
.
Theorem
The zero in a vector space is unique
Proof
Let 𝟎 and 𝟎’ be zeroes in a vector space V then consider
0  0’  0’(0 is a zero) .
However,
0  0’  0 (0’ is a zero) . Hence 0’ = 0 + 0’ = 0. And so zero is unique in V.
Theorem
0a  0
For all 𝑎𝑉.
Proof
It is clear that
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0  0  0 .So
0a   0  0 a  0a  0a .
And from the uniqueness of zero we must infer that
0a  0 .
Theorem
k0=0 for all kR
Proof
Now,
0+0=0 and so 𝑘0 = (0 + 0)𝑘 . Consequently 0 = 0𝑘 + 0𝑘 , and from the uniqueness of 0
it must be that 0𝑘 = 0.
Theorem
a   1 a
For all vectors aV.
Proof
a+(-1)a=(1   1 )a  0a  0
From the definition of a in V, we have that  1 a  a ,.
Theorem
Prove that –a is unique
Proof
Suppose that b and b’ are minuses of a. Then
b  b  0  b   a  b’
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  b  a   b’
 0  b’
 b’
Therefore, – 𝑎 is unique
Theorem
If a  b if and only if a  c  b  c for all a, b,c  V
Proof
If a  b then
 a  c   b  c  (a  c)  (a  c)  0
Hence 𝑎 + 𝑐 = 𝑏 + 𝑐
Conversely, if a  c  b  c for all a, b,c  V ,
then
a=a  0  a  (c+-c)=(a+c) +-c=(b+c)-c=b+(c+-c)=b+0=b
The angle between vectors
In this section we make application of the geometry of a triangle to derive a remarkable
relationship of the dot product and the angle between two vectors.
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
a
In the diagram  is the angle between the vectors 𝑎 and 𝑏 apply the cosine rule in OAB to obtain
b-a2=a2+b2-2ab 𝑐𝑜𝑠
But b-a2   b  a  . b  a 
= b.b  b.a  a.b  a.a
=b2 2a.b +a2
Hence
a2+b2-2a.b=a2+b2-2ab𝑐𝑜𝑠 i.e.
𝑐𝑜𝑠𝜃 =
𝑎. 𝑏
ab
Note that if a and b are parallel vectors then =0.
let b = ka so,
𝑎. 𝑏 =
𝑎. 𝑘𝑎
aka
2
=
𝑘a
2
ka
= ±1 i.e cos=±1 so =0 or 180 degrees .This agrees well with the orientation
of parallel vectors.
If a and b are perpendicular then   900 and cos  0 .On the other hand, if a and b are
perpendicular then a.b  0 so
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0
𝑐𝑜𝑠 = ab = 0 . Hence =900
Example
1. Determine the acute angle between the vectors below
a) 𝑎 = (12) 𝑏 = (−3
)
4
−1
−1
b) 𝑎 = ( 2 ) 𝑏 = ( 0 )
0
2
Solution
𝑎) 𝑐𝑜𝑠 =
𝑐𝑜𝑠 =
𝑎.𝑏
−3+8
=
ab √(12 +22 )(−32 +42 )
5
√5𝑥25
=
5
5√5
=
1
√5
=
√5
5
√5
) = 63.4350
5
 = 𝑐𝑜𝑠 −1 (
−1
−1
b) 𝑎 = ( 2 ) 𝑏 = ( 0 )
0
2
𝑐𝑜𝑠 =
𝑎. 𝑏
ab
−1
−1
𝑎. 𝑏 = ( 2 ) . ( 0 ) = 1 + 0 + 0 = 1
0
2
=
1
√(−12 +22 +02 )(−12 +02 +22
1
=
1
=
1
1
=5
√5.5 √25
1
𝑐𝑜𝑠 = 5 . Hence = 𝑐𝑜𝑠 −1 5=
2. Prove that x + y ≤ x + y for all vectors 𝑥, 𝑦 in R n .
Proof
x  y  ( x  y ).( x  y )  x  2 x. y  y
2
2
2
 x  2 x . y cos  y  x  2 x . y  y  ( x  y ) 2
2
2
2
2
The result follows on taking square-roots of both sides. This relationship is called the triangle
inequality. It alludes to the fact that in a triangle the longest side cannot exceed the sum of the
other sides in length .
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The equation of a straight line
b
b-a
The equation of the line AP is given by
𝑂𝑃 = 𝑎 + 𝑡(𝑏 − 𝑎) where 𝑡 is a scalar
Example
1. Write the value of the line joining (0,1.-1) and (3,0,2) and hence find the parametric
equation of the line.
Solution
0
3
0
0
3
( 1 ) + 𝑡 {(0) − ( 1 )}=( 1 ) + 𝑡 (−1)
−1
2
−1
−1
3
x
0
3
(y) = ( 1 ) + t (−1)
z
−1
3
is the vector equation of the line.
The parametric equations of the line are
𝑥 = 3𝑡, 𝑦 = 1 − 𝑡, 𝑧 = −1 + 3𝑡
2. Find the vector equation and parametric of the line joining the points
𝐴 (2, −1, 3) and 𝐵 (0,5, −1)
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Solution
From the equation 𝑂𝑃 = 𝑎 + 𝑡(𝑏 − 𝑎) we have
2
0
2
2
−2
(−1) + 𝑡 {( 5 ) − (−1)}=(−1) + 𝑡 ( 6 )
3
−1
3
3
−4
𝑥
2
−2
(𝑦) = (−1) + 𝑡 ( 6 )
𝑧
3
−4
The parametric equations are
𝑥 = 2 − 2𝑡
𝑦 = −1 + 6𝑡
𝑧 = 3 − 4𝑡
Equation of a plane
The equation of a plane on R3 is of the form 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 𝑘 where 𝑎 , 𝑏, 𝑐 and k are constants .
𝑎
And vector (𝑏 ) is the normal to the plane. To see this fact note that if (𝑝, 𝑞, 𝑟) is particular point
𝑐
on the plane and (𝑥, 𝑦, 𝑧) is a general point on the same plane then
𝑝−𝑥
𝑎
( 𝑏 ) . (𝑝 − 𝑦 ) = 0
𝑞−𝑧
𝑐
On further expansion we have that 𝑎𝑝 − 𝑎𝑥 + 𝑏𝑝 − 𝑏𝑦 + 𝑐𝑞 − 𝑐𝑧 = 0. Consequently, 𝑎𝑥 +
𝑏𝑥 + 𝑐𝑧 = 𝑎𝑝 + 𝑏𝑝 + 𝑐𝑞 = 𝑘
Remarks
The equation 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 0 is the equation of a plane through the origin.
When 𝑎 = 𝑏 = 0 then we have the equation of a plane parallel to the 𝑋 − 𝑌 plane.
When 𝑏 = 𝑐 = 0 then we have the equation of a plane parallel to the 𝑌 − 𝑍 plane.
Finally, when 𝑎 = 𝑐 = 0 then we have the equation of a plane parallel to the 𝑋 − 𝑍 plane.
Question
1. Find the point of the intersection of the line joining (0, −1,1) and (3, −1,0) with plane
2𝑥 − 3𝑦 + 5𝑧 = 15.
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Question
2. Determine the acute angle between the planes
𝑥 − 2𝑦 + 5𝑧 = 3
2𝑥 + 𝑦 − 3𝑧 = 15.
Distance of a plane from the origin:
Let 𝐴(𝑝, 𝑞, 𝑟) be the nearest point from the origin on the plane 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 𝑘 then A is
located on the normal to the plane.
Thus
𝑎
𝑝
𝑏
𝑂𝐴 = 𝑡 ( ) = (𝑞 ) .
𝑐
𝑟
Substituting in the equation of the plane we obtain
𝑎2 𝑡 + 𝑏 2 𝑡 + 𝑐 2 𝑡 = 𝑘.
From which we find that
𝑎
𝑘
𝑂𝐴 = 𝑎2 +𝑏2+𝑐 2 (𝑏 ).
𝑐
Therefore the shortest Distance of the plane from the origin is
|𝑂𝐴| =
𝑘
√𝑎2 +𝑏 2 +𝑐 2
.
Distance of a plane from a given point:
Let 𝐴(𝑝, 𝑞, 𝑟) be the nearest point from a given point 𝑄(𝑚, 𝑛, 𝑡) on the plane 𝑎𝑥 + 𝑏𝑦 +
𝑐𝑧 = 𝑘 then the A is located on the normal to the plane passing through Q. We make the
transformation 𝑋 = 𝑥 − 𝑚, 𝑌 = 𝑦 − 𝑛 , 𝑍 = 𝑧 − 𝑡 .So that the new origin shifts to (𝑚, 𝑛, 𝑡)
.
The equation of the plane becomes 𝑎(𝑋 + 𝑚) + 𝑏(𝑌 + 𝑛) + 𝑐(𝑍 + 𝑡) = 𝑘
Which on rearrangement becomes 𝑎𝑋 + 𝑏𝑌 + 𝑐𝑍 = 𝑘 − 𝑎𝑚 − 𝑏𝑛 − 𝑐𝑡.
Now from the previous description the distance
𝑄𝐴 =
15
𝑘−𝑎𝑚−𝑏𝑛−𝑐𝑡
√𝑎2 +𝑏2 +𝑐 2
.
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CHAPTER II
Vector subspaces
Definition:
Given a vector space V, a non-empty subset W of V is called a vector subspace of V if W is a vector
space with respect to vector addition and scalar multiplication in V.
Theorem
W is a vector space of V if and only if
i.
a  b  W for all a, b  W
ii.
ka  W for all a  W and k  R
Proof
Suppose that W is a vector space of V then W is a vector space in its own right hence (i) and (ii) will
apply
Conversely
Suppose that (i) and (ii) are satisfied in W then we need to show that the eight remaining
conditions in a vector space apply in W
Since W   then a W
Now 0a  0 W by (i) then
 1 a  a  W by (ii).
All the remaining six conditions apply in W since W is a subset of V.
Examples
a) let V be a vector space then W={0} is a vector subspace of V
b) let V be a vector space and W=V then w is a vector space of V
c) let V=R2
𝑎
𝑊 = {( ) a + 2b = 0} is a vector space of R2
𝑏
𝑥
𝑎
Let ( ) , (𝑦)  𝑊 then
𝑏
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a  2b  0 and x  2y  0 .But,
𝑥
𝑎+𝑥
𝑎
( ) + (𝑦) = (𝑏 + 𝑦) . Now
𝑏
 a  x   2  b  y    a  x   2b  2y   a  2b   x  2y
 00  0
So,
𝑥
𝑎
( ) + (𝑦)  𝑊.
𝑏
Furthermore,
𝑎
𝑘𝑎
𝑘( ) = ( )
𝑏
𝑘𝑏
𝑘𝑎 + 2𝑘𝑏 = 𝑘(𝑎 + 2𝑏) = k 0  0 .
𝑎
Hence 𝑘 ( )  𝑊
𝑏
𝑥
d) 𝑊 = {(𝑦) x + y = 1} is not a subspace of R2
0
Note that 0 = ( ) 𝑊 𝑠𝑖𝑛𝑐𝑒 0 + 0 ≠ 1
0
𝑥
e) 𝑊 = {(𝑦) y = 𝑥 2 } is not a subspace of R2
1
Note ( )  𝑊
1
1
1
2
But ( ) + ( ) = ( )  𝑊 since 2≠22
1
1
2
Linear combinations
In this section we discuss a method of creating subspaces from a finite set of vectors.
Definition
A vector x  c1x1  c2 x 2  c3x 3  cn x n where c1,c2 ,c3 cn are scalars
x1, x 2 .x n are vectors in a vector space V is called a linear combination of the x i ’s
Example
1
2
) , 𝑥2 = ( )
−5
−1
1
11
2
𝑥 = 3𝑥1 + 5𝑥2 = 3 ( ) + 5 ( ) = (
).
−5
−28
−1
11
1
2
(
) is a linear combination of ( ) 𝑎𝑛𝑑 ( ).
−28
−5
−1
𝑙𝑒𝑡 𝑥1 = (
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SCHOOL OF MATHEMATICS
Remark
If the xi’s are orthogonal, the scalars can be found as follows
x  c1x1  c2 x 2  c3x 3  cn x n
We dot both sides by x1 obtain
 x1.x  
c1 (x1.x1)  c2 (x 2.x1 )  cn (x n .x1 ) = c1.(x1.x1 )  0  c1.(x1.x1 )
For x1≠0 we have that
Similarly, 𝑐𝑖 =
(𝑥𝑖 .𝑥)
𝑥𝑖 
2
( x 1 , x)  c1 x1
2
. Hence,
(𝑥1 .𝑥)
𝑥1 
2
= 𝑐1 .
, 𝑖 = 1,2, … , 𝑛.
Example 1
7
1
2
Verify that 𝑥1 = ( ) 𝑎𝑛𝑑 𝑥2 = ( ) are orthogonal and hence write ( ) as a linear
−5
2
−1
combination of the two vectors.
Solution
1
2
( ) . ( ) = 1 × 2 + 2 × (−1) = 0 . So 𝑥1 and 𝑥2 are orthogonal.
2
−1
Let
7
7
1
2
1
1
1
2
1
( ) = 𝑐1 ( ) + 𝑐2 ( ) . So that ( ) . ( ) = 𝑐1 ( ) . ( ) + 𝑐2 ( ) . ( ) and, hence
−5
−5
2
−1
2
2
2
−1
2
(7 − 10) = 𝐶1 (5) .
Hence 𝐶1 =
−3
5
19
1
and a similar computation using ( ) leads to 𝐶2 = 5 . Consequently
2
−3 1
19 2
7
)=
( )+
( )
−5
5 2
5 −1
(
Example 2
1. Show that (0,0,1) is not a linear combination of (1,1,0) and (0,2,0)
Proof
Suppose that
0
1
0
(0) = 𝑐1 (1) + 𝑐2 (2)
1
0
0
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SCHOOL OF MATHEMATICS
𝑐1
0
(0) = (𝑐1 + 2𝑐2 ) , this is absurd as 1≠0.
0
1
0
1
0
Hence (0) is not a linear combination of (1) and (2)
1
0
0
Question
4
0
1
2
2. Verify that (0) , (1) , (−2) are orthogonal and hence write (−2)as a linear combination of
1
0
0
3
the three vectors.
Definition
The subset W of all linear combinations of x1, x 2 ..x n is called the linear span of x1, x 2 ..x n .
We write W  span
x1, x2 ,.xn  for the linear span of x1, x 2 ..xn .
Theorem
The linear span of a set of vectors x1, x 2 ..x n is a vector subspace of the vector space V
Proof
Let c1x1  c2 x 2 .  cn x n and c’1 x1  c’2 x 2 .  c’n x n W
Then  c1x1  c2 x 2 .  cn x n    c’1 x1  c’2 x 2 .  
  c1  c’2  x1   c2  c’2  x 2 .   cn  c’n  x n W.
Note that we have obtained the above by repeated use of the associative and commutative law.
Finally
k  c1x1  c2x 2 .  cn x n   kc1x1  kc2x 2 .  kc2x 2 W.
The subset W of all linear combinations of the vectors x1, x 2 ..x n is called the linear
subspace generated by x1, x 2 ..x n
Remark
Span {0} = {0}
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Span 𝑉 = 𝑉
CHAPTER III
BASIS OF VECTOR SPACES
The objective in this chapter is to obtain the least number of vectors that can generate (span) a
vector space
Definition
A set of vectors x1, x 2 ,.x n in a vector space V is said to be linearly dependent if there exists a
vector in the set  x1, x 2 ,.x n  which is a linear combination of the other vectors.
Example
1
1
0
1
1
0
The set {( ) , ( ) , ( )} is linearly dependent since ( ) = ( ) + ( )
0
0
1
1
1
1
Note that any set of vector which contain zero is linearly dependent since 0=0x 1+0x2+…+0xn
Theorem
A set of vectors x1, x 2 ,.x n in a vector space V is linearly dependent if and only if there exists
scalars c1,c2 ,cn not all zero so that c1x1  c2 x 2 .  cn x n  0
Proof
Suppose that x1, x 2 ,.x n are linearly dependent
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At least one of the vectors is a linear combination of the others. We can assume x 1 is a linear
combination of the others. If not we re-arrange the set so that x1 is a linear combination of the
rest
x1  c2 x 2  c3x 3  cn x n .
So
0  x1  c2 x 2  c3x 3  cn x n
where the first scalar is -1≠0 .
Conversely,
Suppose c1x1  c2 x 2  cn x n  0 and not all the ci ’s are zero. Assume that c1≠0. If c1=0 rearrange the equation so that c1≠0 then
c2 x 2  c3x x .  cn x n  c1x1
i.e.
−𝑐2
𝑐1
𝑥2 +
−𝑐3
𝑐1
𝑐
𝑥3 + ⋯ + − 𝑐𝑛 𝑥𝑛 = 𝑥1
1
Hence x1 is a linear combination of the other vectors and so
 x1, x 2 ,.x n  is a linearly dependent
set.
LINEAR INDEPENDENCE
Definition
A set of vectors x1, x 2 ,.x n is said to be linearly independent if none of the vectors is a linear
combination of the other vectors
Theorem
The vectors x1, x 2 ,.x n are linearly independent if and only if the equation
c1x1  c2 x 2  cn x n  0 has one and only one solution c1  c2   cn  0
.
Proof
Suppose that x1, x 2 ,.x n are linearly independent and c1x1  c2 x 2  cn x n  0 . If any of the
scalars ci’ s is non-zero then x1, x 2 ,.x n are linearly independent contradicting the initial
assumption it must be that c1  c2   cn  0
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Conversely
Suppose that the equation c1x1  c2 x 2  cn x n  0 has one and only one solution c1=c2=……=cn=0
we must show that the xi’s are independent. Suppose that the xi’s are linearly dependent. There
exist scalars c1,c2 ,cn not all zero such that c1x1  c2 x 2  cn x n  0 . This contradicts the initial
assumption as all such scalars are zero.
Examples
2
−1
1. Show that ( ) 𝑎𝑛𝑑 ( ) 𝑎𝑟𝑒 𝑙𝑖𝑛𝑒𝑎𝑟𝑙𝑦 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡
1
2
Proof
0
2
−1
Suppose that 𝑐1 ( ) + 𝑐2 ( ) = ( ) then
0
1
2
[
2𝑐1 − 𝑐2
0
]=( )
𝑐1 + 2𝑐2
0
2𝑐1 − 𝑐2 = 0,𝑐1 + 2𝑐2 = 0
𝑐2 + 4𝑐2 = 0 5𝑐2 = 0𝑐2 = 0
𝑐1 =
𝑐2
2
0
𝑐1 = 2 𝑐1 = 0
1
0
0
Prove that the vectors (1) , (1) 𝑎𝑛𝑑 (1) are linearly independent in R3
0
0
2
Proof
Suppose that
1
0
0
0
𝑐1 (1) + 𝑐2 (1) + 𝑐3 (1) = (0)
0
0
2
0
𝑐1
0
0
0
𝑐
𝑐
( 2 ) + (𝑐2) + ( 3 ) = (0)
0
2𝑐3
0
0
𝑐1 + 0 + 0 = 0 𝑐1 = 0
0 + 0 + 2𝑐3 = 0𝑐3 = 0
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SCHOOL OF MATHEMATICS
𝑐1 + 𝑐2 + 𝑐3 = 0 (but c1=0, c2=0)
0 + 0 + 𝑐3 = 0𝑐3 = 0
2. Prove that the functions 𝑐𝑜𝑠 and 𝑠𝑖𝑛 are linearly independent in the space 𝐶 [0,2] of
continuous functions
Proof
Assume that for some scalars
𝑐1 cos  + 𝑐2 sin  = 0…….i
Differentiating
−𝑐1 𝑠𝑖𝑛 + 𝑐2 𝑐𝑜𝑠 = 0….ii
We multiply (i) by sin and (ii) by 𝑐𝑜𝑠 to obtain
𝑐1 sincos  + 𝑐2 𝑠𝑖𝑛2  = 0……iii
−𝑐1 𝑠𝑖𝑛𝑐𝑜𝑠 + 𝑐2 𝑐𝑜𝑠 2  = 0…..iv
Solving (iii) and (iv) simultaneously and eliminating c1 we obtain
𝑐2 𝑠𝑖𝑛2  + 𝑐2 𝑐𝑜𝑠 2  = 0 𝑐2 (𝑠𝑖𝑛2  + 𝑐𝑜𝑠 2 ) = 0
But 𝑠𝑖𝑛2  + 𝑐𝑜𝑠 2  = 1 , hence
𝑐2 . 1 = 0𝑐2 = 0
From (i) 𝑐1 cos  = 0 (cos is not always zero) .So it must be that 𝑐1 = 0 .
Question
3. Prove that 1 − 𝑥, 𝑥 2 − 1, 𝑥 3 + 2𝑥 − 1 are linearly independent in the vector space P3(x)
Question
4. Prove that 𝑒 −𝑥 , 𝑒 −2𝑥 , 𝑒 −3𝑥 are linearly independent in the space 𝐶 [0,1]
Question
4 Prove that if x1, x 2 ,.x n are orthogonal, then they are linearly independent.
5 Prove that if 𝑢, 𝑣, 𝑤 are linearly independent in a vector space V then 𝑢, 𝑢 + 𝑤, 𝑢 − 𝑤 are
also linearly independent.
Matrices and Matrix operations
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SCHOOL OF MATHEMATICS
We introduce matrices as useful tools of trade as will be evident in the subsequent topics.
A matrix is a rectangular arrangement of numbers. We add matrices according to corresponding
entries. For this reason addition is meaningful only for matrices of the same order.
Thus 𝐴 + 𝐵 = (𝑎𝑖𝑗 ) + (𝑏𝑖𝑗 ) = (𝑎𝑖𝑗 + 𝑏𝑖𝑗 ) 𝑖 = 1,2 … . , 𝑚 𝑗 = 1,2, … . , 𝑛
Matrix multiplication is computed by the dot product of rows of the first matrix with the columns
of the second matrix. More precisely the dot product of the 𝑖𝑡ℎ row in 𝐴 with the 𝑗𝑡ℎ column of 𝐵
yield the 𝑖 𝑗𝑡ℎ entry in 𝐴𝐵 .Matrix multiplication is only possible when the number of columns of
the first matrix is the same as the number of rows of the second matrix.
Elementary row operations
The following are called elementary row operations on a Matrix
i.
ii.
iii.
Interchanging any two rows
Multiplying a row of a Matrix by a non -zero scalar
Adding a scalar multiple of a row to another row
A matrix A is similar to B if B is obtained from A by a finite sequence of elementary operations
Echelon form
A matrix A is said to be in echelon form if for each non-zero row , the first nonzero element 𝑎𝑖𝑗 has
all the element below it as zero ,unless the row is the last row in the matrix.
The echelon form is one of the canonical forms we shall investigate.
Example
The matrices below are in echelon form
2 −1 1
[0 3 ], [0
0 0
0
−1 0
3
5]
0 −1
Each matrix can be reduced to echelon form by use of elementary row operations
Example
1. Reduce the matrix
1 −1 0
[2
3
2 ] to echelon form.
−1 3 −1
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SCHOOL OF MATHEMATICS
Solution
Replace the second row with the sum of −2 times the first row and the second row. Also replace
the third row with the sum of the first row and the third row. Secondly Replace the third row with
the sum of −2times the third row and the second row . Finally replace the third row with the sum
of −5times the third row to the second row
1
[2
−1
−1 0
1 −1 0
1 −1 0
1
3
2 ] ≡ [0 5 2] ≡ [0 5 2] ≡ [0
3 −1
0 2 1
0 1 0
0
−1 0
5
2]
0 −3
The final Matrix is in echelon form
2 Reduce the matrix below to echelon form
2 1
[3 2 ]
1 −1
3 Show that the rows of the matrix below are linearly independent
𝑎 𝑏 𝑐
[0 𝑑 𝑒 ]
0 0 𝑓
Theorem
The nonzero rows of a matrix when reduced to echelon form are linearly independent
The above theorem can be used to provide an alternative method of investigating
independence of vectors in 𝑅 𝑛 . The vectors are written as the rows of a matrix and then an
equivalent matrix is obtained that is in echelon form. If the echelon form does not contain a zero
row the vectors are linearly independent. If on the other hand the echelon form does contain a
zero row then the vectors are linearly dependent
Example
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SCHOOL OF MATHEMATICS
−4
−1
0
1) Investigate the vectors ( 2 ) , (1) , ( 3 ) for linear independence
3
5
0
Solution
−1 2 3
−1
Let the matrix 𝐴 = [ 0 1 5] then 𝐴 ≡ [ 0
−4 3 0
0
2
3
−1 2
1
5 ]≡[ 0 1
−5 −12
0 0
3
5]
13
The echelon form does not contain a Zero row so the three vectors are linearly independent
−1
0
−5
2) Show that the vectors ( 2 ) , (1) , ( 12 ) ,, are linearly dependent and show clearly how one
3
5
25
of the vectors is a linear combination of the other vectors
Solution
−1 2
Let 𝐴 = [ 0
1
−5 12
3
−1 2 3
−1 2
]
≡
[
]
(−5𝑅
+
𝑅
)
≡
[
5
0 1 5
0 1
1
3
25
0 2 10
0 0
3
5] (−2𝑅 ′ 2 + 𝑅 ′ 3 )
0
The echelon form equivalent to A has a zero row so the original vectors are linearly dependent.
Furthermore,
−2𝑅 ′ 2 + 𝑅 ′ 3 = 0 = −2𝑅2 + (−5𝑅1 + 𝑅3 )
Hence,
𝑅3 = 5𝑅1 + 2𝑅2 .
−1
0
−5
Clearly, ( 12 ) = 5 ( 2 ) + 2 (1)
3
5
25
3) Show that the polynomials 1 − 𝑥, 1 − 2𝑥 2 , 𝑥 2 − 𝑥 + 1,1 + 2𝑥 + 𝑥 2 are linearly dependent and
clearly indicate how one of the Polynomials is a linear combination of the other Polynomials
4) Prove that any three non- zero vectors 𝑅 2 are linearly dependent
Theorem
Given that the vectors x1, x 2 ,.x n are linearly independent and x  c1x1  c2 x2 .  cn xn then
this representation is unique
26
SCHOOL OF MATHEMATICS
Proof
Suppose that x  c1x1  c2 x2 .  cn xn and x  c1/ x1 +c2/ x2 cn/ xn
Then subtracting both sides we obtain
0  (c1  c1/ ) x1 +(c1  c2/ ) x2  (c1  cn/ ) xn .
Since the xi’s are linearly independent we must have that:
0  (c1  c1/ )  (c1  c2/ )    (c1  cn/ )
.
Hence
c1  c1/ ,c1  c2/ ,c1  cn/
.
Definition
A set of vectors x1, x 2 ,.x n is called a finite basis of a vector space V if x1, x 2 ,.x n are linearly
independent and generate (span) the vector space V i.e. span{ x1, x 2 ,.x n }=V
Definition
A vector space V is said to be finite dimensional if V has a finite basis. Otherwise, V is infinite
dimensional.
Theorem
Every vector space V that is different from the trivial vector space 𝑉 = {0} has a basis
Theorem
A set of vectors [ x1, x 2 ,.x n ] is a basis of V if and only if the set [ x1, x 2 ,.x n ] is maximal linearly
independent in V
Remark
This means that the vectors x1, x 2 ,.x n are linearly independent and if a new element is
introduced, the resulting set is linearly dependent.
Proof
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SCHOOL OF MATHEMATICS
Suppose x1, x 2 ,.x n is a finite basis of vector space V. we show that the set is maximal linearly
independent
Let xV then since x1, x 2 ,.x n are a basis therefore x is a linear combination of x1, x 2 ,.x n hence
[ x1, x 2 ,.x n , x ,] is a linearly dependent set. Hence
[ x1, x 2 ,.x n ] is maximal linearly independent
Conversely
Suppose that { x1, x 2 ,.x n } is maximal linearly independent. We must show that { x1, x 2 ,.x n } is a
basis for V. It is sufficient to show that { x1, x 2 ,.x n } is a generating set for V.
Let xV then {x, x1, x 2 ,.x n } is a linearly independent set. It must be that x is a linear combination
of x1, x 2 ,.x n , hence { x1, x 2 ,.x n } is a basis for V.
Theorem
A of vectors { x1, x 2 ,.x n } is a basis for vector space V if and only if { x1, x 2 ,.x n } is a minimal
generating (spanning) set of vectors for the vector space V
Remark
The set { x1, x 2 ,.x n } generates the vector space V but if any one of the vectors is deleted from
the list, the remaining vectors do not generate V
Theorem
If a vector space V has a finite basis then each basis of V has the same number of elements
Example
1
0
0
The vectors 𝑖 = (0) , 𝑗 = (1) , 𝑘 = (0) are a basis for R3
0
0
1
Note 𝑖, 𝑗, 𝑘 are orthogonal vectors in R3 hence they are linearly independent
Let
𝑥1
𝑥 = (𝑥2 ) 𝑥 = 𝑥1 𝑖 + 𝑥2 𝑗 + 𝑥3 𝑘
𝑥3
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So span {𝑖, 𝑗, 𝑘} = R3
Definition
If 𝑉 has a finite basis and that n is the number of elements in a basis of V then the dimension of V
is n
Definition
If { x1, x 2 ,.x n } is a finite basis for a Vector space V and
matrix or column vector representation of x by
x  c1x1  c2 x2 .  cn xn then define the
𝑐1
𝑐2
𝑥 = ( ⋮ ).
𝑐𝑛
Example
The Dimension of R4 is 4.
In R4 the vectors
1
0
0
0
0
1
0
0
𝑖 = ( ) , 𝑖2 = ( ) , 𝑖3 = ( ) , 𝑖4 = ( )
0
0
1
0
0
0
0
1
form the standard basis and so R4
has 4 dimensions.
Determine the dimension of the subspace of R3 below
i.
𝑤1 = {(𝑎, 𝑏, 0)a, bR}
ii.
𝑤2 = {(𝑎, 𝑎 − 𝑏, 𝑏)a, bR}
iii.
𝑤3 = {(𝑎, 𝑏, 𝑐)2a + c = b}
Solution
𝑊1 = {(𝑎, 𝑏, 0)a, bR}
𝑥𝜖𝑊1 ↔ 𝑥 = (𝑎, 𝑏, 0) = (𝑎, 0,0) + (0, 𝑏, 0) = 𝑎(1,0,0) + 𝑏(0,1,0) a,bR
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The vector (1,0,0) and (0,1,0) generated W1 since (1,0,0) and (0,1,0) are orthogonal and they are
linearly independent hence form a basis for 𝐷𝑖𝑚(𝑊1 ) = 2
𝑊2 = {(𝑎, (𝑎 − 𝑏), 𝑏)a, bR} ={(𝑎, 0,0) + (0, (𝑎 − 𝑏), 0) + (0,0, 𝑏)}
={𝑎(1,0,0) + 𝑎(0,1,0) − 𝑏(0,1,0) + 𝑏(0,0,1)}
={𝑎(1,1,0) + 𝑏(0, −1,1)}
Hence (1,1,0) and (0, −1,1) generates W2 .
The vector (1, 1, 0) and (0, −1,1) are linearly independent hence they form a basis for 𝑊2 .The
dimension of 𝑊2 is two .
𝑊3 = {(𝑎, 𝑏, 𝑐)2a + c = b}
𝑊3 = {(𝑎, 2𝑎 + 𝑐, 𝑐): 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑠𝑐𝑎𝑙𝑎𝑟𝑠 𝑐}
={(𝑎, 0,0) + (0,2𝑎 + 𝑐, 0) + (0,0, 𝑐)}
={𝑎(1,0,0) + 2𝑎(0,1,0) + 𝑐(0,1,0) + 𝑐(0,0,1)}
={𝑎(1,0,0) + 𝑎(0,2,0) + 𝑐(0,1,0) + 𝑐(0,0,1)}
={𝑎(1,2,0) + 𝑐(0,11)}.
Hence (1,2,0) and (0,1,1) are linearly independent and forms the basis for 𝑊3
The dimension of 𝑊3 is two.
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CHAPTER IV
LINEAR SYSTEMS OF EQUATIONS
A system of equations of the type below is called a linear system of equation of m equations in n
variables.
𝑎11 𝑥1 + 𝑎12 𝑥2 + ⋯ + 𝑎1𝑛 𝑥𝑛 = 𝑏1
𝑎21 𝑥1 + 𝑎22 𝑥2 + ⋯ + 𝑎2𝑛 𝑥𝑛 = 𝑏2
𝑎𝑚1 𝑥1 + 𝑎𝑚2 𝑥2 + ⋯ + 𝑎𝑚𝑛 𝑥𝑛 = 𝑏𝑚
This represents m equation in n unknowns can be written as a matrix equation as follows.
The matrix
𝑎11
𝑎21
𝐴=[ ⋮
𝑎𝑚1
𝑎12
𝑎22
⋮
𝑎𝑚2
⋯ 𝑎1𝑛
⋯ 𝑎2𝑛
⋮
⋮ ]
⋯ 𝑎𝑚𝑛
is called the coefficient matrix .
𝑏1
𝑏
The vector 𝐵 = ( 2 ) is called the constant Matrix. The system can now be summarized in a
⋮
𝑏𝑚
𝑥1
𝑥2
vector equation of the type 𝐴𝑋 = 𝐵 where 𝑋 = ( ⋮ ) is a solution to the system 𝐴𝑋 = 𝐵 , is a
𝑥𝑛
vector in Rn .
So, to solve the system we note that we obtain an equivalent system by any of the following
operations:
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SCHOOL OF MATHEMATICS
i.
ii.
iii.
Interchanging two equations
Multiply an equation by a non-zero scalar
Add a scalar multiple of one equation to another equation
Operations (𝑖) → (𝑖𝑖𝑖) are called elementary row operations similar to those defined on matrices.
The technique is called the Gaussian Elimination method. It involves applying a sequence of row
operation to reduce the number of variables in each subsequent equation and is illustrated below
Gauss elimination method
Solve the system
𝑥 − 2𝑦 + 3𝑧 = 3
(i)
2𝑥 − 4𝑦 + 5𝑧 = 0
( ii)
𝑥 + 3𝑦 − 𝑧 = 5
( iii)
Solution
We apply elementary operations on the system to obtain an equivalent system where each
subsequent equation has fewer variables
Equation (𝑖𝑖) ↔ 2(𝑖) − (𝑖𝑖) and (𝑖𝑖𝑖) ↔ (𝑖) − (𝑖𝑖)
𝑥 − 2𝑦 + 3𝑧 = 3
𝑧=6
−5𝑦 + 4𝑧 = −2
Interchanging rows we have that
𝑥 − 2𝑦 + 3𝑧 = 3
−5𝑦 + 4𝑧 = −2
𝑧=6
And now we substitute in the rest of the equations
−5𝑦 + 4(6) = −2
𝑦=
32
26
5
,𝑥 =
−23
5
,𝑧 =6
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The same steps can be achieved by use of row operations on the augmented matrix [𝐴; 𝐵] . The
objective is to reduce the augmented matrix to echelon form and the recover a more simple
system
1 2 3
 2 4 5

1 3 1
3
1 2 3 3 
1 2 3 3 


'
' 
'
' 
0  2 R1  R2  R2 ; R1  R3  R3 0 0 1 6   R2  R3 0 5 4 2
0 5 4 2
0 0 1 6 
5
From the reduced form we obtain the equations just was found in the earlier method
𝑥 − 2𝑦 + 3𝑧 = 3
−5𝑦 + 4𝑧 = −2
𝑧=6
Question
2𝑥 − 3𝑦 + 𝑧 = 5
5𝑥 − 4𝑦 + 5𝑧 = 1
Solution
𝑥
2 −3 1 𝑦
5
[
][ ] = [ ]
5 −4 5 𝑧
1
The augmented matrix
2
[
5
−5 1 5 2
]=[
−4 5 1 0
−5 1
5
]
7 5 −23
We replace the equation
2x − 5y + z = 5
7𝑦 + 5𝑧 = −23 .
Let 𝑧 =  then 7𝑦 = −5 − 23
𝑦=
𝑥=
33
−5−23
7
−17−11
7
. Here  is an arbitrary value, leading to infinity of solutions
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Question
Solve the system below
𝑥 − 5𝑦 + 7𝑧 = 5
2𝑥 − 3𝑦 + 5𝑧 = 1
4𝑥 − 𝑦 − 𝑧 = 0
Question
Find the solutions of the following systems of equation
1. 𝑥1 + 2𝑥2 − 𝑥3 = 0
2𝑥1 + 4𝑥2 − 𝑥5 = −4
𝑥1 − 2𝑥3 − 𝑥4 = 2
2𝑥1 + 3𝑥2 − 𝑥3 + 𝑥4 − 2𝑥5 = −5
2. 𝑥1 + 𝑥2 = 1
𝑥1 + 𝑥2 = 2
Solve the system below
2𝑥 − 𝑦 = 5
𝑥 + 3𝑦 = 1
4𝑥 − 7𝑦 = 2
Solution
3
1
2 −1 5 1
[1 3 1]=[0
7
3]
4 −7 2 0 −17 −2
1
3
1
1
3
1
=[0 −133 57 ]=[0 −133 57 ]
0 −133 −14 0
0
−71
We recapture the equations from echelon form,
𝑥 + 3𝑦 = 1
−133𝑦 = 57
0 = −71
From the equations we obtain 0 = −71 .This is a contradiction. Hence the system is
inconsistence
Remark
A linear system either has a unique solution or infinity of solutions or is inconsistent.
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The Jordan Gauss reduction Method
The method above can be used to solve a system which has unique solution . The augmented
matrix [𝐴; 𝐵] is reduced to the form [𝐼; 𝑃] .From which the equivalent system is
𝐼𝑋 = 𝑃 Hence 𝑋 = 𝑃
Question
Solve the system below
𝑥 − 5𝑦 + 7𝑧 = 5
2𝑥 − 3𝑦 + 5𝑧 = 1
4𝑥 − 𝑦 − 𝑧 = 0
Recall that the system when reduced to echelon form is
23 


5
1 2 3 3 
1 2 0 15
1 0 0

0 5 4 2   R  4 R  R ' , R  3R  R ' 0 5 0 26   1 R  R '' , R  2 R  R '' 0 1 0 26 
2
3
2
1
3
1
2
1
2
1


 5 2

5
5
0 0 1 6 
0 0 1 6 
0 0 1

6



Leading to the same solution namely,𝑥 =
23
,=
5
26
5
, 𝑧 = 6. This is the same result we obtained
earlier. Of cause the technique involves many more steps hence one needs care to avoid errors.
Question
1
Solve the matrix equation 𝐴𝑋 = 𝐵 where 𝐴 = [
3
Solution
35
2
5 6
] and 𝐵 = [
]
4
7 8
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We use the Jordan-Gauss method to find the Matrix X
The augmented matrix is
[
1 2 5 6
1 2 5 6
1 0 −3 −4
] ≡ 3𝑅1 − 𝑅2 [
] ≡ 𝑅1 − 𝑅2 [
]
3 4 7 8
0 2 8 10
0 2 8 10
1
1 0 −3 −4
≡ 𝑅2 [
]
0 1 4
5
2
Hence the solution is 𝑋 = [ −3 −4]
4
5
Definition
If A is a matrix the number of non-zero rows when it is reduced to echelon form is called
the rank of the matrix.
Example
Find the rank of the matrix below
1
𝐴 = [2
3
−1 0
3 5]
2 5
Solution
1 −1 0
[0 5 5 ]
0 5 5
1 −1 0
[0 5 5 ]
0 0 0
Rank (A)=2
Find the rank of the matrix below
1
𝐴 = [2
1
1 3
5
[0 −7 −7]
0 1
1
1 3
5
[0 −7 −7]
0 0
0
36
3 5
−1 3]
2 4
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The Rank of A is 2 since the reduced matrix equivalent to A has two non- zero rows
A linear system of the type 𝐴𝑥 = 𝐵 has no solution if
Rank (A)Rank of augmented matrix
A square matrix A is said to be a non-singular matrix if there exists a square matrix B such that
𝐴𝐵 = 𝐼 = 𝐵𝐴
The matrix B is called the inverse of A written as 𝐴−1
Theorem If A and B are nonsingular matrices then 𝐴−1 is nonsingular and
(𝐴−1 )−1 = 𝐴
(𝐴 𝐵)−1 = 𝐵 −1 𝐴−1
𝑨𝑩 = 𝑨𝑿 ↔ 𝑩 = 𝑿
i)
ii)
iii)
Proof
𝐴(𝐴−1 ) = 𝐼=(𝐴−1 )𝐴 Hence from definition(𝐴−1 )−1 = 𝐴.
(AB) (𝐵 −1 𝐴−1 ) = ((𝐴𝐵)𝐵−1 )𝐴−1 = (𝐴(𝐵𝐵 −1 ))𝐴−1 = (𝐴𝐼)𝐴−1 = 𝐴𝐴−1 = 𝐼
If 𝑩 = 𝑿 then 𝐴𝐵 = 𝐵𝑋 by substitution.
Conversely, if 𝑨𝑩 = 𝑩𝑿 then,
i)
i)
ii)
𝑩 = 𝑰𝑩 = (𝐴−1 𝐴)𝐵 = 𝐴−1 (𝐴𝐵) = 𝐴−1 (𝐴𝑋) = (𝐴−1 𝐴)𝑋 = 𝐼𝑋 = 𝑋
The method of reduction to echelon can be used to find the inverse of the square matrix. Actually
we solve the equation 𝐴𝑋 = 𝐼 . The augmented matrix [𝐴 ⋮ 𝐼] is reduced to the form [𝐼 ⋮ 𝐵] then
= 𝑋 = 𝐵 . So the matrix 𝐵 = 𝐴−1
Example
Find the inverse of the matrix below
1
𝐴 = [0
1
−1 2
2 1]
2 1
Solution
−1
𝐴𝐴
37
1
= 𝐵 [0
1
−1 2 𝑎11
2 1] [𝑎21
2 1 𝑎31
𝑎12
𝑎22
𝑎32
𝑎13
1
𝑎23 ] = [0
𝑎33
0
0 0
1 0]
0 1
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The augmented matrix is
1 −1 2
[0 2 1
1 2 1
⋮ 1 0
⋮ 0 1
⋮ 0 0
0
0]
1
1 −1 2 ⋮ 1
[0 2
1 ⋮ 0
0 3 −1 ⋮ −1
0 0
1 0]
0 1
1 −1 2 ⋮ 1
[0 6
3 ⋮ 0
0 6 −2 ⋮ −2
0 0
3 0]
0 2
1 −1 2 ⋮ 1
[0 6
3 ⋮ 0
0 6 −5 ⋮ −2
0 0
3 0]
−3 2
1
[0
0
−1 2 ⋮ 1
1 1⁄2 ⋮ 0
0
1 ⋮ 2⁄5
1 −1 0
0
1
0
[0
0
1
1
[0
0 0 ⋮
1 0 ⋮
0
0 1 ⋮
1⁄
5
1
⋮ − ⁄5
⋮ 2⁄5
⋮
0
1⁄
2
3⁄
5
0
0 ]
− 2⁄5
− 6⁄5
1⁄
5
3⁄
5
0
−1
1
1
− ⁄5 ⁄5
3⁄
2⁄
5
5
4⁄
5
1⁄
5
2
− ⁄5]
1
1⁄
5 ]
2
− ⁄5
Hence
𝐴−1
0
1
= [− ⁄5
2⁄
5
−1
1
1⁄
1⁄
5
5 ]
3⁄ − 2⁄
5
5
Question
1
1 Find the inverse of the matrix below 𝐴 = [2
0
−5 3
1 1]
1 4
1 −1 1
2 −1 5
1
1 ] and 𝐵 = [0 1
1]
1 0 −1
2 3 −1
2 Solve the matrix equation 𝐴𝑋 = 𝐵 where 𝐴 = [0
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Definition
A square matrix A can be written in the form 𝐴 = 𝐿𝑈 where L is a lower triangular matrix and U is
an upper triangular matrix . This is called the 𝐿𝑈 Decomposition of the Matrix A
The matrix U can be found by reducing A to echelon form then L can be found by solving the
matrix equation 𝐴 = 𝑋𝑈
1 −5 3
3 Write the matrix below in the form 𝐴 = 𝐿𝑈 where 𝐴 = [2 1 1]
0 1 4
Solution
1 −5 3
1 −5
[2 1 1] ≡ (2𝑅1 − 𝑅2 ) [0 −11
0 1 4
0
1
3
1
5] ≡ [0
4
0
−5
3
−11 5 ]
0
49
So let
1 −5
𝑈 = [0 −11
0
0
𝑎
Then We solve the equation𝐴 = [ 𝑏
𝑑
0
𝑐
𝑒
0 1 −5
0] [0 −11
𝑓 0
0
3
5]
49
3
1 −5 3
5 ] = [2 1 1]
49
0 1 4
1
From which we have that 𝑎 = 1, 𝑏 = 2, 𝑐 = −1, 𝑑 = 0, 𝑒 = − 11 . 𝑓 = 1/11
1
0
−1
Hence 𝐴 = [2
0 −1/11
39
0
1 −5
3
0 ] [0 −11 5 ]
1/11 0
0
49
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CHAPTER V
LINEAR TRANSFORMATIONS
Definition
A matrix of T from a linear space V to a linear space W is called a linear transformation if T(x +
y) = Tx + Ty for all , scalars and vectors 𝑥, 𝑦𝑉
Remark
The scalar multiplication and vector addition on the left hand side are performed in V while the
ones the right hand side are performed in W.
Theorem
If T is a linear transformation of the vector space w, then
i.
ii.
iii.
T(0) = 0
T(x + y) = Tx + Ty
𝑇(−𝑥) = −𝑇(𝑥)
Set  = 1,  = 1. Then 𝑇(𝑥 + 𝑦) = 𝑇(𝑥) + 𝑇(𝑦)
Set  = (−1),  = 0 then
𝑇(−𝑥 + 0𝑦) = −1(𝑇(𝑥) + 0𝑇(𝑦) = −𝑇(𝑥) + 0 = −𝑇(𝑥)
Example
1. Show that 𝑇(𝑥, 𝑦, 𝑧) = (2, 𝑥 − 𝑦, 𝑥 + 𝑧) is not a linear transformation
Proof
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SCHOOL OF MATHEMATICS
𝑇(0,0,0) = (2,0,0) ≠ 0
Prove that 𝑇(𝑥, 𝑦) = (𝑥 − 𝑦, 𝑥 + 2𝑦) is a linear transformation
Proof
Let , be the scalars and (𝑥1 , 𝑦1 ), (𝑥2 , 𝑦2 )𝑅
(𝑥1 , 𝑦1 ) + (𝑥2 , 𝑦2 ) = (𝑥1 + 𝑥2 , 𝑦1 + 𝑦2 )
𝑇((𝑥, 𝑦) + (𝑥2 + 𝑦2 )} = 𝑇(𝑥1 + 𝑥2 , 𝑦1 + 𝑦2)
=((𝑥1 + 𝑥2 ) − (𝑦1 + 𝑦2 ), (𝑥1 + 𝑥2 + 2(𝑦1 + 𝑦2 ))
=((𝑥1 − 𝑦1 ) + (𝑥2 − 𝑦2 ), (𝑥1 + 2𝑦1 ) + (𝑥2 + 2𝑦2 ))
However
T(x, y) + T(𝑥1 , 𝑥2 )
=((𝑥1 − 𝑦1 ), (𝑥1 + 2𝑦1 )) + ((𝑥2 − 𝑦2 ), (𝑥2 + 2𝑦2 ))
=((𝑥1 − 𝑦1 ) + (𝑥2 − 𝑦2 ), (𝑥1 + 2𝑦1 ) + (𝑥2 + 2𝑦2 ))
=(𝑥1 − 𝑦1 ), (𝑥1 − 2𝑦1 ) + (𝑥2 + 𝑦2 ), (𝑥2 + 2𝑦2
=((𝑥1 − 𝑦1 ) + (𝑥2 − 𝑦2 ), (𝑥1 + 2𝑦1 ) + (𝑥2 + 2))
From the computation, T is a linear transformation
Example
Show that 𝑇(𝑥, 𝑦) = (𝑥 2 + 𝑦 2 , 𝑥 − 𝑦) is not a linear transformation
Proof
Note T(0)=0
𝑇(1,1) + 𝑇(1,1) = (12 + 12 , 1 − 1) + (12 + 12 , 1 − 1)
= (2,0) + (2,0)
4,0) ≠ (8,0) = 𝑇(2,2) hence is not a linear transformation
Example
Prove that 𝑇(𝑥, 𝑦) = (𝑥𝑦, 𝑦 − 𝑥) is not a linear transformation
41
SCHOOL OF MATHEMATICS
Proof
𝑇(0) = 0
𝑇((1,1) + (1,1)) = 𝑇(2,2) = [(2𝑥2), (2 − 2)] = (4,0)
𝑇(1,1) + 𝑇(1,1) = [(1𝑥1), (1 − 1) + (1𝑥1), (1 − 1)]
= (1,0) + (1,0)
= (2,0)
(4,0) ≠ (2,0) .T is not a linear transformation.
Definition
Let 𝑇 be a linear transformation of the linear space V into the linear space W then
i.
The kernel of T is defined as ker(𝑇) = {𝑣𝑉T(v) = 0}
ii.
The image of the image of V under T is 𝑇(𝑉) = {𝑤𝑊T(v) = w}
Theorem
i.
ii.
ker(𝑇) is a subspace of V
image of V under T is a subspace of W
Proof of (i)
Let 𝑣1,, 𝑣2 𝑘𝑒𝑟(𝑇). Then 𝑇(𝑣1 ) = 𝑇(𝑣2 ) = 0
Hence 𝑇(𝑣1 + 𝑣2 ) = 𝑇(𝑣1 ) + 𝑇(𝑣2 ) = 0 + 0 = 0
So, 𝑣1 + 𝑣2  ker(𝑇)
Finally let 𝑣ker(𝑇) and  be scalars. Then
𝑇(𝑣) = 𝑇(𝑣) = 0 hence Vker𝑇.
Prove part ii of the above theorem as an exercise
Examples
1 Determine the dimension of the kernel of the transformation below:
𝑇((𝑥, 𝑦. 𝑧) = (𝑥 − 2𝑦 + 𝑧, 2𝑥 + 𝑦+)
2 Determine the dimension of the Kernel and image space of the transformation below:
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𝑇((𝑥, 𝑦, 𝑧, 𝑤) = (𝑥 ∓ 𝑧 + 𝑤, 𝑦 + 𝑤, 𝑥 − 2𝑦 + 𝑧)
Theorem
A linear transformation T from V into W is injective if and only if the Kernel of T is {0}
Proof
If T is injective then T is one to one . But 𝑇(0) = 0. So Kernel of T is {0}.
Conversely suppose that Kernel of Kernel of T is {0} and 𝑇(𝑥) = 𝑇(𝑦) then is 𝑇(𝑥 − 𝑦) = 𝑇(𝑥) −
𝑇(𝑦) = 0 Hence 𝑥 − 𝑦 = 0 and so 𝑥 = 𝑦.
LINEAR TRANSFORMATIONS AND THEIR MATRICES
If V is a finite dimensional vector space then to each linear transformation we associate a matrix.
Also a Matrix thus represents a linear transformation from a finite dimensional space
Let 𝑥1 , 𝑥2 … … . . 𝑥𝑛 be a basis for vector space V and A to be a linear transformation of V into W
with basis 𝑦1 , 𝑦2 … … . . 𝑦𝑚
Each vector xV can be written uniquely as a linear combination of the basis vectors
i.e.
𝑥 = 𝑐1 𝑥1 + 𝑐2 𝑥2 + ⋯ + 𝑐𝑛 𝑥𝑛 Hence,
𝐴𝑥 = 𝐴(𝑐1 𝑥1 + 𝑐2 𝑥2 + ⋯ + 𝑐𝑛 𝑥𝑛 )
= 𝑐1 𝐴𝑥1 + 𝑐2 𝐴𝑥2 + ⋯ + 𝑐𝑛 𝐴𝑥𝑛
= 𝑐1 (𝑎11 𝑦1 + 𝑎21 𝑦2 + 𝑎31 𝑦3 + ⋯ + 𝑎𝑚1 𝑦𝑚 ) + 𝑐2 (𝑎12 𝑦1 + 𝑎22 𝑦2 + ⋯ +
𝑎𝑚2 𝑦𝑚 )+. . +𝑐𝑛 (𝑎1𝑛 𝑦1 + 𝑎2𝑛 𝑦2 + ⋯ + 𝑎𝑚𝑛 𝑦𝑚 )
= (𝑐1 𝑎11+𝑐2 𝑎12+…+𝑐𝑛 𝑎1𝑛 ) 𝑦1 +(𝑐1 𝑎21 + 𝑐2 𝑎22 + ⋯ + 𝑐𝑛 𝑎2𝑛 ) 𝑥2 +…+(𝑐1 𝑎𝑛1 + 𝑐2 𝑎𝑛2+ …
+𝑐𝑛 𝑎𝑛𝑚 ) 𝑥𝑛 .
An inspection of the above expression confirms that 𝐴𝑥 is a matrix product where
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𝑐1
. . . 𝑎1𝑚
𝑐2
… 𝑎2𝑚 ] and 𝑥 = ( ) . Note that the basis vectors 𝑥1 , 𝑥2 … … . . 𝑥𝑛 can
⋮
𝑎𝑛𝑚
𝑐𝑛
1
0
0
0
1
0
be written as column vectors namely ( ) , ( ) … ( ) where it is understood to mean that
⋮
⋮
⋮
0
0
1
𝑎11
𝐴 = [ 𝑎21
𝑎𝑛1
𝑎12
𝑎22
𝑎𝑛2
𝑥1 = 1𝑥1 + 0𝑥2 + ⋯ 0𝑥𝑛 , 𝑥2 = 0𝑥1 + 1𝑥2 + ⋯ 0𝑥𝑛 .. 𝑥𝑛 = 0𝑥1 + 0𝑥2 + ⋯ 1𝑥𝑛
𝑎11
𝑎12
. . . 𝑎1𝑚 1
𝑎11 𝑎12 . . . 𝑎1𝑚 0
𝑎
𝑎
… 𝑎2𝑚 ] (0) = ( 21 ) and 𝐴𝑥2 = [𝑎21 𝑎22 … 𝑎2𝑚 ] (1) = ( 22 )
⋮
⋮
⋮
⋮
𝑎𝑛𝑚
𝑎𝑛1 𝑎𝑛2
𝑎𝑛𝑚
𝑎
𝑎
0
𝑛1
0
22
and etc. It is clear the columns of the Matrix are precisely the images of the basis vectors.
𝑎11
So 𝐴𝑥1 = [𝑎21
𝑎𝑛1
𝑎12
𝑎22
𝑎𝑛2
Example
1 Find the matrix of the transformation of
1
0
𝑇(𝑥, 𝑦) = (𝑥 − 𝑦, 2𝑥 − 3𝑦) 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑎𝑠𝑖𝑠 ( ) , ( )
0
1
Solution
1
1
𝑇( ) = ( )
0
2
−1
0
𝑇( ) = ( )
−3
1
Hence the matrix of the transformation is
𝑇=[
1 −1
]
2 −3
2 Find the matrix of transformation if T is defined by
𝑇(𝑥, 𝑦, 𝑧) = (2𝑥 − 𝑦, 𝑥 + 3𝑧, 𝑥 − 2𝑦 = 𝑧 in the standard basis
1
0
0
𝐵 = {(0) , (1) , (0)}
0
0
1
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Solution
1
2
𝑇 (0) = (1)
0
1
0
−1
𝑇 (1 ) = ( 0 )
0
−2
0
0
𝑇 (0) = (3)
1
1
The matrix of transformation is
2 −1 0
𝑇 = [1 0 3].
1 −2 1
Obviously, a change in the basis will result in different matrix representation of a linear function F
3 Write the matrix representation of 𝑇(𝑥, 𝑦) = (𝑥 − 2𝑦. 2𝑥 − 3𝑦) in the (u) standard basis (ii)
𝐵 = {(2,1), (−1,2)}
Solution
(i)
We first find the image of the basis vectors under T as follows
1
1
𝑇( ) = ( )
0
2
−2
0
𝑇( ) = ( )
−3
1
1
Hence the matrix is [
2
−2
]
−3
1
(ii)
1 2
2 −1
2
0
Now 𝑇 ( ) = ( ) = ( ) + ( ) = (52)
5
5 2
1
1
1
5
And
−14
−14 2
−3
−1
−1
𝑇( ) = ( ) =
( )−( )=( 5 )
−8
2
2
5 1
−1
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Hence the matrix is
1
−14
[52
5
5
−1
]
4 Let D be the differential linear transformation defined on the space V of all polynomials of
degree three. Find the matrix for D in the basis (i) 𝐵 = {1, 𝑥, 𝑥 2 , 𝑥 3 (ii) 𝐵 ′ = {1, 1 − 𝑥. 1 −
𝑥2 , 1 − 𝑥3}
5 Let V be the Euclidean space and the linear transformation T be represented by the Matrix
1
0 0
[−1 1 0 ]
0 −1 1
in the standard basis .(i) Write the transformation T in the Cartesian f form (ii) Hence find matrix
for T using the basis 𝐵 = {(1,1,1), (0,1,1), (0.0.1)}
If T is a linear transformation of a linear space V into a linear space W and S is a linear
transformation of W into a linear space X than we obtain a composite map of ST of V into X.
Now ST is defined by 𝑆𝑇(𝑥) = 𝑆(𝑇(𝑋)) for values of x in V.
Theorem If T𝑻: 𝑽 → 𝑾and 𝑆: 𝑊 → 𝑋are linear transformations then 𝑆𝑇: 𝑉 → 𝑋 is a linear
transformation.
Proof
𝑆𝑇(𝑥 + 𝑦) = 𝑆(𝑇(𝑥 + 𝑦)) = 𝑆(𝑇(𝑥) + 𝑇(𝑦)) = 𝑆(𝑇(𝑥)) + 𝑆(𝑇(𝑦)) = 𝑆𝑇(𝑥) +
𝑆𝑇(𝑦) .
Theorem. A linear transformation 𝑇: 𝑉 → 𝑊 is a one to one mapping if and only if ker(𝑇) = {0}
Proof
Suppose that ker(𝑇) = {0} and 𝑇(𝑥) = 𝑇(𝑦) then𝑇(𝑥) − 𝑇(𝑦) = 𝑇(𝑥 − 𝑦) = 0. So it
must be that 𝑥 − 𝑦 = 0 leading to 𝑥 = 𝑦.
Conversely, let 𝑇 be one to one and 𝑇(𝑥) = 0 then since 𝑇(0) = 0 𝑥 = 0. Soker(𝑇) = {0}
Definition The mapping 𝐼: 𝑉 → 𝑉 defined by 𝐼(𝑥) = 𝑥 is called the identity mapping.
The identity mapping is a linear transformation that is both one to one and onto.
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Definition If 𝑇: 𝑉 → 𝑊 is a one to one onto mapping then the mapping 𝑆: 𝑊 → 𝑉 defined such
that for 𝑇(𝑥) = 𝑦 ,then 𝑆(𝑦) = 𝑥 is called the inverse function of 𝑇
Note that 𝑆𝑇(𝑥) = 𝑆(𝑇(𝑥)) = 𝑥 .So 𝑆𝑇 = 𝐼.
We write 𝑇 −1 for the inverse function of 𝑇.
Theorem
If 𝑇: 𝑉 → 𝑊 is a linear transformation that admits an inverse then the Inverse 𝑇 −1 is a linear
transformation.
Proof
Let 𝑦1 , 𝑦2 be elements of 𝑊 such that there exists 𝑥1 , 𝑥2 with 𝑇(𝑥1 ) = 𝑦1 , 𝑇(𝑥2 ) =
𝑦2 Then 𝑇(𝑥1 ) = 𝑦1 and from the linearity of T we have T(𝑥1 + 𝑥2 ) = T𝑥1 +
T𝑥2 = 𝑦1 + 𝑦2
Hence from the definition of the inverse𝑇 −1 (𝑦1 + 𝑦2 ) = 𝑥1 + 𝑥2 = 𝑇 −1 (𝑦1 ) + 𝑇 −1 (𝑦2 )
Definition A vector space V is said to be isomorphic to a vector space W if there exists an
invertible linear transformation T from V onto W
Theorem
If 𝑥1 , 𝑥2 … … . . 𝑥𝑛 are linearly independent vectors of V and T is a one to one linear map of V into
W then 𝑇𝑥1 , 𝑇𝑥2 … … . . 𝑇𝑥𝑛 are linearly independent in W.
Proof
Suppose that c1Tx1  c2Tx 2  cn Tx n  0
Then from the linearity of T c1Tx1  c2Tx 2  cn Tx n  T (c1x1 c2 x 2  ...  cn x n )  0
Now the Kernel of T is consist of the Zero vector alone, it must be that
c1x1  c2 x 2  cn x n  0
Since the vectors are linearly independent in V we must have that
c1 =c2    cn  0
.
Remark. It follows from the above result that Isomorphic spaces have the same dimension
Corollary.
Each finite dimensional space V is isomorphic to
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Rn
Proof
Let
𝑥1 , 𝑥2 … … . . 𝑥𝑛 be a basis for vector space V then each vector in V is of the form
n
𝑥 = 𝑐1 𝑥1 + 𝑐2 𝑥2 + ⋯ + 𝑐𝑛 𝑥𝑛 . The mapping 𝑇: 𝑉 → R defined by 𝑇(𝑥) = (c1,c2 ,,cn )
Is an isomorphism. The fact that T is linear can be easily verified. Also, note that if
T ( x )  0 . Then (c1,c2 ,,cn )  0 . Hence c1 =c2    cn  0 and so the Kernel of 𝑇 = {0} .
Theorem 𝑅 𝑛 is not isomorphic to 𝑅 𝑚 for 𝑚
≠ 𝑛
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