Download lecture 1 Definition : A real vector space is a set of elements V

lecture 1
Definition : A real vector space is a set of elements V together with
two operation  and  satisfying the following properties :
If u and v any elements of V, then :
1. u  v is in V ( i.e V is closed under the operation  ) .
2. u  v = v  u, for u and v in V .
3. u  ( v  w ) = ( u  v)  w, for all u , v and w in V .
4. There is an element 0 in V such that :
u  0 = 0  u, for u in V ( 0 is called zero vector ) .
5. For each u in V, then the element –u in V such that :
u  -u = 0 ( -u is called negative of u ) .
6. If u any elements of V and c is any real number, then :
c ʘu is in V ( i.e V is closed under the operation  ,the real number c is
called scalar ) .
7. c ʘ ( u  v ) = c ʘu  c  v for all the real number c and all u and v
in V .
8. ( c  d ) ʘu = c ʘu  d ʘu for all the real numbers c and d, and all
u in V .
9. c ʘ ( d ʘu ) = (cd ) ʘu for all the real numbers c and d, and all u
in V .
10. 1 ʘu = u for all u in V .
The operation  is called vector addition ; the operation ʘ is called
vector scalar multiplication .
Example (1)
let F [a , b] be the set of all real – valued functions that are defined on
the interval [a , b] . If f and g are in V, we define f  g by :
( f  g )(t) = f(t) +g(t)
If f is in F [a , b] and c is a scalar, we define c  f by :
(c ʘ f )(t) = c f (t)
Show that F [a , b] is a vector space ?
Solution :
F [a , b] = { f(a,b), a,b  R }
1. ( f  g)(t) = f(t) + g(t)
2. ( f  g)(t) = f(t) + g(t) = g(t) +f(t) = ( g + f )(t)
3. f(t)  ( g  k )(t) =(f  (g +k))(t) = f (t)+(g+k)(t) =f(t)+g(t)+k(t)
4. f (t)  0 = 0  f (t) = f (t)
5. f(t)  - f (t) = 0
6. (c ʘ f )(t) = c f (t)
7. c ʘ ( f  g)(t) =c ( f  g)(t) = c ( f(t) + g(t) ) = cf ( t ) +c g ( t )
= cʘf (t)  cʘg (t)
8. (c +d ) ʘ f(t) = ( c+d ) f(t) = ( c + d )f(t) = c f(t) + d f(t) = cʘ f(t) +
dʘ f(t) = cʘ f(t)  dʘ f(t)
9. c ʘ ( d ʘ f (t) ) = c ( d ʘ f (t) )= c (df (t) ) =c d f(t) =(cd)f(t)
=( c d ) ʘ f(t)
10. 1 ʘf(t) = f(t)
Then F is vector space
Example (2):
Let V = P ={p1(t) , p2(t) , ……… , pn(t) }
and if p1(t) = antn + an-1tn-1 + ………..+ a1t + a0
and p2(t) = bntn +bn-1tn-1 + …………. + b1t + b0
W define p1(t)  p2(t) as
p1(t)  p2(t) =
(an + bn )tn + (an-1 + bn-1 )tn-1 +……..+ (a1 + b1 )t + (a0 + b0 )
And we also define c ʘ p(t) as :
c ʘ p(t) = (c an )tn +(c an-1 )tn-1 + ……… + (c a1)t + ca0
Show that V is vector space or not
Solution :
1. p1(t)  p2(t) =
(an + bn )tn + (an-1 + bn-1 )tn-1 + …….. + (a1 + b1 )t +(a0 + b0)
2. p1(t)  p2(t)
= (an + bn )tn + (an-1 + bn-1 )tn-1 + …….. + (a1 + b1 )t + (a0 + b0)
= (bn + an )tn + (bn-1 + an-1 )tn-1 + …….. + (b1 + a1 )t + (b0 + a0 )
= p2(t)  p1(t)
3. Let p3(t) = cntn +cn-1tn-1 + …………. + c1t + c0
p1(t)  (p2(t)  p3(t)) =
p1(t)  ((bn + cn )tn + (bn-1 + cn-1 )tn-1 + …….. + (b1 + c1 )t +(b0 + c0 )
=(an +(bn + cn )tn + (an-1 +( bn-1 + cn-1) )tn-1 + …….. +
(a1+( b1 + c1 ))t + (a0 +( b0 + c0 ))
= ((an +bn) + cn )tn + ((an-1 + bn-1) + cn-1 )tn-1 + …….. +
( (a1+ b1) + c1 )t + (a0 + b0 )+ c0 )
= (p1(t)  p2(t))  p3(t)
4. P(x)  0 = (an + 0)tn + (an-1 + 0 )tn-1 + ………. + (a1 +0 )t + (a0 + 0)
= antn + an-1tn-1 + ………..+ a1t + a0
= P(x)
5. P(x)  - P(x) = (an + (-an)) tn + (an-1 + (an-1 )) tn-1 + ………… +
(a1 + (-a1 )) t + (a0 +(- a0)) = 0 +0 +………+0 + 0 = 0
6. c ʘ p(t) = (c an )tn +(c an-1 )tn-1 + ……… + (c a1)t + ca0
7. C ʘ (p1(t)  p2(t)) = c ʘ ((an + bn )tn + (an-1 + bn-1 )tn-1 + …….. + (a1
+ b1 )t +(a0 + b0)) = c (an + bn )tn + c (an-1 + bn-1 )tn-1+ ………. +
c (a1 + b1 )t + c (a0 + b0) =can tn+ +cbntn +can-1tn-1+c bn-1tn-1
+ ………+ ca1t + c b1t + ca0 + cb0
= ( can tn + can-1tn-1 + ……… + ca1t + ca0 ) +( cbntn + c bn-1tn-1
+ …….. + c b1t + cb0 )
= c ʘ P1(x)  cʘ P2(x)
8. ( c + d ) ʘ P(x) =
( c + d ) antn + ( c + d ) an-1tn-1 + ………..+( c + d ) a1t + ( c + d ) a0
= can tn + dan tn + can-1tn-1 + dan-1tn-1 +……… + ca1t +da1t + ca0 + a0
= (can tn +can-1tn-1 + ……… + ca1t + ca0 ) + (dan tn +dan-1tn-1
+ ……… + da1t + da0 )
= c ʘ P(x)  dʘ P(x)
9. c ʘ ( dʘ P(x)) = cʘ (dan tn +dan-1tn-1 + ……… + da1t + da0 )
= c (dan tn +dan-1tn-1 + ……… + da1t + da0 )
= cdan tn +cdan-1tn-1 + ……… +c da1t + cda0
= (cd)an tn +(cd)an-1tn-1 + ……… +(cd)a1t + (cd)a0
= (cd) P(x)
10.1 ʘ P(x) = 1 . antn + 1 .an-1tn-1 + ………..+1 . a1t +1 . a0
= antn + an-1tn-1 + ………..+ a1t + a0
=
P(x)
Then V is the vector space .
Example (3) :
Let Vis the set of all ovdered triples real number
and define
the operations ⊕ and ʘ by
⊕
, , )=
cʘ
,
=
Show that V is vector space or not ?
Solution (3) :
1.
⊕
, , )=
,
2.
⊕(
, , )⊕
, , ))=
=(
+ +
,
3.
⊕ ( 0 , 0 ,0 ) =
4.
+
5. c ʘ
⊕
)
=
= ( 0, 0 ,0 )
)
+ ,
+ , + )
6. c ʘ ((x,y,z )⊕(x', , )) = c ʘ (
=(c
,
=
,
,
7. (c +d) ʘ(x , y ,z )
c ʘ(x , y , z ) ⊕ d ʘ(x , y ,z )
= ( cx , y , z ) ⊕
=
( c + d ) ʘ ( x , y , z ) ≠ c ʘ ( x , y , z ) ⊕ d ʘ( x , y , z )
 V is not vector space .
Lecture (2)
Theorem (1)
If V is a vector space then :
1. 0 ʘ u = 0 for every u in V
2. c ʘ 0 = 0 for every u in V
3. If c ʘ u = 0 then c = 0 or u = 0
4. ( - 1 ) ʘ u = -u for every u in V
Proof :
We have :
0ʘ u = ( 0+ 0 ) ʘ u
0 ʘ u = 0 u ⊕0 u
We adding – 0 ʘ u to both sides :
0ʘ u ⊕ ( -0 ʘ u) = ( 0 ʘ u ⊕ 0 u ) ⊕ (- 0 u )
0 = (0 ʘ u ⊕ 0 ʘ u ) ⊕ ( - 0 ʘ u )
0=0⊕ u0
0=0 ʘu
Exercise
Suppose that c ʘ u =0 and c ≠ 0 we have :
1
c
u = 1ʘ u =( c ) ʘ u = 1/c ʘ (c u) = 1/c ʘ 0 = 0
( -1) ʘ u ⊕u = (-1) ʘu ⊕ (1) ʘ u = (-1+1) ʘ u = 0 u = 0
-u is unique we conclude that (-1) ʘ u = -u
Subspace
Definition :
Let V be a vector space and w anon empty subset of V . Then w is a
subspace of V if w is a vector space with respect to the operations in V
then w is called a subspace of V .
Example (1)
Every vector space has at least two subspace itself and the subspace {0}
consisting only of the zero vector is called the zero subspace .
Theorem (2)
Let V be a vector space with operations ⊕ and ʘ and let w be a non
empty subset of V . Then w is a subspace of v if and only if the following
conditions hold :
1 ) If u and v are any vectors in w then u ⊕ v is in w
2 ) If c is any real number and u is any vector in w, then c ʘ u is in w .
Proof (2 )
→ w is subspace of V
If w is subspace of v then it is vector space and :
(a) If u and v are any vectors in w then u ⊕ v is in w
b) If c is any real number and u is any vector in w, then c ʘ u is in w
of defined subspace hold : these are precisely (1) and (2) of the theorem .
← conversely, Suppose that (1) and (2) hold
To prove w is subspace of V .
First, from (2) we have that :
(- 1) ʘ u is in w for any u in w
From (1) we have that u (-1) ʘ u is in w
But u (-1) ʘ u =0, so 0 is in w
Then u ʘ 0 = u for any u in w .
finally properties 2 -10 hold in w because they hold in V . Hence w is a
subspace of V .
Remark
1. Observe that the subspace consisting only of the zero vector is a non
empty subspace
2. If a subset w of a vector space V does not conation the zero vector
then w is not a subspace of V .
Example (2)
a b 0 

0 c d 
Consider the set w consisting of all 2  3 matrices of the form 
where a,b,c and d are arbitrary real numbers . Then w a subset of the
vector space M23 show that w is a subspace of M23 .
Solution (2)
a b 0 
Consider u  

0 c d 
a  a '
b  b'
 0
cc
u⊕v= 
a b
0
'
a '
and v  
0
b'
c
'
0

d'
0 

d  d'
ka kb 0 
kc kd 
kʘu=k 
 =
0 c d   0
Then by theorem (2) w is subspace of M23
Example (3)
Which of the following subset of R2 with the usual operations of vector
addition and scalar multiplication are subspace ?
 x
1. w1 is the set of all vectors of the form   , where x  0
 y
 x
2. w2 is the set of all vectors of the form   , x  0 , y  0 where
 y
 x
3. w3 is the set of all vectors of the form   , where x=0
 y
Solution(3)
1. w1 is the right half of xy – plane it is not subspace of R2 because
 x
  3x 
the scalar multiply -3   = 
 is not in w1 .
 y
 3 y 
2. w2 is the first quadrant of the xy – plane the same vector and scalar
multiple -3 = is not in w2 .
w2 is not subspace .
3. w3 is the y – axis in the xy – plane
0
0
 1
 2
let u    and v   
b
b
 0 

b1  b2 
be vectors in w3 then u u  v  
0  0 
c  u  c    
b1  cb1 
Which is in w3 so property (1) and (2) hence w3 is a vector space of R2.
Example (4)
Consider the homogeneous Ax = 0 where A is an m matrix .
A solution consists of a vector x in Rn , let w be the subset of Rn
consisting of all solutions of homogeneous system . To check that w is
subspace of Rn .
Solution (4)
Let x and y be solutions then :
Ax = 0 , Ay = 0
A ( x+ y ) = Ax + Ay = 0 + 0 = 0
c is scalar then :
A ( cx ) = c Ax = c .0 = 0
So cx is olso solution hence w is a subspace of Rn .
Definition
Let v1 , v2 ,……… , vk be vectors in vector space V . A vector v in V is
called a linear combination of v1 , v2 ,……… ,vk
if v = c1v1 + c2v2 + …….. + ck vk
for some real numbers c1 , c2 , ……. , ck
Example (1)
In R3 let v1 = ( 1 , 2 , 1 ) , v2 = ( 1 , 0 , 2 ) and v3 = ( 1 , 1 , 0 ) the
vector v = ( 2 , 1 , 5) fined linear combination of v1 , v2 and v3
Solution (1)
If can find real numbers c1 , c2 and c3 so that :
c1 v1 +c2v2 +c3v3 = v
c1 (1 , 2 ,1 ) + c2 ( 1 , 0 ,2 ) + c3 ( 1 , 1 , 0 ) = ( 2 , 1 , 5 )
(c1 , 2c1 , c1 ) + ( c2 ,0 ,2c2 ) + (c3 , c3 ,0 ) = ( 2 ,1 , 5 )
( c1 + c2 +c3 ,2c1 +0+c3 ,c1+2c2+0) = ( 2,1,5)
c1 + c2 +c3 =2 …….. (1)
2c1 +0 +c3 =1 ……..(2)
c1 +2c2 +0 =5 ………(3)
By equation (2)
c3 = 1-2c1
By equation (3)
c2 =(5-c1)/2
Then in equation (1)
c1 +(5-c1 )/2 +1-2c1 =2
2c1 +5 - c1+ 2 - 4c1 = 4
2c1 + + 7 -5c1 = 4
-3c1 = -3
c1 = 1
c2 =(5-c1)/2 = (5-1)/2 =2
c3 =1-2c1 = 1-2 = -1
Lecture (3) :
If S = { v1 , v2 , ……….. , vn } is a set of vectors in a vector space V . Then
the set of all vectors in V that are linear combinations of the vectors in S
is denoted by span S or span { v1 , v2 , ………. , vn } .
Example (1) :
consider the set S of 2  3 matrices given by :
1 0 0 0 1 0 0 0 0 0 0 0
S  {
, 
, 
, 
}
0 0 0 0 0 0 0 1 0 0 0 1
Then span S is the set in M23 consisting of all vectors of the form
1 0 0
0 1 0
0 0 0
0 0 0 c1
c1 
 c2 
 c3 
 c4 




0 0 0
0 0 0
0 1 0
0 0 1  0
c2
c3
0
c4 
That is span S is the subset of M23 consisting of all matrices of the form
c1
0

c2
c3
0
where c1 , c2 , c3 and c4 are real numbers .
c 4 
Theorem (3)
Let S = { v1 , v2 , ……….. , vn } be a set of vectors in a vector space V .
Then span S is a subspace of V.
Example (1)
Let v1 = 2t2 + t + 2 , v2 = t2 -2t , v3 = 5t2 – 5t +2 , v4 = - t2-3t – 2
determine if the vector u = t2 + t + 2 belongs to span { v1 , v2 , v3 , v4 }
Solution (1)
If we can find scalar c1 , c2 , c3 and c4 so that :
c1v1 + c2v2 + c3v3 +c4v4 = u
c1 (2t2 + t +2 )+c2 ( t2 -2t )+ c3 ( 5t2 -5t + 2 )+c4 ( t2 – 3t – 2 ) = t2 + t + 2
2c1t2 + c1t + 2c1 + c2t2-2c2t + 5c3t2 – 5c3t + 2c3 +c4t2 – 3c4t – 2c4
= t2 + t +2
(2c1 + c2 +5c3 +c4 ) t2 + ( c1 – 2c2 -5c3 -3c4 )t + ( 2c1 + 2c3 -2c4 ) = t2 + t
+2
2c1 + c2 +5c3 + c4 =1………. (1)
c1 -2c2 -5c3 -3c4 = 1 ……….. (2)
2c1 +2c3 -2c4 = 2 ……………(3)
we devoid equation ( 3) by 2 we get
c1 + c3 –c4 = 1
c1 = 1- c3 + c4
Put in Eq. (2)
1 – c3 + c4 -2c2 -5c3 -3c4 =1
2c2 +6c3 + 2c4 = 0
c2 + 3c3 + c4 = 0 ………. (4)
Put c1 = 1- c3 + c4 in Eq. (1)
2 – 2c3 +2c4 + c2 +5c3 + c4 = 1
c2 +3c3 + 3c4 = -1 ………. (5)
c2 + 3c3 +c4 = 0
c2 +3c3 +3c4 = -1
_______________
-2c4 = 1
c4 =
1
2
 c1 =
1
1
c3 2
2
1
2
c1 =  c 3
To put in Eq. (3)
1 – 2c3 + 1 = 0
-2c3 = -2
c3 = 1
 c1 
1
2
c 2  3c3  c 4  3 
1 5

2
2
which indicates that the system is in consistent that is it has no solution .
hence u dose not belong to span { v1 , v2 ,….. ,vn } .
Remark
In general to determine if a specific vector V belong to span S, we
investigate the consistency of an appropriate linear system .
Subspaces in Rn (optioal )
Example (1)
Let V = R3 and let W = { w1 , w2 , w3 , w4 } when
0 
w1 = 0  , w2 =
 0 
1 
0  , w =
3
 
 0 
0 
1 
1  and w = 1 
4
 
 
 0 
 0 
Determine if W is subspace of V .
Solution (1)
Apply Th. 1 using bit scalars and binary arithmetic we have :
w1 + w1 = w1
w1 + w2 = w2
w1 + w3 = w3
w1 + w4 = w4
w2 + w2 = w1
w2 + w3 = w4
w3 + w3 = w1
w3 + w4 = w2
w4 + w4 = w1
So W is closed under vector addition
Also 0wi = w1 and 1wj = wj for j = 1 , 2 , 3 , 4
Hence w is closed under multiplication . Thus w is subspace of B3
Example (2)
3
In B let
1 
v1  1 
0
0 
, v 2  1
1
1
1 


, v3  1 determine if the vector u  0
1
0
belongs to span {v1 , v2 , v3 }.
Solution (2)
If we can find (bit) scalar c1 , c2 , c3 so that
c1v1 + c2v2 + c3v3 = u
Then u belongs to span { v1 , v2 , v3 } substituting for u , v1 , v2 and v3 we
have :
1 
0 
1
1 






c1 1  + c 2 1 + c3 1 =  0 
0
1
1
 0 
c1   0  c3  1
c   c   c   0
 1  2   3  
 0  c 2  c3  0
 c1  c3  1
c  c  c   0 
2
3
 1
 
 c 2  c3  0
c1 +c3 =1 ………. (1)
c1 = 1-c3
c1 + c2 + c3 = 0 ……. (2)
c2 + c3 = 0 ………… (3)
c2 = -c3 to put in Eq. (2) then :
1 - c3 - c3 + c3 = 0
c3 = 1
 c2 = - 1
c1 = 0
 u is in span { v1 , v2 , v3 } .
Lecture (4) :
Linear Independence
Definition:
The vectors v1 , v2 ,…….. , vn in a vector space V are said
to be linearly dependent if there exist constants c1 ,c2 , …….. , cn not all
zero such that : c1v1 + c2v2 + …….. + cnvn = 0
Other wise, v1 , v2 ,…….. , vn are called linear independent . That is v1 , v2
,…….. , vn are linear independent if whenever
c1v1 + c2v2 + …….. + cnvn = 0 we must have c1 = c2 = …….. = cn = 0
Example (1)
 1
 2
1
 0 
Determine whether the vectors   and   and found the
0
1 
 
 
0
1 
homogeneous linear system Ax = 0 linear dependent or linear
independent .
Solution (1)
Forming Equation
 1
1
c1   + c2
0
 
0
  2  0 
 0  0 
  
 1  0 
   
 1  0 
 c1   2c 2  0
 c   0  0 
 
 1 
 0   c 2  0 
  

 
 0   c 2  0 
 c1  2c 2  0

 0 
c1

 

 0 
c2

  
c2

 0 
-c1 –2c2 =0
c1 = 0
c2 = 0
Whose only solution is c1 = c2 = 0 . Hence the given vectors are linearly
independent .
Example (2)
Consider the vectors :
p1(t) = t2 + t +2 ,
p2(t) = 2t2 + t , p3(t) = 3t2 +2t + 2
To find out whether S = { p1(t) , p2(t) , p3(t) } is linear dependent or
linear independent .
Solution (2)
Let c1 p1(t) + c2 p2(t) + c3 p3(t) = 0
c1 ( t2 + t + 2 ) + c2( 2t2 + t ) + c2 ( 3t2 +2t + 2 ) = 0
c1 t2 + c1t + 2 c1 + 2 c2 t2 + c2 t + 3 c3 t2 +2 c3 t + 2 c3 = 0t2 + 0t +0
(c1 +2 c2 + 3 c3 ) t2 + (c1 + c2 + 2 c2 ) t + ( 2c1 + 2 c3 ) = 0t2 +0t + 0
c1 +2 c2 + 3 c3 = 0 …………… ( 1 )
c1 + c2 + 2 c2 = 0 ……………..( 2 )
2c1 + 2 c3 = 0 …………………( 3 )
In Eq. ( 3 )
c1 = - c3
To put in Eq. (1) we found :
- c3 +2 c2 + 3c3 =0
And put in Eq. (2)
- c3 + c2 + 2 c3 = 0
c2 + c3 = 0
2 c2 + 2 c3 = 0
c2 + c3 = 0
If c3 = 1 then c2 = -1
 c1 = -1
 c1 = c2  c3  0
 p1(t) , p2(t) and p3(t) are linear dependent
Example (3)
Are the vectors v1 = ( 1 , 0 , 1 , 2 ) , v2 = ( 0 , 1 , 1 , 2 ) ,
v3 = ( 1 , 1 , 1 , 3 )
in R4 linearly dependent or linearly independent .
Solution (3)
Let c1 v1 + c2 v2 + c3 v3 = 0
c1( 1 , 0 , 1 , 2 ) + c2 ( 0 , 1 , 1 , 2 ) + c3 ( 1 , 1 , 1 , 3 ) = 0
(c1 , 0 , c1 , 2c1 ) +( 0 , c2 , c2 , 2c2 ) + (c3 , c3 , c3 , 3c3 ) =( 0 ,0 ,0 ,0 )
(c1 +c3 , c2 + c3 , c1 + c2 + c3 , 2c1 +2c2 +3c3 ) = ( 0 , 0 , 0 )
c1 + c3 = 0 ………..……. (1)
c2 + c3 = 0 …………..… (2)
c1 + c2 + c3 = 0 ………..(3)
2c1 + 2c2 + 3c3 =0…….(4)
In Eq. (1)
c3 = - c1
In Eq.(2)
c3 = - c2
 c1 = c2  c3 = 0
 c1 = c2 = c3 = 0
Then v1 , v2 and v3 are linear independent .
Theorem (3)
The nonzero vectors v1 , v2 ……… , vn in a vectors space V dependent if
and only if one of the vectors vj , j  2 is a linear combination of the `
vectors v1 , v2 , …… .,vj-1 .
Proof
If vi is linear combination of v1 , v2 , …… , vj-1
vj = c1 v1 + c2 v2 + ……… + cj-1 vj-1
Then :
c1 v1 + c2 v2 + ………. + ( - 1 ) vj + 0 vj+1 + ……… + 0vn = 0
 cj = ( - 1)  0
 v1 , v2 , ……….. , vn are linearly dependent .
Conversely :
Suppose that v1 , v2 , …………. , vn are linearly dependent then there exist
scalar c1 , c2 ,……. , cn not all zero, such that :
c1 v1 + c2 v2 + ………. + cnvn = 0
Let j be the largest subscript for which cj  0 if j > 1
Then :
vj  
c j 1
c1
c
v1  2 v 2  ............ 
v j 1
cj
cj
cj
If j = 1 then c1 v1 = 0  v1 = 0, a contradiction of the hypothesis that
none of the vectors is the zero vector .
 one of the vectors vj is linear combination of the preceding vectors
v1 , v2 , ……… , vj-1
Example (1)
Consider the vectors v1 = ( 1 , 2 , -1 ) , v2 = ( 1 , -1`, 1 ) , v3 = ( -3 ,2 ,-1 )
v4 = ( 2 , 0 , 0 ) in R3
Is S = { v1 , v2 , v3 , v4 } linearly dependent or linearly independent ?
Solution (1)
Let c1 v1 + c2 v2 + c3 v3 +c4 v4 = 0
c1 (1 , 2 , -1) + c2 (1 , -1 , 1) + c3 ( -3 ,2 ,-1 ) + c4 ( 2 , 0 , 0 ) = ( 0 , 0 , 0 )
(c1 , 2c1 , - c1)+(c2 , - c2 , c2 )+( -3 c3 , 2 c3 , - c3 )+( 2 c4 , 0 , 0 ) = ( 0 , 0 ,0 )
(c1 + c2 -3 c3 +2 c4 , 2 c1 - c2 +2 c3 , - c1 + c2 - c3 )= ( 0 , 0 , 0 )
c1 + c2 -3 c3 +2 c4 =0 …………(1)
2 c1 - c2 +2 c3 = 0 ………….....(2)
- c1 + c2 - c3 = 0 …………….. (3)
By multiplication 2 in Eq. (3) we get :
- 2c1 +2 c2 -2 c3 = 0 ………… (4)
We addition Eq. (1) and Eq. (4)
c1 + c2 -3 c3 +2 c4 = 0
- 2c1 +2 c2 -2 c3 = 0
___________________
c2 = 0
In Eq. (2) we get :
c1 = - c3
If c3 = 1  c1 = -1
c4 = 2
Basis And Dimension
Basis
Definition : The vectors v1 , v2
, …… , vn a vector space V are said to
form a basis for V if :
1. v1 , v2 , …… , vn span V .
2. v1 , v2 , …… , vn are linear independent .
Remark
If v1 , v2 , …… , vn form a basis for a vector space V then they must be
non zero and distinct and so we write them as a set { v1 , v2 , …… , vn }.
Example (1)
The vectors e1 = (1 , 0 ) and e2 = ( 0 , 1 ) form basis for R2, the vectors e1
, e2 and e3 form basis for R3 and in general , the vectors e1 , e2 , ….. , en
form a basis for Rn . Each of these sets of vectors is called the natural
basis or standard basis for R2 , R3 and Rn respectively .
Example (2)
Show that the set S = { v1 , v2 , v3 , v4 } ,where v1 = ( 1 , 0 , 1 , 0 )
v2 = ( 0 , 1 , -1 , 2 ) , v3 = ( 0 , 2 , 2 , 1) and v4 = ( 1 , 0 , 0 ,1 ) is basis for
R4
Solution (2)
To show that set S = { v1 , v2 , v3 , v4 } is linear independent, we form the
Equation c1 v1 + c2 v2 + c3 v3 + c4 v4 = 0
c1( 1,0,1,0 ) + c2( 0,1,-1,2 ) +c3( 0,2,2,1) +c4( 1,0,0,1 ) = ( 0,0,0,0)
(c1,0,c1,o ) + ( 0,c2 -c2 ,2c2 ) + ( 0,2c3 ,2c3 ,c3 ) + (c4 ,0,0,c4 )= ( 0,0,0,0 )
(c1 + c4 , c2
+ 2 c3
, c1 - c2 +2c3 , 2c2 +c3 + c4 ) = ( 0, 0, 0, 0 )
c1 + c4 = 0 ………….. (1)
c2 + 2c3 = 0 ……..….. (2)
 c1 = - c4
 c2 = -2 c3
c1 - c2 +2c3 = 0 ……(3)
2c2 +c3 + c4 = 0……(4)
 c1 = c2 = c3 = c4 = 0
 S linearly independent .
To show that S span R4
Let v = ( a , b , c , d )
k1v1 + k2v2 + k3v3 + k4v4 = ( a , b , c , d )
substituting for v1 , v2 , v3 , v4 and v , we find a solution for k1 , k2 , k3 ,and
k4 to the resulting linear system for any a , b , c , and d . Hence S span
R4 and is basis for R4
Example (2)
We must show that the set S ={ t2 + 1 , t – 1 , 2t + 2 } is a basis for the
vector space .
Solution (2)
We must show that S spans V and is linearly independent
c1( t2 + 1 ) + c2 (t – 1) + c3 ( 2t + 2 ) = at2 + bt + c
c1 t2 + c1 + c2 t - c2 + 2 c3 t + 2 c3 = at2 + bt + c
c1 t2 + (c2 + 2 c3 ) t + (c1 - c2 + 2 c3 ) = at2 + bt + c
 c1 = a …………………..……….……… (1)
c2 + 2 c3 = b  c2 = b - 2 c3 ………. (2)
c1 - c2 + 2 c3 = c ……………..………..... (3)
To put Eq. (1) and Eq.(2) in Eq.(3) we get :
a – (b - 2 c3 ) + 2 c3 = c
a – b + 4 c3 = c  4c3 = c + b – a  c3 
cba
4
In Eq. (2)
 c2 = b-2
cba abc

4
2
Hence S span V
To show that S is linearly independent
Let : c1 (t2 + 1 ) + c2 ( t – 1) + c3 (2t + 2 ) = 0t2 + 0t +0
c1t2 + c1 + c2t – c2 + 2 c3t +2 c3 = 0t2 + 0t +0
c1t2 + (c2 + 2 c3 )t + (c1 - c2 +2 c3 ) = 0t2 + 0t +0
 c1 = 0
c2 + 2 c3 = 0
 c2 = -2 c3
c1 - c2 +2 c3 = 0  0 + 2 c3 + 2 c3 =0
c3 = 0
c2 = 0
 S is basis .
Theorem (4)
If S = { v1 , v2 , ………. , vn } is a basis for a vector space V .
Then every vector in V can be written in one and only one way as linear
combination of the vectors in S .
Proof (4)
 S is basis
 S is span  every vector v in V can be written as a
linear combination of the vectors in S .
 Let v = c1 v1 + c2 v2 + ………. + cnvn
and v = a1 v1 + a2 v2 + ………. + anvn
 c1 v1 + c2 v2 + ………. + cnvn = a1 v1 + a2 v2 + ………. + anvn
 c1 v1 + c2 v2 + ………. + cnvn – ( a1 v1 + a2 v2 + ………. + anvn ) = 0
(c1 - a1 ) v1 + (c2 - a2 ) v2 + ……….. + (cn - an ) vn = 0

S basis  S linear independent  S linear combination
 c1 - a1 = 0  c1 = a1
c2 - a2 = 0  c2 = a2



cn – an = 0  cn = an
 There is only one away to express v as a linear combination of the
vectors in S .
Theorem (5)
Let S = { v1 , v2 , ………. , vn } be a set of onozero vectors in a vectorand
space V and let W = span S . Then some subset of S is a basis for W .
Example
Let S = { v1 , v2 , v3 , v4 , v5 } be a set of vectors in R4 where
v1 = ( 1 , 2 , -2 , 1 ) , v2 = ( -3 , 0 , -4 , 3 ) , v3 = ( 2 , 1 , 1 , -1 )
v4 = ( -3 , 3 , -9 , 6 ) and v5 = ( 9 , 3 , 7 , -6 ) find a subset of that is a
basis for W = span S .
Solution
step 1 :
c1 ( 1 , 2 , -2 , 1 ) + c2 ( -3 , 0 , -4 , 3 ) + c3 ( 2 , 1 , 1 , -1 ) +
c4 ( -3 , 3 , -9 , 6 ) + c5 ( 9 , 3 , 7 , -6 ) = ( 0 , 0 , 0 , 0 )
step 2 :
Equating corresponding components, we obtain the homogeneous system
c1 - 3 c2 + 2 c3 -3 c4 + 9 c5 = 0
c2
+ c3 + 3 c4 + 3 c5 = 0
-2 c1 - 4 c2 + c3 – 9 c4 + 7 c5 = 0
c1 + 3 c2 - c3 + 6 c4 – 6 c5 = 0

 1  3 2  3 9 0 1


0
1
3
3 0 
1
 0
  2  4 1  9 7 0 

 0
3  1 6  6 0 
 1
0
1
2
1
1 
2
0 0
3
2
3
2
0
0
0
0
0

0
0

0

0
0 
3
2
5

2
0
step (3) ;
The leading a pear in columns land 2 so { v1 , v2 }is a basis for W=span S
Theorem (6)
If S = { v1 , v2 , ………. , vn } is a basis for a vector space V and T { w1 ,
w2 , …….. , wr } is a linearly independent set of vectors in V, then r  n
Proof
Let T1 = {w1 , v1 , v2 , ………. , vn }

S span V   S span T1

W is linear combination of the vectors S we find that T1 is linearly
dependent
by theorem ( )some vj is a linear combination of the preceding vectors in
T1
Delete that particular vector vj
Let = { w1 , v1 , ……….. , vj-1 , vj+1 ,……. , vn }
Note that S1 span V
Let T2 = { w2 , w1 , v1 , …… , vj-1 , vj+1 , …… , vn }
Then T2 is linearly dependent
Some vector in T2 is linear combination of preceding vectors in T2

T is linear independent this cannot be w1 so it is i  j .
If the V vectors are all eliminated be for we can run out of w vectors 
then the resulting set of w vectors, a subset of T, is a linearly dependent,
which implies that T is also linearly dependent .

we have reached a contradiction,
1  1 1 0

0  1 0 1
1  1 0 0

0 0 0 0

0 0 0  1
 
0 0 0  0

1 0 0  0
 
0 1 0 0
0 0

1 0 1 0 0 0
0 1 0  1 0 0

0 0 0 0 1 0
0 0 1
1
The leading Is appear in columns 1 , 2 , 3 and 6, we conclude that
{ v1 , v2 , e1 , e4 } is basis for R4 containing v1 and v2
Theorem (7)
Let V be an n – dimensional vector space and S = {v1 , v2 , ……. , vn }
be a set of n vectors in V
 If S is linearly independent then it is a basis for V .
 If S span V then it is a basis for V .
Example
In example span Sis subspace of R4 , so dim w 4

S contains five vectors, we conclude by corollary that S is not a basis
for W .
In example 2

dim R4 = 4 and the set S contains four vectors it is possible for S to
be basis for R4 .
If S linearly independent or span R4, it is a basis ; other wise it is not a
basis .thus , we need only check on of the conditions in Th . 7 not both .
Inner Product Space
Definition
: Let V be a real vector space . An inner product on V is a
function that assigns to each order pair of vectors u and v in V a real
number ( u , v ) satisfying the following properties :
1. ( u , v )  0 , ( u , v ) = 0 if and only if u = 0 .
2. ( u , v ) = ( v , u ) for any u , v in V .
3. ( u + v , w ) = ( u , w ) + ( v , w ) .
4. (cu , v ) = c ( u , v ) for u , v in V and c real number .
Its clear that c ( u , v ) = ( u , cv )
because ( u , cv ) = ( cv , u) = c ( v , u )
Also, ( u , v + w ) = ( u , v ) + ( u , w ) .
Example :
We defined the standard inner product or dot product on Rn as the
function that assigns to each ordered pair of vectors :
u1 
u 
u   2
 
 
u n 
,
v1 
v 
v   2  in Rn the number denoted by ( u , v ), given by :
 
 
v n 
( u , v ) = u1v1 + u2v2 +…… + unvn .
of course, we must verify that this function satisfied the properties of
definition .
Remark :
If we view the vectors u and v in Rn as n× 1 matrices, then we can write
the standard inner product of u and v in terms of matrix multiplication
as ( u , v ) = uT v
Example:
Let v be any finite – dimension vector space and let S = { u1 , u2 ,…. , un }
be an ordered basis for V . if :
V = a1u1 + a2u2 + ……… +anun
and
W = b1u1 + b2u2 + ………. + bnun
we define :
( V , W ) = ( [v] s , [w]s ) = a1b1 + a2b2 + ……. + anbn
We must satisfying the properties of inner product
Example :
u 
Let u   1 
u
 2
v 
and v   1  be vectors in R2
v
 2
( u , v ) = u12 -2 u1 u2 +3u22 show that this an inner product .
Solution:
( u , v ) = u12 – 2 u1 u2 +u22 + 2u22 = ( u1 – u2 )2 +2u22  0
moreover, if ( u , u ) = 0  u1 = u2 and u2 = 0 so, u1 = 0
As follows ( u , u ) = u2 – 2uu + 3u2 = u2 – 2u2 + 3u2 = u2 -2u2 +3u2 = 0
Definition : A real vector space that has an inner product defined on it
is called an inner product space "a Euclidean space "
In an inner product space we define the length of a vector u by
u  (u, v)
we have u  0 if u  0 we can show that 0  0
Theorem [Cauchy –Schwarz Inequality ]
If u and v are any two vectors in an inner product space V, then :
(u, v)  u . v
Example
 3
1 
Let u  2  and v  2  be in the a Euclidean space R3 with the
2 
 3
standard inner product .Then :
( u , v ) = (1)(-3) + (2)(2) +(-3)(2)
= -3 + 4 -6 = -5
u  (u, v)  (1) 2  (2) 2  (3) 2  14
v  (u , v)  (3) 2  (2) 2  (2) 2  17
(u , v)  u . v
Corollary [ Triangle Inequality ]
If u and v are any vectors in an inner product space V, then :
uv
2
 (u  v, u  v)  (u, v)  2(u, v)  (v, v)
uv
2
 u  2(u, v)  v
2
2
 ( u  v )2
so, u  v  u  v
Orthonormal basis
Definition : Let V be an inner product space two vectors
u and v are
orthonormal if ( u , v ) = 0
Example
Let v be the a Euclidean space R4 with the started inner product if
1 
0 
u 
0 
 
1 
0 
2
, v 
3 
 
0 
Then:
( u , v ) = (1)(0) + (0)(2) +(0)(3) + (1)(0) = 0
Then, u , v are orthogonal .
Definition : Let V an inner product space . A set S of vectors in V is
called " orthogonal " if any two distinct vectors in S are orthogonal also,
each vectors in S is of unit length then S is called " orthonormal" .
Remark :
The vector that length equal to 1 called unit vector and it of the form
u
1
1
1
x so, u  (u, v)  ( x, x) 
x
x
x
( x, x)
x.x
x
x.x
1
Theorem
Let S = { u1 , u2 , ……. , un } be an orthogonal set of non zero vectors Rn
in an inner product V, then S is linearly independent .
proof
Consider the equation
c1u1 + c2 u2 + …… + cnun = 0 ………… (1)
Taking the dot product of both side of (1) with ui , 1 i  k , we have :
(c1u1 + c2 u2 + …… + cnun ).ui = 0 . ui …………… (2)
by properties ( )
the left side of (2) is :
c1 ( u1 . ui ) + c2 ( u2 . ui ) + …….. + cn ( un . ui ) = 0
And the right side is 0 . since uj . ui = 0 if i  j ,(2) become :
0 = ci (ui .ui ) = ci u i
By ( )
2
………….. (3)
ui  0 , since ui  0
Hence (3) implies that ci = 0 , 1 i  k , and S is linearly independent
Example
Find the orthogonal set if :
 2
, x 2  0 
1 
1 
x1  0 
2
and
0 
x 3  1  , then { x1 , x2 , x3 } is an orthogonal
0 
set and :
(x1 , x2 ) =0
(x1 , x3 ) = 0
(x2 , x3 ) = 0
so, the vectors :
u1 
x1
 1 
 5


x1  0 
 2 




5


Whose x1  5 ,
x2
u

2
,
x2
x2  5
 2 
 5
 
 0 
 1 
 
 5 
0
x3
 
, u 3  x  1 
3
0
, x3  1
so, u1 , u2 , x3 are unit vectors that :
u1  u 2  x3  1
so, { u1 , u2 , x3 } is an orthonormal .
Example
1 
 
Let x1  3 
2
 1
 
and x 2  1  find the orthonormal set .
 1
Solution
( x1 , x2 ) = 0
x1  1  9  4  14
,
x2  1  1  1  3



x1

u1 

x1




1 

14 
3 

14 
2 

14 
x
2
, u2  x
2
 1
 
 3
 1 
 
 3
 1
 
 3
so, { u1 , u2 } is orthonormal set .
Example
1 
 3
Let x1  1 
 
2 
4 
1 
 
, x 2   1
 
0 
( x1 , x2 ) = 0
x1  1  9  1  4  15
x 2  16  1  1  0  18





x
u1  1  
x1






1 
15 
3 
15 
1 
15 
2 
15 
 4 
 18 


 1 
x
u 2  2  18 

x2 
 1 
 18 


0 
So, { u1 , u2 } is an orthonormal set .
Definition:
F0r every Euclidean space
Theorem
Let S = { u1 , u2 , ….. , un } be an orthonormal basis for a Euclidean
space V and let v any vector in V, then
v = c1v1 + c2v2 + …….. + cnvn
where
ci = ( v , u i )
, i = 1 , 2 , …….. , n
Example
Let S = { u1 , u2 , u3 }be a basis for R3
2
u1   
 
Theorem [Grom – Schmidt Process]
Let V be an inner product space and w 0 an n- dimensional subspace of
V, then there exist an orthonormal basis T = { w1 , w2 , ….. , wn } for w .
Gram –Shmidit Process :
Let S = { u1 , u2 , ….. , un }be any basis for w where w is subspace of V
then :
v1 = u 1
and
v2  u 2  a1v1  u 2 
(u 2 , v1 )
v1
(v1 , v1 )
and
v3  u 3 
(u3 , v1 )
(u , v )
v1  3 2
(v1 , v1 )
where
a1  
(u 2 , v1 )
(v1 , v1 )