Tran Viet Dung
LECTURE ON MATHEMATICS II
For HEDSPI students
Hanoi 2008
1
CONTENTS
Chapter I. MATRICES AND DETERMINANTS................................................ 4
1 Matrices ..............................................................................................4
1.1. Definitions and examples.......................................................................... 4
1.2. Matrix addition , scalar multiplication .................................................. 6
2 Determinants. ...................................................................................11
2.1. Definitions and examples. ...................................................................... 11
2.2. Laplace expansion . .................................................................................. 12
3 Inverse matrix anf rank of a matrix ....................................................15
3.1. Inverse Matrix ........................................................................................... 15
3.2. Rank of a matrix ....................................................................................... 19
ChapterII.
SYSTEM OF LINEAR EQUATIONS ........................................... 22
1 Cramer’s Systems ..............................................................................22
1.1. Basic concepts............................................................................................ 22
1.2. Cramer’s systems of equations .............................................................. 23
1.3. Homogeneous system of n equations in n variables ........................ 25
2 Solution of a general system ...............................................................27
of linear equations ................................................................................27
2.1. Condition for the existence of solution. ............................................... 27
2.2.
Method of solving system of linear equations. ................................... 28
ChapterIII. VECTOR SPACES ............................................................................... 31
1. Definitions and simple properties .....................................................31
of vector spaces ....................................................................................31
1.1. Basic concepts, examples. ....................................................................... 31
1.2. Simple properties...................................................................................... 33
2 Subspaces and spanning sets..............................................................35
2.1. Subspaces. .................................................................................................. 35
2.2. Spanning sets, linear combinations...................................................... 37
3 Bases and dimension .........................................................................38
3.1. Independence and dependence. ............................................................. 38
3.2. Bases , dimension ..................................................................................... 40
3.3. Rank of a system of vectors. .................................................................. 45
3.4. Change of basis ........................................................................................ 46
Chapter IV
LINEAR MAPPINGS ...................................................................... 50
1 Definitions and elementary properties of linear mappings....................50
1.1. Definitions and examples. ...................................................................... 50
2
1.2. Images and kernel of a linear mapping. .............................................. 51
1.3. Operations on linear mappings. ............................................................ 53
2 Matrix of a linear mapping .................................................................53
2.1. Matrix with respect to pair of bases. .................................................... 53
2.2. Matrix of a linear operation ................................................................... 55
3 Diagonalization .................................................................................56
3.1. Similar matrices........................................................................................ 56
3.2. Eigenvalues and eigenvectors of a matrix .......................................... 57
3.3. Eigenvalues and eigenvectors of a linear operation ......................... 59
ChapterV. BILINEAR FORMS, QUADRATIC FORMS, EUCLIDEAN
SPACES. ................................................................................................... 61
1 Bilinear forms ...................................................................................61
1.1. Linear forms on a vector space .............................................................. 61
1.2. Bilinear forms ............................................................................................ 61
1.3. Matrix of a bilinear form......................................................................... 62
2 Quadratic forms.................................................................................64
2.1. Definitions and examples. ...................................................................... 64
2.2. Canonical form of a quadratic form...................................................... 65
2.3. Lagrange’s Method. .................................................................................. 66
2.4. Jacobi’s Method. ........................................................................................ 67
3 Euclidean Spaces. ..............................................................................68
3.1. Inner product on a vector space. ........................................................... 68
3.2. Norms and othogonal vectors................................................................. 70
3.3. Orthonormal basis. ................................................................................... 72
3.4. Orthogonal subspaces, projection. ........................................................ 74
3.5. Orthogonal diagonalization .................................................................... 75
3
Chapter I.
MATRICES AND DETERMINANTS
1. Matrices
1.1. Definitions and examples
Definition1
A matrix A of size m×n is a table of m rows n columns contaning
numbers aij in the form:
 a11
a
21
A= 
 ..

a m1
a12
a 22
...
am2
... a1n 
... a 2 n 
.
... ... 

... a mn 
The element aij lies in row i and column j. If the elements are real numbers
then A is said a real matrix, if they are complex then A is a complex matrix.
 
One can denote matrix A shortly by A = aij
mn
.
4
If the size of A is n×n, the matrix A is called square matrix of order n.
Example 1
1 2 3 
For A = 
 = aij
6 4 5
 
, the size is 2×3 ,
a11 =1, a12 = 1 ; a13 =2 ; a21 = 6 ; a22 = 4 ; a23 = 5.
Definition 2
 
Let A= aij be a square matrix of degree n.
a) Elements a11, a22, ..., ann are said to lie on the main diagonal of
the matrix A.
b) The matrix A is said to be a diagonal matrix if a ij = 0 for all i,j
with ij.
c) The matrix A is said to be a upper triangular matrix if a ij = 0 for
all ij.
d) The matrix A is said to be a lower triangular matrix if a ij = 0 for
all i,j , ij.
Example 2
a11 0
0 a
22
a) A = 
 ..
...

0
0
...
a11 a12
0 a
22
b) B= 
 ..
...

0
0
... a1n 
... a 2 n 
is an upper triangular matrix
... ... 

... a nn 
0 
... 0 
is a diagonal matrix
... ... 

... a nn 
5
 a11 0
a
a 22
21
c) C = 
 ..
...

a n1 a n 2
0 
0 
is a lower triangular matrix.
... ... 

... a nn 
...
...
Definition 3
a) Two matrices A and B are called equal ( writen A = B ) if their sizes
are the same and corresponding elements are equal.
b) A matrix is called the zero matrix if all elements are zeros. Denote the
zero matrix by 0.
 
 
c) For a matrix A = aij of size m×n, a matrix B = bij of size n×m is
called the transpose matrix
of A
if bij = aji for all i,j. Denote by A t the
transpose matrix of A .
Example 3
1 2 3 
For A = 
 ,
6 4 5
1 6 
A t = 2 4 ;
3 5
1  2 0


For A = 3 4 5 ,
6 7 8
At
 1 3 6


=  2 4 7 
 0 5 8 
Definition 4
 
A square matrix A = aij is called symmetric if aij aji for all i,j.
Note that A is symmetric iff A t = A
1.2. Matrix addition , scalar multiplication
1.2.1 Matrix addition
6
Definition 5
a) If A and B are matrices of the same size, their sum A+B is the
matrix formed by adding corresponding elements.
 
b  then A + B = a  b .
b) for a matrix A = a  , the negative matrix (-A) of A is defined by
-A =  a 
Thus , if A= aij , B =
ij
ij
ij
ij
ij
Example 4
 1 1 1 
1 2 3 
 2 3 4
if A = 
,B =
then A+B = 

 and

0  2 3
6 4 5
 6 2 8
 1  2  3
-A = 

 6  4  5
Theorem1 If A, B, C are any matrices of the same size then
a) ( A+ B ) + C = A+( B + C ), ( associative law )
b) A + B = B + A , ( commutative law )
c) for the zero matrix 0 of the size as A , A+0 = A,
d) A + (-A) = 0
e) ( A + B )t = At + Bt .
If A and B are two matrices with the same size, the difference of A and B
denoted by A-B is the matrix A+(-B).
Example 5
 1 1 1 
1 2 3 
0 1 2
6 4 5 -  2 0 2 = 8 4 3 .






1.2.2 Scalar multiplication .
Definition 6
 
Given a matrix A = aij
of size mn and a number k . The scalar
multiple kA is the matrix obtaned from A by multipying each element of A by

k as follows: kA = k.aij

7
Example 7
1 2 3 
Given A = 
 , B=
  3 0 4
2 6 0
 8 10 5 , then


 1 3 0 
1
B =
.
2
  4 5 5 / 2
2 4 6
2A = 
 ;
 6 0 8
Theorem2
1) For zero number 0, 0.A is the zero matrix: 0A = 0.
2) k.A = 0 iff k=0 or A=0.

 

Proof. 1) Let A = aij , we have 0.A = 0.aij = 0 .
Thus, 0A is the zero matrix
 
2) Assume that k.A = 0. So for A = aij we have k.aij =0 for
all i,j. If k  0 then aij = 0 and A is the zero matrix. Theorem is proved. 
Theorem 3
Let
A, B be two arbirary matrices with a fixed size mn. Let k, p
denote arbitrary real numbers. Then
1)
2)
3)
4)
5)
k( A+ B ) = kA +kB,
( k + p )A = kA + pA ,
(kp)A = k(pA),
1.A = A ,
(-1)A = -A .
 
 


Proof. 1) If A = aij , B = bij then A+B = aij  bij .
    

 
For A = a , (k + p ) A = (k  p)a  = ka  pa 
= ka  +  pa  = kA + pA
k(A + B ) = k (aij  bij ) = kaij  kbij = kaij + kbij = kA +kB.
Hence ,
2)
ij
ij
ij
ij
ij
ij
3) 4) 5) are proved analogously.
8
1.2.3 Matrix multiplication
Matrix multiplication is a little more
complicated
than matrix
addition or scalar multiplication, but it is well worth the extra effort.
Definition 7
If A =
a  is
ij
a matrix of size mn and B =
the product A.B of A and B is the matrix C =
b  is a matrix of size np ,
c 
ij
ij
of size mp defined by :
cij = ai1b1j + ai2b2j + ... +ainbnj ;for i = 1, ... m; j =1, ... p.
(1.1)
or shortly,
n
cij =  aik bkj ;
i = 1, ..., m ; j = 1, ... , p
(1.2)
k 1
Note that to obtain cij we need use row i of matrix A and column j of
matrix B.
Remark
a)Product AB exists if and onlly if the number of columns of A is
equal to the number of rows of B.
b) If A is a square matrix then AA exists and denoted by A 2
c) For a square matrix A, denote A.A. ... A = A n
Example 8
b11 b12 
 a11 a12 a13 
 c11 c12 


a) A = 
, B = b21 b22  , then AB = C = 
,

c 21 c 22 

a 21 a 22 a 23 
b31 b32 
c11 = a11b11 + a12b21 + a13b31 ;
c21 = a21b11 + a22b21 + a23b31
c12 = a11b12 + a12b22 + a13b32
;
c22 = a21b12 + a22b22 + a23b32.
9
b)
1 2 
1 2 3
 10  5 
0  1 4 3 1  =  1  13


 1  3 


c)
2
2
1 2 3  1
 
0  1 4  1  =  5 but  1 

   1  
  1
 
1 2 3
0  1 4 not exist.


Definition 8.
An identity matrix E is a diagonal matrix in which every diagonal
element is 1. If the size of E is nn then E is said to be the identity matrix of
order n.
Warning
If the order of the factors in a product of matrices is changed , the
product may change ( or may not exist ).
Theorem 4
Assume that k is an arbitrary scalar, and A, B, C are matrices the
indicated operations can be performed . Then
1) EA = A; BE = B , where E is identity,
2) (AB)C = A( BC),
3) A (B+C ) = AB + AC,
A( B-C) = AB - AC
4) (B+C ) A = BA +CA ,
( B – C ) A = BA – CA
5) k(AB ) = ( kA ) B = A(kB ),
6) (AB )t = AtBt .
Proof. we prove properties 3 and 6, leaving rest as exercises.
property3) . Let A = aij  , B = bij  , C = cij  . Then A+B = aij  bij  ,
10
 
 dij =
A(B+C) = D = d ij
 
n
a
k 1
ik
(bkj  c kj )
n
=
a
k 1
n
ik
bkj
  . So A(B+C ) = AB +AC.
A = a , B = b  ,At = a '  , B t = b ',
+
a
k 1
ik
c kj
=
eij + hij
here AB = eij ; AC = hij
property6)
 
AB = C = cij , cij =
n
a
k 1
cij’ = cji =
n
a
k 1
jk
bki =
ik
bkj
n
b
k 1
ij
ij
ij
jk
;
ij
aij’= aji, bij’ = bji
 
(AB )t = C t = cij ' ,
' a ki ' 
Ct = B tAt . 
Remark.
If A is a square matrix then AA exists and denoted by A 2. In general, A.A....A
( m times ) denoted by A m.
2. Determinants
2.1. Definitions and examples
Definition1
 
Let A aij be a square matrix, Sn the set of substitutions of n elemrnts
{1, 2, ..., n } . The determinant of matrix A is a number det(A) defined by the
formula
det ( A ) =
 sign( )a 
 S n
1 (1)
...a n ( n ) .
(2.1)
a11 ... a1n
 a11 ... a1n 
 ... ... ... 
We also dente det(A ) by A, for A = 
 , det A = ... ... ... .
a n1 ... a nn 
a n1 ... a nn
11
Example 1
a) A = a11  of size 11 , det (A) = a11,
 a11 a12 
b) A = 
 , det (A) = a11 a22 – a12.a21
a 21 a 22 
a11
c)
a12
a13
a 23 = a11a22a33 + a12a23a31 + a13a21a32
a33
a 21 a 22
a31
a32
- a13a22a31 - a12a21a33 - a11a23a32 .
1 2
= 1.4 – 2.3 = -2
3 4
d)
1
0
2
e) 1  2 3 =1.(-2).1 + 2.1.5 - 2.(-2).4 – 1.3.5 = -2 + 10 +16 -15 =7
4 5 1
2.2. Laplace expansion
 a11 ... a1n 


Let A =  ... ... ...  . For each aij denote by Mij the matrix of order (na n1 ... a nn 
1) obtaned from A by deleting row i and column j. Put A ij = (-1)i+j det(Mij ) and
call algebraic complement of aij .
Theorem 1
For A = aij , det(A) can be expressed by
1) det(A) = ai1Ai1 + ... + ainAin, ( expansion along row i )
2) det( A) = a1jA1j + ... + anj Anj .( expansion along column )
Examle 2
a11
a12
a) a 21 a 22
a31 a32
a13
a
a 23 = a11. 22
a32
a33
a 23
a33
_ a12 .
a 21 a 23
a31
a33
+ a13 .
a 21 a 22
a31
a32
12
( expansion along the first row )
a11
a12
b) a 21 a 22
a31
a32
a13
a
a 23 = a11. 22
a32
a33
a 23
a33
_ a 21.
a12
a13
a32
a33
+ a31.
a12
a13
a 22
a 23
( expansion along the first column )
1 2 3
5 1
2 3
2 3
c) 0 5 1 = 1.
- 0.
+ 3.
=21+3.(-13)= -18
1 4
5 1
1 4
3 1 4
Theorem2
Let A be a square matrix. The following properties hold:
1) det ( A ) = det ( At )
2) Assume that A’ obtained from A by interchanging two rows
( columns ) of A. Then det(A’ ) = - det (A)
3) If A has two rows equal to each other then det(A) =0
4) If A has a zero row (or zero column ) then det( A) = 0,
5)If multiply a row( column ) of A by a scalar k then the
determinant of new matrix A’ is equal to k.det(A).
6)
Let the matrix A’ obtained from A by adding to a row by a
product of a scalar and another row . Then det( A) = det (A’ ),
a11
...
7) b1  c1
...
a n1
a12
...
b2  c 2
...
an 2
...
a1n
a11
a12
... a1n
a11
a12
... a1n
...
...
...
... bn  c n = b1
...
...
...
a nn
a n1
...
...
b2
...
an2
... ...
...
... bn + c1
...
... ...
... a nn a n1
...
c2
...
an2
... ...
... c n
... ...
... a nn
8) Determinant of a triangular matrix equals the product of diagonal
elements:
det( A) = a11a22 ...ann
9) de( AB ) = det(A) de(B) for A, B of degree n.
13
Example
2 4 6 8
1 3 5 7
a)
= 0 because row 2 equals row 4
2 5 9 0
1 3 5 7
2 4 6 8
1 3 5 7
b)
2 5 9 0
1 3 6 9
=
2 4 6 8
1 3 5 7
2 5 9 0
0 0 1 2
The second matrix received from the
first matrix by adding to row 4 by product of (-2) and row 2. So determinants
of them are equal.
c)
12 24 16 8
1 3 5 7
2
1
5
3
9
6
0
9
=4
3 8 4 4
1 3 5 7
2 5 9 0
1 3 6 9
. The first row of the first matrix A is
equal to 4 times the first row of the second matrix A’ . So
det( A) = 4 det(A’).
d)
=
=
1
1
2 1 1
3 5 4
2 5 9
2 3 6
1
0
0
9
2 1 1
1 4 5
0 1 7
2 3 6
2
9
1 2 1 1
0 1 4 5
0 1 7
0 7 8
2
7
=
1
0
2 1 1
1 4 5
2 5 9
2 3 6
0
9
( add to second row by (-1)time row 1. )
( add to row 3 by (-2) times row 1 )
( add to row 4 by 2 times row 1 )
14
=
=
1 2 1 1
0 1 4 5
0 0 3 3
0 7 8 7
1 2
0 1
( add to row 3 by ( -1 ) times row 2 )
1
5
1
4
0 0
3
3
0 0  20  28
=3.4
1 2
0 1
1
4
1
5
0 0 1 1
0 0 5 7
( add to row 4 by ( -7 ) times row 2 )
= 3.4
1 2 1
0 1 4
1
5
0 0 1 1
0 0 0  12
= 3.4. (-12) = -144
3. Inverse matrix anf rank of a matrix
3.1. Inverse Matrix
3.1.1 Definition and examples
Definition 1
Let A be a square matrix of order n . If there is a square B of degree n
such that AB =BA = E , here E is identity then B is said to be the inverse
matrix of A denoted by B = A -1 and A is said to be invertible.
15
Example 1
 3 2
-1
a) A= 
 ,A =
4
3


 3  2
 4 3  because


3 2  3  2  3  2 3 2 1 0
 4 3    4 3  =   4 3   4 3  = 0 1  = E



 
 

1 0 0
1


b) A = 0 2 0 is invetible and A -1 = 0
0 0 3
0
0
1
2
0
0
0 
1 
3
Theorem 1
Suppose A, B are square matrices with the same degree .
1) If A is invertible then A -1 is invertible and ( A -1)-1 = A,
2) If A, B are invertible then AB is invetible and ( AB )-1 = B-1 A-1
3) Inverse matrix of identity matrix E is E.
Proof. 1) , 3 ) are easy to see from the definition,
2) ( AB )( B -1A-1) = A (BB -1)A -1 = A (EA -1) = AA -1 = E.
analogously, ( B -1A-1)(AB) = E. Thus B -1A-1 = (AB)-1
The theorem is proved .
3.1.2 Condition for a invertible matrix
Theorem 2
Matrix A is invertible if and only if det (A )  0 .
Proof. Necessary : If A is invertible then A.A -1 = E . It follows
16
det(A) .det( A -1) = det( E ) = 1 and det(A )  0.
Sufficient :
 
Let A = aij and det(A)  0. For a ij , the algebraic
 
complement is A ij . Denote C = Aij and A * = C t . We can see A -1 =
1
A*
det(A)
3.1.3 Method of finding inverse matrice s.
a)Using algebraic complements
From the proof of the theorem 2, we can find the inverse matrix of A as
follows.
Step 1 Compute det( A ). If det(A)  0 then there exists A -1.
 
Step 2 Compute comlement A ij, Write C = Aij , A* = C t ( transpose of C ).
Step 3 Obtain A -1 =
1
A*.
det(A)
1 2 3 


Example 2 Find the inserve matrix of A = 2  1 0 ..
3 1 4
det ( A ) = -4 –2.8 +3.5 = -5  0
A11 =
1 0
= -4 ;
1 4
A 12 = -
2 0
= -8 ;
3 4
A21 = -
2 3
=-5;
1 4
A 22 =
1 3
= -5 ;
3 4
A31 =
2 3
=3;
1 0
A32 = -
1 3
=6;
2 0
2 1
3 1
A 13 =
A 23 = A 33 =
=5 ;
1 2
= -5
3 1
1 2
= -5
2 1
 4  8 5 
 4  5 3 
  5  5  5

*
t 
We have C = 
 ; A = C =  8  5 6 
 3
 5  5  5
6  5
17
The inserve matrix
A -1 =
 4  5 11 
1 
 8  5 6 

5
 5
4  5
b)Method of elemetary operations
Three elementary operations on rows of a matrix are:
1. Interchane two rows
2. Multiply one row by a nonzero number,
3. Add a multiple of a row to a different row.
By using above operations one can obtain the inserve matrix of A as
follows
Let A be invertible .Denote A the matrix of size n2n obtained by
putting the identity matrix E near A :
A = [ A  E ] . Use the above
operations on rows of A such that carry A to the identity matrix E, then E
becomes A -1 :
A = [ A  E ]  [ E  A-1]
Example 3
2 3
Let A = 
 , det(A) = - 2  0 , A is invertible.
4 5
 2 3 1 0
A= 

4 5 0 1
Add to row 2 by (-2) times row 1 , we have
 2 3 1 0


0  1  2 1 
Add to row 1 by 3 times row 2, we receive
2 0  5 3


0  1  2 1
Multiply row 1 by
1
and multiply row 2 by (-1) :
2
18
1 0 5 2 3 2 


0 1 2  1
5 2 3 2 
Thus , A became identity . Then A -1 = 

 2  1
3.2. Rank of a matrix
3.2.1 Definitions and examples
Definition 2
Given a matrix A of size mn. A subdeterminant of degree k of A is the
determinant of a matrix of order k obtained from A by deleting ( m- k) rows
and ( n-k) columns .
Definition 3
The largest degree of nonzero subdeterminants of matrix A is said to be
the rank of A and denoted by rank(A) or r(A).
Thus, rank(A) = r if and only if there is a nonzero subdeterminant of
degre r and every subdeterminant of degree larger than r is zero.
Note that if size of A is mn , rank(A)  min{m,n}
Example 4
1 2 3 4 


a) A = 0 6 2 0 , every subdeterminant of degree 3 is zero
0 0 0 0
there is a subdeterminat of degree 2 nonzero
1 2
= 6 0 , rank(A) =2
0 6
19
1 2 3 
b) B = 2  1 4 , det(B) = 0  rank(B)  3. We have
3 1 7 
a nonzero subdeterminant of degree 2 :
1 2
= -5  0, rank(B ) =2
2 1
3.2.2 Echelon matrices.
Definition 4
A matrix is said to be echelon if it satisfies the following conditions
1) All zero rows are at the bottom,
2) The first nonzero element from the left in each nozero row is to
the right of the first nonzero element of the above row.
Example 5
9
0
A =
0

0
1 0 0 4
3  1 2 7
is an echelon matrix. One zero row is the fourth
0 0 8 1

0 0 0 0
row ( bottom ). The first nonzero element of row 3 is number 8 being to the
right of number 3 which is the first nonzero element of row 2. Number 3 is
to the right of number 9 being the first nonzero element of row 1.
Remark
Rank of an echelon matrix is equal to the number of nonzoro rows
In Example 5 rank(A) = 3
Theorem 3
The rank of a matrix is not exchanged if apply elementary operations
20
3.2.3 Method of finding the rank of a matrix .
In several cases , using definition to compute is very difficult because
we have to compute many subdeterminants of the matrix. We often apply
Theorem 3 to translate a matrix to an echelon matrix then obtain the rank.
We recall elementary operations:
1. Interchange two rows
2. Multiply one row by a nonzero number,
3. Add a multiple of a row to a different row.
Example 6
1
2
A= 
1

3
1 1
1 2 
.
2 3  1 1

2 3 2 3
1 1
1 2
Using elementary operations we translate
1
2
A= 
1

3
1 1
1 1
0  1

1 2
 
0 1
2 3  1 1


2 3 2 3
0  1
1 1
1 2
1 1 
1 1
0  1

1 0
 
0 0
2  2 0


0 1 0
0 0
1
0
1 1 
 1 0 
= A’
2  3 0

0 0 0
1
0
So, A’ is echelon and rank( A ) = rank (A’) = 3
21
Chapter II
SYSTEM OF LINEAR EQUATIONS
1. Cramer’s Systems
1.1. Basic concepts.
Definition 1
A system of m linear equations in n variables is a system of the form
 a11 x1  a12 x 2  ...  a1n x n  b1
 a x  a x  ...  a x  b
 21 1
22 2
2n n
2

..........
..........
..........
..........
..

a m1 x1  a m 2 x 2  ...  a mn x n  bm
here a ij in a field K are coefficients,
( 1.1)
bi in K are constant terms and
x1 , x 2 ,... x n are variables . The field K may be real or complex.
Sequence ( s1, ... , sn ) of n numbers is called a solution of the
system ( 1.1) if
a11s1  ...  a1n x n  b1

 ... ... ... ..........
 a  ...  a x  b
mn n
m
 m1
 x1  x2  x3  3
For example (-2, 5, 0) is a solution of the system : 
2 x1  x 2  3x3  1
22
Definition 2
Given a system of the form (1.1). The matrix
 a11 ... a1n 
A =  ... ... ...  is called the coefficient matrix of the system,
a m1 ... a mn 
 b1 
 x1 


B=  ...  is the constant column, X =  ...  is the column of
bm 
 x n 
variables. The matrix form of system (1.1) is
AX = B
(1.2).
Definition 3
If in system (1.1) bi =0 for all i then it is called a homogenous
system of linear equations.
What conditions for the existence of solution of a system of linear
equations ?
How can solve a given system ?
At first we consider Cramer’ systems.
1.2. Cramer’s systems of equations
Definition 4
A system of n linear equations in n variables of form
 a11 x1  ...  a1n x n  b1

 .... .... .... ....
a x  ...  a x  b
nn n
n
 n1 1
(1.3)
is called a Cramer’s system if the determinant of the coefficient matrix is
 
nonzero :det( aij )  0.
23
Theorem 1
Assume that the system
 a11 x1  ...  a1n x n  b1

 .... .... .... ....
a x  ...  a x  b
nn n
n
 n1 1
is a Cramer’s system. Then the system has an unique solution ( x1, ..., xn)
defined by formula :
xj 
det( A j )
det( A)
; j = 1, ..., n .
where each A j is the matrix obtained from A by replacing column j of A
by column B.
The above formula of solution is called Cramer’s rule.
Proof. The matrix form of the system is
AX = B .
(*)
From det(A)  0 , A is invertible. Multiply both sides of (*) by A -1 on
the left we have
Recall that
X=
X = A -1 B.
A -1 =
A*
, where A* is the adjoint matrix of A. Thus
det( A)
A* B
. By Laplace expansion row j of the numerator is det(A j).
det( A)
Theorem is proved.
Example 1
Solve the system of equations
 x  2y  3
.

3x  4 y  8
24
Solution.
unique solution x =
1 2
Det(A) = det( 
 ) = -2  0 . The system has an
3 4
det( A1 )
;
det( A)
1 3
det(A 2) = det ( 
 ) = -1.
3 8
3 2
det( A2 )
. Det( A 1) = det ( 
 ) = -4 ,
8
4
det( A)


y=
Hence
(x=2, y =1/2 ) is the solution of the
above system.
Example 2
Solve the system of equations
 x yz 6

2 x  y  z  3
3x  y  z  8

1
Solution.
1
1
det(A) = 2  1 1 = 4  0 implies the system is a
3 1 1
Cramer’s system. By Cramer’s rule we have the solution (x,y,z) as
x=
det( A1 )
det( A)
y=
det( A3 )
det( A2 )
, z=
;
det( A)
det( A)
1 1 6
det(A 1) = 3  1 1 = 4; det(A 2) = 2 3 1 = 8 ; det (A 3) = 2  1 3 =12
3 1 8
8 1 1
3 8 1
6
1
1
1 6 1
Hence x= 1; y = 2; z = 3.
1.3. Homogeneous system of n equations in n
variables
Let us consider the system
25
 a11 x1  ...  a1n x n  0

 .... .... .... ....
a x  ...  a x  0
nn n
 n1 1
(1.5 )
It is easy to see ( x 1,..., xn ) = ( 0,..., 0 ) is a solution that called the trivial
solution.
Remark
If det( A )  0 then the system ( 1. 5 ) is Cramer’s and the unique
solution is just the trivial solution.
A nonzero solution of ( 1.5) is called a nontrivial solution.
What condition for the existence of nontrivial solution of ( 1. 5) ?
The answer is given by the following theorem.
Theorem 2
The homogeneous system ( 1.5 ) has a nontrivial solution if and
only if the determinant of the coefficient matrix det(A) is equal to zero.
Example 3
 x yz 0

The system  2 x  y  z  0 has nontrivial solutions because det(A )
4 x  y  3 z  0

1 1 1
= 2  1 1 = 0 . We can see ( x, y, z ) = ( 2, 1, -3 ) is a nontrivial solution.
4 1 3
Example 4
Find the value of parameter a such that the following system of
equations has nontrivial solutions
ax  y  z  0

 x  ay  z  0
 x  y  az  0

26
a 1 1
Solution
det(A) = 1 a 1 = ( a+ 2 )( a-1 )2.
1 1 a
det( A ) = 0 iff a = - 2 or a = - 1. That are needed values of a.
2. Solution of a general system
of linear equations
2.1. Condition for the existence of solution.
Given a system of m equations in n variables
 a11 x1  ...  a1n x n  b1

 .... .... .... ....
a x  ...  a x  b
mn n
m
 m1 1
 a11 ... a1n

...
The matrix A = [AB] =  ... ...
a m1 ... a mn
( 2 .1 )
b1 
... 
bm 
is called the
augmented matrix of the system ( 2.1).
Theorem.3
The system (2.1) has a solution if and only if
rank( A ) = rank( A ).
27
Proof. Apply elemetary operations on rows of A to lead A to an
.. ..
a'1n
 a '11 ..
0 a '
.. ...
a' 2 n
22

 ... ...
... ...
...

echelon matrix A ’ = 0 0 ... a ' rj ... a' rn

... ...
0
 0 0
...
...
...

.... ...
 0 ...
...
... ...

b'1 
b' 2 
.. 
b' r 

b' r 1 
... 

0 
The given system is equivalent to the system with the augmented
matrix A ’. If r = rank( A)  rank( A ) = r+1 then b’r+1 0 .
Then the
(r+1)th equation of the new system has no solution. Thus the system has
no solution as the given system .
If r = rank( A ) = rank( A ) then b’r+1 = 0.
In the new system we can solve
r variables ( corresponding to the first
nonzero elements on rows of A ’ ) dependent on ( n –r ) remain
variables.Thus system has at least one solution. Theorem is proved. 
By the Theorem 1 we have conclusions:
a) If rank( A )  rank( A ) , system has no solution
b) If
rank( A ) = rank( A ) = n ( number of variables ), system has
an unique solution.
c) If rank( A ) = rank( A ) = r  n, System has an infinite number
of solutions dependent on ( n- r) parameters.
2.2. Method of solving system of linear equations
From the proof of Theorem 1 we can receive a method of solving the
system (2.1) as follows
Step1. Write the augment matrix A of the system
28
Step 2. Apply elementary operations on rows of A to lead
A to an
echelon matrix A ’.
Step 3. Compute rank(A ); rank( A )
 If rank( A )  rank( A ), the system has no solution

If rank(A) = rank( A ) = r , write the system corresponding to
the matrix A ’, continue step 4.
Step 4. Stand r variables corresponding to the first nonzero elements
on rows of A ’ and consider ather variables as parameters. Solve r
variables on parameters.
Example 1
Solve the system
 x1  x 2  x3  x 4  1
 2 x  x  2 x  3x  2
 1
2
3
4

 x1  2 x 2  3x3  x 4  3
4 x1  4 x 2  6 x3  3x 4  6
Solution. Augment matrix of the system
1

2
A = 
1

4
1

1 2 3 2
. Using elementary
2 3  1 3

4 6 3 6
1 1
1
operations lead A to an echelon matrix as follows
1

2
A= 
1

4
1

1 2 3 2

2 3  1 3

4 6 3 2
1 1
1
1 1

0  1
0 1

0 0
1
1
0 1
2 2
2 1
1

0

2

2
29
1 1

0  1
0 0

0 0
1

0 1 0

2  1 2

2  1 2
1
1
1 1

0  1
0 0

0 0
1

0 1 0
= A’
2  1 2

0 0 0
1
1
The corresponding system is
 x1  x 2  x3  x 4  1

x4  0
  x2 

2 x3  x 4  2

consider x 4 as a parameter t , we have
 x1
x
 2

 x3
 x 4

 52 t
t

 1  12 t

t
30
Chapter III
VECTOR SPACES
1. Definitions and simple properties
of vector spaces
1.1. Basic concepts, examples.
Definition 1
Let K be a field (of real numbers R or complex numbers C ) . A vector
space on K consists of a nonempty set V of elements
( called vectors )that can be added , that can be multiplied by a number in K
( called scalar ) and for which certain axioms hold. For vectors x, y in V their
sum x + y is a vector in V and scalar product of a vector x by number k  K
denoted as kx such that the following axioms are assumed hold.
A1.
A2 .
A3 .
x + y = y + x , for every x, y in V
( x+ y ) =z = x + ( y + z ), for x, y , z in V
There exists an element  ( called zero vector ) in V such
 + x = x +  = x for all x in V.
that
A4.
For each vector x there exists a vec tor (-x) such that
x + (-x) = (-x) + x =  .
A5.
k( x + y ) = kx + ky for all x, y in V, k in K
A6 .
( k1 + k2 ) x = k1x + k2x for all k1, k2 in K , x in V.
31
A7.
(k1.k2)x = k1(k2.x)
for k1, k2 in K, x in V
A8.
1.x = x , for x in V.
The vector –x in axiom A 4 is called the negative vector of x. If K is the
field R of real numbers then V is called a real vector space . If K = C of
complex numbers then V is called a complex vector space.
In this lecture we consider real spaces if nothing added. All results can
be applied for the complex case.
Example1
Show that the set V of vectors in the space with the vector addition and
scalar multiplication defined as in Geometry is a real vector space.
Solution. It is easy to check all axioms are satisfied.
Example 2
Consider R n = { (x1, x2, …, xn )  xi R , i =1, …, n}.
For x =(x 1, x2, …, xn ) ; y =(y 1, y2, …, yn ), kR
Put x + y := (x 1 + y1, x2 + y2, …, xn + yn );
kx : = (kx 1, kx2, …, kxn ) .
Then R n
is a real vector space. The zero vector is  = ( 0, …, 0 ).
The negative vector (–x) of x is as (- x1, - x2, …, -xn ).
Example 3
Denote by P[x] the set of all polynomials of real coefficients with
polynomial addition and multiplication by real numbers. Then P[x] is a real
vector space .
Example 4
Denote by Pn[x] the set of
degree
equal to or less
polynomials of real coefficients and with
than n. With the addition of polynomials and
multiplication of a polynomial by a number , P n[x] is a real vector space.
32
Example 5
C n = { (x 1, x2, …, xn ) xi  C, i = 1, … , n } is a complex vector space
using addition and scalar mutiplication similar to operations in Example 2.
Example 6
The set Mmn of real matrices of size mn is a vector space using
matrix addition and scalar multiplication.
Example 7
The set F[a,b] of all functions on the interval [a,b] is a vector space if
pointwise addition and scalar multiplication of functions are used. The zero
vector is the function  defined by
 (x) = 0 for all x in [a, b].
The negative vector of vector f is the function (-f) defined by
( -f) (x) = - f(x) for all x in [a, b].
Example 8
The set C[a, b] of all continuous functions on [a, b] is a vector space
using operations as in Example 7.
1.2. Simple properties.
One often consider real vector spaces. From now vector spaces are real
if nothing added.
Theorem1
1) Let x, y, z be vectors in vector space V.
If x + z = y + z then
x = y.
2) The zero vector of a vector space is unique.
33
Proof. 1) Let x + z = y + z . Add to both sides the vector ( -z ):
( x + z) + (-z) = (y + z ) + (-z)  x+( z +(-z) ) = y + ( z + ( -z ))
 x+ =y+ 
2). If

x = y.
 ,  ’ are zeros then
 + ’ =  ’
 =  +  ’ ( because  ’ is zero ) and
  = ’.
The proof is completed. 
Theorem 2 .
Let v be an vector in a vector space V, k be a real number,  the zero
vector of V. Following properties hold.
Proof.
1)
0.v =  ,
2)
k.  =  ,
3)
If kv =  then either k = 0 or v =  ,
4)
(-1) v = - v,
5)
(-k) v = k(-v) = -kv.
1) 1v +0v = (1+0)v = 1v +   0v =  .
2)
k  = k(0v) =( k.0)v = 0v =  .
3)
Assume that kv =  . If k  0 then multiplying both sides of
the equality by k-1 we have (k-1k)v = k-1  . It follows 1v = 
and hence v =  .
4)
5)
( -1) v + 1v = ( -1 +1 ) v = 0 v =  . Thus
(-1) v = - v.
(-k) v = (-1) (kv) = - ( kv ). On the other hand
(-k)v = (k(-1)) v = k ( -1v) = k (- v). Propertiy 5) is proved.
The proof is completed. 
34
2. Subspaces and spanning sets
2.1. Subspaces
Definition1
Given a vector space V.
A nonempty subset U of V is called a
subspace of the vector space V if U is itself a vector space where U uses the
vector addition and scalar multiplication of V.
Note that if U is a subspace of V, it is clear that the sum of two vectors
in U is a vector again in U and that any scalar multiple of a vector in U is
again in U . In short, that U is closed under the vector addition and scalar
multiplication .The nice part is that the converse is true . If U is closed
under the vector addition and scalar multiplication then U is a subspace of V.
Theorem 1
Let U be a nonempty subset of a vector space V. Then U is a subspace
of V if and only if it satisfies the following conditions:
1). If u1, u2 lie in U then u1+ u2 lies in U.
2). If u lies in U, then ku lies in U for all k in R .
Proof. If U is a subspace then the above conditions are satisfied.
Assume that
the above conditions are satisfied. We will prove
axioms of a vector space.
A 1)
The equality u 1+ u2
=
u2 + u1 holds for all u1, u2 in U because this
holds for all u1, u2 in V.
A 2)
(u1 + u2 ) +u3 = u1 + ( u2 + u3 ) holds for all u 1, u2, u3 in U because
this equality holds for all u 1, u2, u3 in V.
A 3). take u in U , 0 in R then 0.u =  lies in U and  + v = v+  = v for
all v in U.
35
A 4). For u  U, (-1)u = - u  U satisfies u+(-u) = (-u) + u =  .
Axioms A 5,A6, A 7, A 8 are obtained analogously.
Thus U is a vector space and hence is a subspace of V.
Example1
If V is a vector space , then U = {  } is a subspace of V and V is a
subspace of V. They are called trivial suspaces of V.
Example 2
Show that U ={ (x 1, x2, 0 ) x1, x2  R } is a subspace of R 3.
Solution. For x =(x 1, x2, 0 ) , y =(y 1, y2, 0 ) in U we have
x + y = (x1 + y1 , x2 + y2 , 0 ) lies in U and
kx = (kx 1, kx2, k.0 ) also lies
in U. By Theorem1, U is a subspace of R 3.
Example 3
Show that the set U ={ (x 1, x2 , x3 )  R 3  x3 = x1 + 2x2 } is a subspace
of R 3.
Solution. If x = ( x 1, x2, x3 )  U, y = (y1, y2 , y3 ) U then
x3 = x1 + 2x2
and y 3 = y1 + 2y 2. It follows x 3 + y3 = (x1 + y1 ) + 2(x 2 + y2 ).
Hence x + y = ( x 1 + y1 , x2 + y2 , x3 + y3 ) is in U.
If x =( x 1, x2, x3 )  U, k R then kx = ( kx1, kx2 , kx3 ) and
kx3 = kx 1 + 2kx2. Hence kx lies in U. Thus, U is a subspace of R 3.
Example 4
Let Pn[x] be the set of polynomials of degree equal or less than n.
Prove that the set U = { p(x)  Pn[x]  p(3) = 0 } is a subspace of Pn[x].
36
2.2. Spanning sets, linear combinations
Definition 2
Let { v 1 , v2 , … vn } be a system of vectors in a vector space V. A
vector v is called a linear combination of the system if it can be expressed in
the form :
v = k1.v1 + k2 v2 + …+ knvn
where k1 , … , kn are scalars called
coefficients of v 1, …, vn .
Example 5
In R 3 given vectors v 1 =( 1, 1, 0 ), v 2 = ( 0, 1, 1 ) . For k1, k2 in R the
vector v = k1v1 + k2v2 = ( k1, k1 + k2 , k2 ) is a linear combination of v 1, v2.
In case k1 = 1, k2 =2, v = ( 1 , 3, 2 ) is a linear combination of vectors v 1, v2.
Denote by Span{v1,…, vn } the set of all linear combinations of vectors
v1, …, vn. We have the following theorem
Theorem 2
For vectors v 1, …, vn in a vector space V, Span{v1,…, vn } is a
subspace of the vector space V.
Proof. If x Span{v1,…, vn } , y Span{v1,…, vn } then
x = x 1v1 + … xnvn, y = y 1v1 + … + ynvn.
It follows that x + y = ( x 1 +y1 )v1 +… +( xn+yn )vn is in Span{v1,…, vn }
and kx = k (x 1v1 + … xnvn ) = kx 1v1 + … + kxnvn is in Span{v1,…, vn } . So ,
Span{v1,…, vn } is a subspace of V by Theorem1.
37
Definition 3
W =Span{v1,…, vn } is called the subspace spaned by v 1, … vn and the
system {v1,…, vn } is called a spanning set of W.
Example 6
In Rn consider the system { e1,…, en } where e1 = ( 1, 0, … , 0 );
e2 = ( 0, 1, …,0); …; en = ( 0, …, 1). Then each x = (x1, … , xn ) can be
expessed as
x = x1e1 +x2e2 +… + xnen , hence x  Span{v1,…, vn }.
Thus R n = Span{v1,…, vn }.
Example 7
In R 2 consider vectors v1 = ( 1, 2 ) ; v 2 = ( 2, 1 ). Show that
R2 =
Solution.
the linear combination
Span{v1, v2 } .
For
b = ( b1 , b2 )  R2, we define coefficients k1, k2 in
v = k1v1 + k2v2 .
This implies to solve the system of equations
 k1  2k 2  b1

2k1  k 2  b2
Then
k1 = (2b2 – b1)/2 ; k2 = ( 2b1 – b2 ) / 2
3. Bases and dimension
3.1. Independence and dependence.
Definition 1
1) In a vector space V, a system of vectors { v 1, …, vn } is called a
linearly independent system if it satisfies the following condition.
If
k1v1 +k2v2 +… + knvn = 
k1 = k2 = … = kn = 0.
then
(3.1)
38
2) A system of vectors that is not linearly independent is
called linearly dependent.
Note that a system { v 1, …, vn } is linearly dependent if satisfies the
following condition: there are numbers k1, …, kn not all zeros such that
k1v1 +k2v2 +… + knvn =  .
(3.2)
Example 1
In R 3 given vectors v 1 = ( 1, 1, 1 ); v 2 = ( 1, 1, 0 ); v 3 = ( 2, 1, 0 ). Show
that the system of vectors v 1, v2 , v3 is linearly independent.
Solution.Suppose that k1v1 + k2v2 + k3v3 =  . Then
 k1  k 2  2 k 3  0

 k1  k 2  k 3  0 .
 k
0
 1
This implies k1 = 0, k2 = 0, k3 = 0. The system is linearly independent.
Example 2
Show that in R 3 the system { v 1, v2, v3 } where v 1 = (1, 1, 0 );
v2 = ( 0, 1, 1 ) ; v 3 = ( 1, 2, 1 ) is linear dependent.
Solution. k1v1 + k2v2 + k3v3 =  
k3  0
 k1 

 k1  k 2  2 k 3  0

k 2  k3  0


(k1, k2 , k3 ) = t ( -1, -1, 1 ), tR
So the system is linearly dependent.
Theorem 1
In a vector space V, following statements are true:
1
A subset of a linearly independent system is a linerly
independent system.
2
A system containing a linearly dependent system is a
linearly dependent system.
3
A system containing zero vector is linearly dependent.
39
Proof. 1) Let { v 1, …vk, …vn } be a linearly independent system,
1v1 + 2v2 +… +kvk =  
{ v1, …, vk} is a subset . Assume that
1v1 + 2v2 +… +k vk + k+1vk+1 +… + nvn =  where k+1 =…n = 0 .
(3.3)
From the linear independence of given system and expression (3.3) it
follows
1 = … =k = 0
. Thus the system { v1, …, vk} is linearly
independent.
2) Let
{ v 1, …vk, …v n }
dependent system { v 1, …, vk}. If
is a system containing
a linearly
{ v 1, …vk , …vn } is linearly independent
then by statement 1) we have { v 1, …, vk } is linearly independent . This is a
contradition. So { v 1, …vk, …v n } is linearly dependent.
3) Assume { v 1, …vk , …vn } is a system contaning zero vector . Let
vk = . Then we have ( 1, …, k, …, n ) = ( 0, …, ,1, …,0) that satisfies
1v1 + 2v2 +… +kvk + k+1vk+1 +… + nvn =  .
Thus, the system { v 1, …vk , …vn } is linearly dependent..
The proof is completed.
Theorem 2.
Let {u1, …, um} be an linearly independent such that each vector u i can
be expresed as a combination of a system { v 1, …vn}. Then m  n.
3.2. Bases , dimension
Definition 2
A system {v 1, …, vn } is called a spanning system of space V if each vV
is a linear combination of vectors of the system.
40
Definition 3
A system {v 1, …, vn } is called a basis of vector space V if it is a
spanning system and is linearly independent .
Example 3
In R n , Show that B = { e1, …, en } with e1 = ( 1, 0, …, 0 );
e2 = ( 0,1,…, 0 ) ; …, en = ( 0, …, 1 ) is a basis of R n.
Solution. It is to see that if 1e1` + 2e2 + … + nen = 
then
1 =…=n = 0. The independence of B is proved. Now let
x = ( x 1, …, xn ) Rn . Then x = x 1e1 + x2e2 +…+xnen. So B is a spanning
system of R n and is a basis of R n. It is called the canonical basis of R n
Example 4.
In Pn[x] we set B = { 1; x; …, x n}. Then B is a basis of the vector space
Pn[x].
Solution. If 1.1 + 2.x + …+n+1xn+1 = 0 then 1 =…= n+1 = 0. The
system is independent.
Now take an arbitrary polynomial p  Pn[x] . Then
p = a0 + a1.x + …+ anxn.
Thus, p is a linearly combination
of the system B . Therefore, B
becomes a basis of the space P n[x]. That is called the canonical basis
of
Pn[x].
41
Example 5
In R 2 consider B = { v 1, v2 } where v 1 = ( 1, 2 ); v 2 = ( 2, 3 ). Show that B
is a basis of R 2.
Solution. If 1v1 +2v2 =  then
 1  22  0

21  32  0
This follows 1 = 2 = 0 and therefore, the system B is linearly independent.
Now take an arbitrary b = ( b 1, b2 ) in R 2 . Then there are 1, 2 in R
satisffies
1v1 + 2v2 = b. Actually, this is equivalent to the esistence of
 1  22  b1
solution of system 
.
21  32  b2
The determinant
1 2
= -1  0 
2 3
System has a unique solution (1, 2 ) and B= { v1,v2 } is a spanning set of R 2.
Thus B= { v1,v2 } is a basis of R2.
Theorem 3
If a space V has a basis containing n vectors then an arbitrary basis of
V contains n vectors.
Proof. Assume that B = { e1, …, en }, B’ = { e’1, …, e’n } are bases of V.
Then each each vector ei can be expressed as a linear combination of
{ e’1, …,e’n }. By Theorem 2, n  m. Analogously, m n . So m = n.
Definition 4
a) If the vector space V has a basis containing n vectors then the
number n is called the dimension of V and denoted by n = dimV.
42
b) In case V = { } , the dimension of V is zero and denoted dimV = 0.
c)If dim V = n or dimV = 0 then V is called a finite dimension space.
d)If dimension of V is not finite , V is called an infinite dimension
space.
Example 6
a) dim (R n) = n
b) dim( Pn[x] ) = n+ 1,
c) dim P[x] = .
Theorem 4
a) In an n dimensional vector space V every system of
linearly
independent n vectors is a basis.
b) In an n dimensional vector space V, every system of linearly
independent m vectors ( m  n ) can be added by n-m vectors to become a
basis.
Theorem 5
Let B = { e1, …,en } be a basis of a vector space V and v is an arbitrary
vector then v can be expressed uniquely in the form
v = x1e1+x2e2 +… + xnen
, xi  R
(3.4).
Proof. B is a basis, it is a spanning system of V . Hence a vector v can
be expressed in the form (3.4) :
v = x 1e1+x2e2 +… + xnen
, xi  R
43
Now we will prove the uniqueness of the expession. Let v can be
expressed in another form:
v = y 1e1 + … + x nen ,
yi  R
( x 1 – y1 ) e1 +.. +(x n – yn )en =  . Then x 1 –y1 = … = x n –yn = 0 and
We get
therefore x 1 = y1 , …, xn = yn.
Theorem is proved. 
Definition 5
For a basis B = { e1, …,en } of the vector space V, if a vector v is
expressed as v = x 1e1 + …+xnen then the sequence ( x 1, …, xn ) is called the
coordinate of v with respect to the basis B . Denote (v)B = ( x 1, …, xn ).
Theorem 6
Let B be a basis of a vector space V, u, v in V. Assume that
(u)B = ( x1,…,xn ), (v)B = ( y1, …,yn ) . Then
a)
( u + v ) B = ( x1 +y1 ,…, xn +yn );
b)
( ku ) B
= ( kx 1 , …, kxn ) for k  R.
Proof. a) u + v = ( x 1e1 +…+xnen ) + ( y1e1 + … +y nen ) =
= ( x 1 +y1 )e1 + …+ (xn+yn)en.
This concludes a).
ku = k ( x 1e1 +…+ynen ) = (kx 1e1 +…+ kynen )
 ( ku ) B
= ( kx 1 , …, kxn ). 
44
3.3. Rank of a system of vectors.
Definition 6
Given a system { v 1, …,vm } in a vector space V. Rank of the system is
the dimension of Span{ v 1, …, vn } denoted by rank{v 1,…, vn }.
Definition 7
Let B = { e1, …,en } be a basis of a space V, {v 1,…,vn} be a system of
vectors in V . If vectors v 1,…, vn are expressed

 v1  a11e1  a12e2  ...  a1n en

 v 2  a 21e1  a 22e2  ...  a 2 n en

............................

v m  a m1e1  a m 2 e2  ...  a mn en
 a11
a
21
then the matrix A = 
 ...

a m1
a12
a 22
...
am2
(3. 5 )
... a1n 
... a 2 n 
is called the matrix of coordinate
... ... 

... a mn 
rows of the system {v 1,…,vn} wit respect to the basis B.
Example 7
In R 3 consider {v 1, v2 }, v1 = ( 1, 2, 1), v 2 = ( 2, 1, 0 ). Prove that
rank{v 1,v2 } = 2.
Solution. {v1, v2 } is linearly independent and it is a basis of
span{v1,v2 } . Thus rank{v 1, v2 } = dim span{v1,v2 } = 2.
Example 8
45
In P2[x] given a system {v 1, v2, v3} where v 1 = 1+x+x 2;
v2 = 1+2x+3x 2 ; v3 = 1 + 3x+ 5x 2.
Define the matrix of coordinate of the
system with respect to the canonical basis { 1; x; x 2 }.
Solution. The matrix of coordinate rows of system is the matrix
1 1 1
A = 1 2 3 .
1 3 5
The relation between rank of systems of vectors and rank of matrices
is given by the followwing theorem.
Theorem 7
In a finite dimensional vector space the rank of a system of vectors {v 1,
…, vm} is equal to the rank of matrix of coordinate of that system with
respect to a basis.
Corollary 8
Let A be the matrix of system { v 1,…,vn } with respect to a basis in an
n dimensions space V. Then the system is linearly independent if and only if
det(A)  0.
3.4. Change of basis
Let B = {e1,...,en } , B’ = {e1’,...,en’ } be two bases ,and v be a vector in a
vector space V . We can consider coordinates of v respect to B ,B’. We find the
relation between those coordinates.
46
Each vectors of B’ can be expressed as a combination of vectors of B
 e1'  b11e1  ...  b1n en

................................ ,
e'  b e  ...  b e
n1 1
nn n
 n
( 3. 6)
 
Put B = bij , and the matrix A = B t is called the matrix change of basis from
B to B’ ( in short, change matrix ).
Note:
1)Each change matrix is a convertible matrix
2) If A is the change matrix from B to B’ then A -1 is the change matrix
from B’ to B.
Theorem
Let v be a vector in V and B ,B’ be bases and A the change matrix from
B to B’ . If (v)B = (x1, ..., xn ), (v)B’ = (x1’, ..., xn’ ) then
 x1 ' 
 x1 
 
 ... 
  = A  ...  and
 x n '
 x n 
 x1 ' 
 x1 
 ... 
-1  
  = A  ...  .
 x n '
 x n 
( 3.7 )
 a11 ... a1n 


Proof. Let B = {e1,...,en } , B’ = {e1’,...,en’ }, A =  ... ... ...  then
a n1 ... a nn 
 e'1  a11e1  ...  a n1en

 ..............................
e'  a e  ...  a e
1n 1
nn n
 n
v = x1e1 +...+x nen ;
,
(3.8)
( 3. 9 )
47
v = x 1’e1’+... + x’nen’ 
v = ( a11x’1 + ...+a1nx’n )e1 + ...+ (an1x’1+...+ann x’n) en
(3. 10 ).
From ( 3.9 ) and ( 3.10 ) obtain
 x1  a11 x'1 ...  a1n x' n

 .............................. or
 x  a x' ...  a x'
n1 1
nn n
 n
 a11 ... a1n   x1 ' 
 x1 
 ...  =  ... ... ....   ... 

 
 
a n1 , , , a nn   x n '
 x n 
 x1 ' 
 x1 


and therefore,  ...  = A -1  ...  .
 x n '
 x n 
Theorem is proved.

Example 9
In R 3 given canonical basis B and a basis B’ {e’1, e2’,e3’ };
e’1= (1,1,1 ), e2’ =( 1, 1, 0), e3’ = ( 1, -1, 0 ).
(v)B’ = ( 2, 3, 4 ), ( u) B = ( 1, 2, 1 ).
Compute ( v)B, (u)B’.
Solution. We have the expression
 e'1  1e1  1e2  1e3

e' 2  1e1  1e2  0e3
e'  1e  1e  0e
1
2
3
 3
1 1 1 


The change matrix A = 1 1  1 from B to B’.
1 0 0 
0

A -1=  12
 12
1
1
2
1
2
1
 1 ,
0 
(v)B =(x1,x2, x3 ), ( u)B’ = ( y’1, y’2, y’3 ). By formula (3.7).
48
 x1  1 1 1   2  9 
 x  = 1 1  1  3  = 1  ,
 2 
   
 x3  1 0 0   4   2 
 y '1 
 y'  =
 2
 y ' 3 
0
1
2
 12
1
1
2
1
2
1
 1
0 
1   1 
2 =  1  .
  2
1   21 
49
Chapter IV
LINEAR MAPPINGS
1. Definitions and elementary properties of
linear mappings
1.1. Definitions and examples.
Definition 1
Let V and V’ be vector spaces on R, a mapping
f : VV’ is called a linear mapping if it satisfies :
1) f ( x+ y) = f(x) + f(y) ,  x, y  V
2) f( kx) = kf(x) ,  x V, k R
Example1
a) f : V  V given by f(x) =  ,  xV is a linear mapping.
b) f : V  V given by
f(x) = x , xV is a linear mapping.
c) f : R 3 R2 , f( x 1, x2, x3) = ( x 1 +x2 +x3 ; x1 – x2 –2x3 ) is a linear
mapping.
50
Theorem 1
For a linear mapping f : VV’ :
1) f ( ) =
2) f ( -x ) = - f(x),  x in V,
3) f( x –y ) = f(x) – f(y)
4) f ( k1v1 + k2v2 + ... + knvn) = k1f(v1) +k2 f(v2) + ...+ knf(vn ),
where ki R, vi  V.
1.2. Images and kernel of a linear mapping.
Definition 2
Let f : VV’ be a linear mapping.
a) the set Ker(f) = { x V f(x) =  } is called the kernel of f,
b) The set Imf = { f (x)  x V } is called the image of f
Theorem 2
Let f : VV’ be a linear mapping. Then Ker(f) is a subspace of V and Imf
is a subspace of V’.
Theorem 3
Let f : VV’ be a linear mapping . Then f is injective if and only if
Kerf = { }.
Proof. Necessary . Assume f is injective, f(x) =  = f ()  x =
51
 Kerf = { }.
Surfficient. Assume Kerf = { }. If x, y in V such that
f(x) = f(y)  f(x) – f( y) =   f( x –y ) =   x –y lies in Kerf ={ }.
So x –y =  and x = y , f is injective .
Theorem 4
For a linear mapping f from an n-dimensional space V to another
space V’, the following equality holds
dim(V ) = dim(Imf) + dim(Ker f).
Proof. Hint. Let { e1, ..., er } be a base of Kerf, { w1,..., ws } a base of Imf.
Assume u1, ..., us are in V such that f(u i ) = wi. Then we show that
{ e1,..., er, u1,...,us } is a base of V.
Thus, dim V = r + s = dim(Imf )+ dim(Kerf).
Definition 3
Given a linear mapping f : VV’.
a) f is called an isomorphism if it is a bijection.
b) If f is a isomorphism then V, V’ are called isomorphic
Theorem 5
52
1) Let dim V = dim V’ = n. A linear mapping f from V to V’ is an
isomorphism if and only if f is injective or onto.
2)Two finite dimensionl spaces are isomorphic if and only if their
dimensions are the same.
1.3. Operations on linear mappings.
Definition 4
Let f, g : VV’ be linear mappings. The sum of f and g is a mapping
( f +g ) : VV’defined by ( f+g ) (x) = f(x) + g(x). The scalar product of f by
kR is a mapping kf : VV’ defined by (kf)(x) = kf(x).
Note. f + g and kf are linear mappings.
Theorem 6
Let f: VV’, g : V’V’’ be linear mappings . Then the product
gof : : VV’’ is a linear mapping .
2. Matrix of a linear mapping
2.1. Matrix with respect to pair of bases.
Definition1
53
Let V, V’ be finite dimensional spaces with corresponding bases B =
{ e1, ..., en }, B’ = { e’1, ..., e’m },
Assume that f :
VV’ is a linear mappings
such that
 f (e1 )  b11e'1 ...  b1m e' m

 .....................................
 f (en )  b e' ...  b e'
n1 1
nm m

( 2.1)
b11 ... b1m 
Put B =  ... .... ...  . The matrix A = B t is called the matrix of linear
bn1 ... bnm 
mapping f with respect to pair of bases B ,B’ .
Thus, matrix A is the matrix of collumns of cordinates of vectors f(e i) with
respect to B’.
Remark.
 y1 
 x1 
 
 ... 
If vV, [v]B =   , [f(v)]B’ =  ...  , A is the matrix of f then we have
 y m 
 xn 
 y1 
 ... 
 = A
 y m 
 x1 
 ... 
 
 x n 
(2.2)
Example 1
Given a linear mapping f : R 3  R2 defined by the formula
f(x1,x2, x3 ) = (x 1 +x2 +2x3 ; 2x1 –x2 – x3 ). Find the matrix of f with respect to
pair of canonical bases of R 3, R 2.
Solution. B = { e1, e2, e3 }, B’ = {e’1, e’2 } are canonical bases of R 3, R2.
54
f(e1) = f (1,0,0) = ( 1, 2 ) = 1e’1 + 2e’2
f(e2) = f(0,1,0) = ( 1, -1 ) = 1e’1 – 1e’2
,
f(e3) = f(0,0,1) = ( 2, -1 ) = 2e’1 – e’2 .
So the matrix of f is
2
1 1
A= 

 2 1 1 
Theorem 1
Let A be the matrix of alinear mapping f with respect to pair of bases
B ,B’ . Then dim(Imf) = rank(A)
Dim(Imf) is also called the rank of f
2.2. Matrix of a linear operation
Definition2
a) A linear mapping from V to itself is called a linear operation on V.
b)Let f be a linear operation on V. The matrix of f with respectto pair
of bases B, B’ is called the matrix of f with respect to B.
Remark
Matrix of a linear operation is square.
Example 2
f : R 3  R3 defined by
55
f(x1,x2,x3 ) = (x 1 + x2 , x2 + x3, x1 +2x2 +3x3 ).
The matrix of f with
respect to the canonial basis is
1 1 0
A = 0 1 1 .
1 2 3
Theorem 1
Denote by A the matrix of a linear operation f with respect to a base B
and B the matrix of f with respect to another basis B’. Let P be the matrix of
change of basis fom B to B’. Then B = P -1AP
3. Diagonalization
3.1. Similar matrices.
Definition 1
Let A, B be square matrices of degree n. A is called similar to B if
there is an invertible matrix P such that B = P -1AP. Denote A  B.
Proposition 1
1) A  A
2)If A  B then B  A.
3)If A  B and B  C then A  C
56
Proposition 2
Two matrices of a linear operation with respect to two bases are similar
to each other.
3.2. Eigenvalues and eigenvectors of a matrix
Definition 2
Let A be an nn matrix . A number  R is called an eigenvalue of A if
there is a column X such that
AX = X.
(3.1)
This column X is called an eigenvector corresponding to ..
Example 1
1 2 
1 2   2  4
2
2
For A = 
, X=   we have AX = 
=   =2   =2X




3  4
3  4 1   2
1 
1 
So, X is an eigenvector corresponding to the eigenvalue 2.
Now we will find eigenvalues and eigenvectors of a given matrix A.
Condition AX =X is equivalent to the condition
( A - E)X = , where E is identity matrix
(3.2).
Condition for the existence of nonzero column X satisfying (3.2) is eqivalent to
det(A-E) = 0.
(3.3)
Definition 3
For an nn matrix A, the polynomial on  det(A-E) is called the
characteristic polynomial of A. The equation
det(A-E) = 0
is called the
charateristic equation of A.
57
Theorem3
Let A be an nn matrix . Then
1) The eigenvalues of A are the roots of the charateristic equation
det(A-E) = 0
2) The eigenvectors of A corresponding to an eigenvalue  are the
non-zero solutions of the homogeneous equation
( A - E) X = 0
(3.4).
Example 2
1 0 0 
Given a matrix A = 2 3 1 . Find eigenvalues and eigenvectors .
3 1 3
Solution.  det(A-E) =
1 
0
0
2
3
1
3
1
3
= (1-)(4-)( 2-)
Thus, det(A-E)= 0   =1,  =4,  = 2 . They are eigenvalues.
 for  =1 ,
0 x1  0 x 2  0 x3  0

 2 x1  2 x 2  x3  0 
 3x  x  2 x  0
2
3
 1
2 x1  2 x 2  x3  0

  x1  3x 2  0
 x1  3t

  x 2  t , tR
 x  4t
 3
 3
 
For  =1 , eigenvectors X = t  1  , t  0.
 4 
 For  = 4,
 x1  0
 3x1  0 x 2  0 x3  0
3x1  0 x 2  0 3  0 

  x 2  t , tR
 2 x1  x 2  x3  0  
2
x

x

x

0
1
2
3

x  t
 3x  x  x  0
1
2
3
 3

58
0 
Eigenvectors X = t 1  , t  0.
1 
0
 For  = 2, eigenvectors X =  1  , t  0.
  1
3.3.
Eigenvalues and eigenvectors of a linear operation
Definition 4
Let f be a linear operation on a vector space V. A number  R is
called an eigenvalue of f if there is a nonzero vector vV such that f(v) = v.
Such vector v is called an eigenvector corresponding to .
Theorem 4
Given a basis B of the space V. Denote by A the matrix of a linear
operation f on V with respect to B .Then
1) Eigenvalues of A are eigenvalues of f
2)A vector v is an eigenvector of f corresponding an eigenvalue  if
and only if vB is an eigenvector of A corresponding to .
Proof. Clearly, f(v) = v  A v B =  v B . This implies the proof.
3.4 Diagonalization
Theorem 5
Let f be a linear operation on a finite dimensional space V. Then the
matrix of f with respect to a basis is diagonal iff all vectors of this bais are
eigenvectors.
59
Proof.  Let B = { v 1, ..., v n } be a basis such that each v i is an
 f (v1 )  v1

eigenvector to eigenvalue i , i =1, ..., n. Then  ................
 f (v )   v
n
n n

1 0
0 
2
Hence the matrix of f is D = 
 ... ...

0 0
0
0 
. That is diagonal.
... ... 

...  n 
...
...
 Conversely, if B = { v1, ..., v n } is a basis to which the matrix of f is
1 0
0 
2
a diagonal matrix D = 
 ... ...

0 0
0
0 
then each v i is an eigenvector
... ... 

...  n 
...
...
corresponding to i, for i = 1,..., n. 
Definition 5
Let A be an nn matrix. If there is a matrix P suc that P -1AP = D,
where D is diagonal then A is called diagonalizable.
Theorem 6
Matrix A is diagonalizable iff A has n linearly independent eigenvectors.
Theorem 7
1)Eigenvectors corresponding to distint eigenvalues are linearly
independent.
2). If A has n distint eigenvalues then A is diagonalizable.
60
Chapter V
BILINEAR FORMS, QUADRATIC
FORMS, EUCLIDEAN SPACES
1. Bilinear forms
1.1. Linear forms on a vector space
Definition 1
A linear form f on a vector space V is a linear mapping
f : VR.
Example 1
f : R 3  R given by f(x 1,x2,x3 ) = x1+ 2x2 – 4x3 is a linear form on R 3.
Theorem1
The set of all linear forms on a vector space becomes a vector space
with addition and mutiplication operations defined by:
( f+g)(x) =f(x) +g(x) ;
(kg)(x) = k.f(x)
Definition 2.
The vector space of all linear forms on a vector space V is called the
dual vector space of V and denoted by V * .
1.2. Bilinear forms
Definition 3
A bilinear form  on a vector space V is a mapping
61
 : VV  R such that following properties hold:
1)  ( x+x’, y ) = (x,y) +  (x’,y ) for x, x’, y  V,
2) ( kx, y ) = k( x, y ) for x, y  V, k R ,
3)  ( x, y + y’ ) = ( x, y ) + ( x, y’ ) for x, y, y’  V,
4) ( x, ky ) = k ( x, y ) for x, y  V, k R
(x,y) is bilinear if  is linear on x for y fixed and linear on y for x
Note
fixed.
Example 2
 : R 2R2 
R
given by (x, y) = x 1y1 +x1y2 +2x2y1 + 3x 2y2,
where x = (x1, x2 ) , y = (y 1 , y2 ) in R 2 is a bilinear form.
Example 3
Let f, g are linear form on a vector space V . Denote  the mapping
 : VV  R defined by ( x,y) = f(x).g(y). Then  is a bilinear form on V.
1.3. Matrix of a bilinear form  
In space V given a basis B = { e1,..., en }. Let  be a bilinear form on V.
Let x = x 1e1 +...+ x nen , y = y 1e1 + ...+ y nen. Then
( x, y ) = ( x1e1 +...+ x nen , y1e1 + ...+ ynen ) =
n
 x y  (e , e
i , j 1
i
j
i
j
)
Put aij = (ei,ej ). We have
( x, y ) =
n
a
i , j 1
ij
xi y j
(1.1)
 x1 
 y1 


 
Let A= aij , X =  ...  , Y=  ...  . Then
 x n 
 y n 
62
( x, y ) = x  A  y 
t
(1.2 )
Definition 3
 
Matrix A= aij
is called the matrix of the bilinear form respect to the
basis B.
Example 4
Given
a
bilinear
form

:
R3
R3
R3
defined
by
(x, y) = x1y1 + x1y2 +2x2y1+3x2y2 +x2y3 +x3y1 +x3y2 –x3y3,
where x = ( x 1, x2, x3 ), y = (y1, y2 , y3 ). Find the matrix of  with respect to the
canonical basis B = { e1, e2, e3 }.
 a11 a12 a13 
Solution. The matrix A = a21 a22 a23  of  is defined as follows a11 =
a31 a32 a33 
(e1,e1)= 1; a12 = (e1,e2 )=1 ;
a21 = 2
; a22 = 3; a23 = 1 ;
a13 = (e1,e3) = 0;
a31 = 1 ; a32 = 1; a33 = -1.
1 1 0 


So A = 2 3 1  .
1 1 1
Theorem 2
Let  be a bilinear form on a vector space V. Assume that A, B are
matrices of  with B, B’ ,respectively and P the change matrix from B to B’ .
Then B = P tAP.
63
Definition 4
A bilinear form on a vector space V is
called
symmetric
if
(x, y) = ( y, x) for all xV, y V.
Theorem3
Let  be a bilinear form on V, and A be the matrix of of  with respect to
a basis. Then  is symmetric if and only if A is symmetric.
2. Quadratic forms
2.1. Definitions and examples
Definition 1
Let  : VV  R be a symmetrix form. The map  : VR defined by :
(x) = (x,x) is called a quadratic form on V.
Example 1
: R 2R2 R given by ( x, y ) = x 1y1 + 2x1y2 +3x2y2 , is a symmetrix
bilinear form. Then (x) = (x,x) = x 12 +4x1x2 +3x22 is the quadratic form
defined by .
Definition 2
Let  be a quadratic defined by a symmetrix bilinear form  and A be the
matrix of  with respect to a basis B. Then A is also called the matrix of the
quadratic  with respect to B.
64
Thus, each matrix of a quadratic form is a symmetrix matrix.
 
Let A= aij be the matrix of a quadratic , (x)B = (x1, ..., x n ). Then
( x) =
n
a x x
i , j 1
ij i
j
(2.1).
This is called the coordinate formula of .
2.2. Canonical form of a quadratic form.
Definition 3
Let  be a quadratic form on V, B = { e1, ..., en } be a basis such that the
coordinate formula of  is expressed by:
( x) = a11x12 +...+ annxn2
(2.2).
Then the formula (2.2) is called a canonical form of  .
Remark
The coordinate formula of a quadratic is canonical if and only if the
matrix is diagonal:
a11 ... 0 


A =  ... ... ...  .
 0 ... ann 
Definition4
a)A quadratic form  on V is called positively determinate if
(x) 0 for every x  0.
65
b)
A quadratic form  on V is called negatively determinate if
(x)  0 for every x  0.
2.3. Lagrange’s Method
Let B = { e1,..., en } be a basis and the coordinate formula of a quadratic
(x) =
n
a x x
i , j 1
ij i
j
(2.3)
Lagrange’s Method to move  to a canonical form is presented as follows
Case 1.
There exists a diagonal coefficient a ii  0, assume a11 0.
Then (x) =
1
(a11x1  ...  a1n xn ) 2 + 1(x2,...,xn ).
a11
Change of basis by the formula:
 y1  a11x1  ...  a1n xn

y2  x2


...


yn  xn
Then (x) =
(2.4)
1 2
y1 + 1(y2,...,yn).
a11
Continue to 1 and so on we can receive a canonical form of given quadratix
form  :
(x) = 1z12 + ...+nzn2 , where (x)B’ = ( z1, ..., zn ) is the coordinate of
x with respect to the latest basis in above steps.
Case 2. aii = 0 for all i =1,..., n and there is a coefficient a ij  0.
66
xi  yi  y j


Put  x j  yi  y j
,
 x  y , k  i, k  j
k
 k
then
aijxixj = aij( yi2- yj2 ) and the coefficient of yi2 is nonzero as in case 1.
2.4. Jacobi’s Method
 a11 ... a1n 
Assum that the matrix of  is A =  ... ... ...  .
an1 ... ann 
Put 0 =1, 1= det a11
 a11 ... a1k 
 a11 ... a1n 


,..., k = det  ... ... ...  ,..., n=  ... ... ... 
ak1 ... akk 
an1 ... ann 
k is called the main determinant of degree k of A.
Theorem 1.
Assume that all k  0.Then there is a basis B’= { e1’,..., en’ } such that for
a vector v, (v)B’ = ( y1,..., yn ),
 (v) = 1y12 +... + nyn2,
where 1 = 1/0 ,..., n = n/n-1
Theorem 2.( Sylvestre )
Let A be the matrix of a quadratic form , k be the main determinant
of degree k for every k = 0, .., n. Then
67
a)  is positively determinate if and only if k 0, for all k,
b)  is negeitively determinate if and only if (-1)kk 0, for all k.
Example 1
1 1 0
Matrix of  is A = 1 3 1 . Then 1= 1, 2 = 2 , 3 = 9 and therefore, 
0 1 5
is positively determinate.
Example 2
 1 1 0
Matrix of a quadratic form  is B =  1  3 2 . Then 1= -1, 2 = 2 ,
 0
2 2
3 = -8 and therefore,  is negeitively determinate.
3. Euclidean Spaces
3.1. Inner product on a vector space
Definition 1
Let V be a vector space on R, for x, y  V, the inner product
 x, y is a real number such that
1)
 x, y = y, x, for x, y  V,
2)
 x+x’, y = x, y + x’, y
for x, x’, y  V,
68
3)
 kx, y = k x, y for x, y  V, k R,
4)
 x, x  0 for xV,  x, x = 0 iff x = .
Example 1
For V = R n, x=( x 1, ..., xn ), y = ( y 1,..., yn ), put
 x, y = x1y1 + x2y2 +...+ xnyn. Then  ,  is an inner product and it is called
the euclidean inner product of R n.
Example 2
b
For f, g V = C [a,b], put  f, g =
 f ( x).g ( x)dx . Then  x, y
a
is an inner product of C [a,b].
Remark
Put  x, y =  (x,y) . If  x, y is an inner product then  (x,y) is a
symmetric bilinear form such that the corresponding quadratic form is
positive definite. Conversely, if  (x,y) is a symmetric bilinear form such that
the corresponding quadratic form is positive definite then  x, y =  (x,y) is
an inner product.
Example 3
for x = (x 1, x2, x3 ), y = ( y 1, y2, y3 ), put  x, y = x1y1 + x1y2 + 3x2y2 +
6x3y3 . Then  (x,y) =  x, y is a symmetric bilinear form and its matrix
69
1 1 0
A = 1 3 0 has 1 = 1, 2 = 2, 3 = 12. Therefore, the corresponding
0 0 6
quadratic form is positive definite and  x, y is an inner product.
Definition.
A n- dimensional space with an inner product is called Euclidean space
of dimension n.
3.2. Norms and othogonal vectors
Definition 2
1) Let V be an inner product space. Vectors x, y in V are othogonal
if  x, y = 0.
2) The norm of a vector v ( or length ) is defined by v = v, v
d(v,w) = v  w .
The distance between v, w is
Theorem1( Schwarz Inequality )
If v, w are in inner product space V then
 v, w2  v
2
w
2,
Moreover, equality occurs if and only if one of v and w is a scalar multiple
of the other.
70
Theorem 2
Let V be an inner product space, the norm has the following properties:
1)
v  0 ,vV,
2) v = 0  v =,
kv = k v ,  kR, v V,
3)
4) u  v  u + v ,  u V, v V.
Theorem 3
For an inner product space V, the distance d has the folowing
properties :
1) d(v,w)  0, for v, w  V,
2) d( v,w ) = 0  v = w,
3) d(v, w) = d (w,v), for v, w  V,
4)
d( v,w)  d(v, u) + d(u,w), for u, v, w  V,
Definition 3
1) Two vectors u, v in V is called orthogonal if  u, v = 0 and denoted
by u v.
2) A system of vectors { v 1,..., v m} in V is called an orthogonal system if
vi vj , i j .
71
3) A system of vectors { v1,..., v m} is called an orthonormal system if
it is orthogonal and vi =1 for i = 1, ..., m.
Theorem 4
An orthogonal system that not contain zero vector is linearly independent.
3.3. Orthonormal basis
Definition 4
A basis of V is called an orthonormal if it is an orthonormal system.
Example 2
V= R n with the canonical inner product, B ={ e 1,..., en },
e1 = ( 1, 0,..., 0 ), e2 = ( 0, 1,..., 0 ), ..., en = ( 0,0 ,..., 1). Then B is an
orthonormal basis.
Example 3
V= R 2, B ={ u1,u2 }, u1 = (
1 3
,
), u2 =(2 2
3 1
, ). Then B is an
2 2
orthonormal basis.
Theorem 5
Let B = { e1,..., en } be an orthonormal basis of V, u, v V. Assume that
(u)B = ( x1, ..., xn ), (v)B = ( y1, ...,yn ). Then
a)  u, v = x1y1 +x2y2 + ...+ xnyn
b) u =
n
x
i 1
2
i
72
assume V has an inner product  ,. We can obtain an
Now
orthonormal system from a system of linearly indpendent vectors by GramSchmidt algorithm.
Theorem 5
Let S = { v1,..., vn } be a linearly independent system in a space with an
inner product. Then there exists an orthornormal system { e1,..., en } such that
Span{ e1,..., ek } = span{ v 1,..., vk }, for k =1, ..., n.
Proof. Put e1 =
v1
,
v1
e 2 = v2 -  v2 , e1e1, e2 =
e2
, ...,
e2
ek = vk -  vk , e1e1 - ... -  vk , ek-1ek-1; ek =
Then {e1, ..., en }
ek
.
ek
is an orthonormal sytem satisfying the condition of
the theorem. The above algorithm is called Gram-Schmidt algorithm.
Example 4
In R 3 given v 1 = ( 1,1,0 ), v 2 = (0,1,1), v 3 = ( 1, 1, 1). Orthonormalization
of this system, we have
e1 =
1
2
( 1,1,0 ), e 2 = v2 – v2, e1e1 =
1
(-1, 1, 2 ),
2
73
e2 =
1
( -1, 1, 2), e3 = v3 – v3, e1e1 – v3, e2e2 =
6
e3 =
1
1
( 1, -1, 1),
3
( 1, -1, 1). System {e1, e2, e3 } is orthonormal.
3
3.4. Orthogonal subspaces, projection
Definition 5
Let U, W be subspaces of V . U is called orthogonal to W if u w for all
u U, w W and then denote UW. Let U be a subspace of V, v V. If v = u +
w, u U, w  U then u is called the projection of v onU, denote
u = projU(v) .
Theorem 6
Let{ e1, e2,..., em } be a orthonormal basis of a subspace U of V. Then
for a vector v V we have
projU(v) = v,e1e1 +v,e2e2 + ... + v,emem.
Example 5
In R 3 system { e1, e2 }, where e1 =
1
1
( 1, 2, 2); e2 =
( 2, -1, 0 ) is an
3
5
orthonormal basis of U = span{e1,e2 }. For a vector v = (1, 3, 2 )
projU(v) = v,e1e1 +v,e2e2 =
11
1
( 1, 2, 2 ) +
( 2, -1, 0 )
9
5
74
=
1
( 37, 119, 110 ).
45
3.5. Orthogonal diagonalization
3.5.1 Orthogonal matrix
Definition 6
An nn matrix P is called an orthogonal matrix if PP t = E, where E is
identity, Pt is the transpose of P.
Note that, If P is orthogonal then P t = P-1.
Example 6
 1
a) P =   2 3
 2
3
2
1
2

 is an orthogonal matrix,

b)The idetity matrix E is orthogonal.
Theorem 6
Let B = { e1, ...,en } be an orthonormal basis of an Euclidean space V,
B’ ={ e’1, ...,e’n } be other basis, P be the matrix of change of basis from B to
B’. Then the matrix P is orthogonal if and only if the basis B’ is orthonormal .
3.5.2 Orthogonal diagonalization
75
Definition 7
Let A be an nn matrix. If there is an orthogonal matrix P such that
PtAP = D , where D is a diagonal matrix then A is called orthogonal
diagonalizable and P is called orthogonal diagonalizing A.
Theorem 7
The matrix A is orthogonal diagonalizable if and only if A is
symmetrix.
To find a matrix P which is orthogonal diagonalizing the matrix A we
use following results
Theorem 8
For a symmetrix matrix A, eigenvectors correspongding to distin ct
eigenvalues are orthogonal.
Theorem 9
For each symmetrix matrix A, all eigenvales are real numbers.
Theorem 10
For each symmetrix matrix A, there is a basis containing orthonormal
eigenvectors.
For a root  of multiple k of the characteristic equation det( A-E) =
0, dimension of the eigenspace is k.
To diagonalize a symmetric matrix A we can use the following Algorithm.
76
Algorithm.
Step1. Compute eigenvalues of A : 1,2,..., m of A and express
det( A - E ) = (  1 ) d1 ...(  m ) d m .
Step 2. For each eigenvalue i , find a system B i of di
orthonormal
eigenvectors. Put B = B 1...B m ={e1,...,en }. B is an orthonormal basis of R n.
Step 3.
The matrix P containing n columns e 1, ..., en is an orthogonal
matrix and diagonalizing for A.
 '1
0
t
P AP = D, D = 
 ...

0
0
 '2
...
0
0 
... 0 
,
... ... 

...  ' n 
...
where ’i is corresponding to ei for i = 1,..., n.
Example 7
3 1 0 


Given A = 1 3 0 . Find the matrix P orthogonal diagonalizing for
0 0 2
the matrix A.
Solution.  Eigenvalues are 1= 2 = 2, 3 = 4

For 1= 2 = 2, orthonormal eigenvectors e1 =

2
2
; 
2
2

; 0,
e2 =( 0, 0, 1 ).
 for
3= 4, e3 =

2
2
;
2
2

; 0,
77
 22

 P =  22
 0


2 0 0



t
 is orthogonal and P AP = 0 2 0 =D
0 0 4
0 
2
2
2
2
0
0
1
Now we consider the problem of orthogonal diagonalization of a quadratic form as
follow.
Given a quadric form w in an Euclidean space. Find an orthornormal basis such that
to this basis the quadric form in the canonical form .
Note
. Assume that if w is in canonical form with respect to orthonormal basis B then
the matrix of w with respect to this basis is a diagonal matrix..
Method of diagonalizing a quadratic form.
Assume that
the matrix of w w.r to orthonormal basis B is A and
orthonormal basis such that
the
B’ is the
matrix of w is diagonal matrix D. Denote by P the
matrix of change from B to B’ . Then P is orthogonal and P -1 AP = D
Hence we have an algorith to find B’
Algorithm
Step 1.
Find
Quadric surfaces
78

Quadrics in the Euclidean
plane are those of dimension
D = 1, which is to say that they
are curves. Such quadrics are
the same as conic sections, and
are typically known as conics
rather than quadrics.
Ellipse (e=1/2), parabola (e=1) and hyperbola (e=2) with fixed focus F and
directrix.
In Euclidean space, quadrics have dimension D = 2, and are known as quadric surfaces.
By making a suitable Euclidean change of variables, any quadric in Euclidean space can
be put into a certain normal form by choosing as the coordinate directions the principal
axes of the quadric. In three-dimensional Euclidean space there are 16 such normal
forms. Of these 16 forms, five are nondegenerate, and the remaining are degenerate
forms. Degenerate forms include planes, lines, points or even no points at all.[2]
Non-degenerate quadric surfaces
79
Ellipsoid
Spheroid (special case of ellipsoid)
Sphere (special case of spheroid)
Elliptic paraboloid
Circular paraboloid (special case of elliptic
paraboloid)
Hyperbolic paraboloid
Hyperboloid of one sheet
80
Hyperboloid of two sheets
Degenerate quadric surfaces
Cone
Circular Cone (special case of cone)
Elliptic cylinder
Circular cylinder (special case of elliptic cylinder)
Hyperbolic cylinder
81
Parabolic cylinder
82