Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Chapter 3 Linear Algebra
Lecture 7 Matrix
1 Introduction
- Algebra & Geometry
- Vector
- Change of coordinates (transformation)
H. W. (Due Apr. 16th)
Chapter 3
3-2. 8, 9
3-3. 1
3-5. 12, 16, 21, 37
3-6. 6, 15
3-7. 23, 24
“The problems similar to the above ones will appear in the midterm exam.”
2. Matrix; row reduction (행렬 ; 행줄이기)
- A matrix is just a rectangular array of quantities, usually enclosed in
large parentheses.
 1 5  2
 : 2  3 (2 by 2) matrix
A  
 3 0 6 
Aij : i (row number), j (columnnumber)
A11  1, A12  5, A13  2, A21  3, A22  0, A23  6.
- Transpose of a matrix
 1  3


T
A  5
0  : 3  2 matrix
 2 6 


A 
T
ij
 Aji
- Sets of Linear Equations
 z  2
2 x

6 x  5 y  3z  7
2 x  y
 4


 2 0  1  x   2 

   
Mr  k ,  6 5
3   y    7 ,
 2 1 0   z   4

   
 2 0  1


where M   6 5
3 ,
2 1 0 


3
M
j 1
ij
x j  ki ,
i  1,2,3.
 x
 2
 
 
r   y , k   7 .
z
 4
 
 
- Augmented matrix
 z  2
2 x

6 x  5 y  3z  7
2 x  y
 4

 2 0 1 2


A  6 5
3 7
 2 1 0 4


(a) Eliminate the x terms in the other two equations by using the first equation.
ex. 2) – 1) x 3  2) , 3) – 1)  3)
 z  2
2 x

5 y  6z  1


 y  z  2

 2 0 1 2


6 1
0 5
 0 1 1 2


(b) For convenience, interchange the second and third equations
 z  2
2 x

 y  z  2


5 y  6z  1

 2 0 1 2


 0 1 1 2
0 5
6 1 

(c) Eliminate the y terms by using the second equation.
ex. 3) – 2) x 5  3)
2 x






y 
z

2
z  2
11z  11
2 0 1 2 


0 1 1 2 
 0 0 11 11


(d) Eliminate the z terms by using the third equation.
ex. 1) + 3) / 11  1) , 2) – 3) / 11  2)
2 x






y
3
 1
11z  11
2 0 0 3 


0 1 0 1 
 0 0 11 11


(e) finalizing
2 x





y
3
 1
z  1
1 0 0 3/ 2


0 1 0 1 
0 0 1 1 


- Allowed rules
i. Interchange two rows
ii. Multiply (or divide) a row by a (nonzero) constant
iii. Add a multiple of one row to another; this includes subtracting, that is,
using a negative multiple.
Ex. 2
 x  y  4z  5

2 x  3 y  8 z  4
 x  2 y  4z  9

5 
1 1 4 5
1 1 4 5 
1 1 4

 with the1st row 
 with the2nd row 

 2  3 8 4     0  1 0  6      0  1 0  6 
1  2 4 9
0 1 0 4 
 0 0 0  20






We can not get an answer. The equations are inconsistent.
- Rank of a Matrix
- The number of nonzero rows remaining when a matrix has been row reduced
is called the rank of a matrix.
- For example 2, the rank of A is 3, but the rank of M is 2. In this case, the
equations are inconsistent [(rank of M) < (rank of A)].
 x  y  4z  5

2 x  3 y  8 z  4
 x  2 y  4z  9

5 
1 1 4


A   0  1 0  6 ,
 0 0 0  20


rank: 3

5 
1 1 4


0 1 0  6 
 0 0 0  20


1 1 4


M  0 1 0
0 0 0


rank: 2
‘Inconsistent’
a. If (rank M) < (rank A), the equations are inconsistent and there is no solution.
b. If (rank M) = (rank A) = n (number of unknowns), there is one solution.
c. If (rank M) = (rank A) < n , then R unknown can be found in terms of the
remaining n – R unknowns.
3. Determinants; Cramer’s rule (행렬식 ; Cramer의 규칙)
- We have said that a matrix is simply a display of a set of numbers; it does not
have numerical value. For a square matrix, however, there is a useful number
called the determinant of the matrix.
- Evaluating determinants
i) 2 by 2
a b 
,
A  
c d 
a b
det A 
 ad  bc.
c d
ii) nth order matrix
a11
a12
a13  a1n
a21
a31
a22
a32
a23  a2 n
a33  a3n

an1

an 2

an 3  ann
- When removing the row and the column containing the element a_ij, we
have the remaining determinant, M_ij, called a minor of a_ij.
ex)
a11
a12
a13  a1n
a21
M 33  a31
a22
a32
a23  a2 n
a33  a3n

an1 an 2


an3  ann
(n-1) by (n-1) determinant
- Sign





 
  
   etc
  
etc

 
 
- cofactor :
 1i  j  M ij
- Finally, multiply each element of one row (or one column) by its cofactor
and add the results.
det A 
i j



1
aij  M ij

i ( or j )
“Laplace’s development”
ex)
1 5 2
7
2
3
1
 a23  4,
4
5
M 23 
1 5
2
1
i) method 1 (third column)
1 5 2
7
2
3
1
7 3
1 5
4 2
4
2 1
2
5
1
5
1 5
7
3
 2  1  4  11  5  38  148.
ii) method 2 (first row)
1 5 2
7
2
3
1
3 4
7 4
7 3
4 1
5
2
 11  135  2  148.
1 5
2 5
2 1
5
- Useful facts about determinants
1. If each element of one row (or one column) of a determinant is multiplied by a
number k, the value of the determinant is multiplied by k.
2. The value of a determinant is zero if,
(a) all elements of one row are zero
(b) two rows ( or two columns) are identical
(c) two rows (or two columns) are proportional.
3. If two rows ( or two columns) of a determinant are interchanged, the value of
the determinant changes sign.
4. The value of a determinant is unchanged if
(a) rows are written as columns are columns as rows
(b) we add to each element of one row, k times the corresponding element of
another row, where k is any number (and a similar statement for columns).
Example 2. Find the equation of a plane through the three given points (0,0,0),
(1,2,5), and (2,-1,0)
x y
0 0
1 2
2 1
z
0
5
0
1
1
0
1
1
4 3
Example 4. Evaluate the determinant
9 7
D
4 0
3 1
0
3
0
3
0
2
2
4
1
3
1
0
2 1
0
3 2
0 3 2
4 3
  1  3 7  4   1  3 7  4
0 1
3 1 4
0 6 0
1 4 0
3
7
0
  1   1  6 
0 2
 36
3 4
1. Subtract 4 times the fourth column from the first column, and subtract 2 times the
fourth column from the third column
2. Do a Laplace development using the third row
3. Add the second row to the third row.
4. Do a Laplace development using the third row.
- Cramer’s rule
a1 x  b1 y  c1
 a1
 

 a2
a2 x  b2 y  c2
c1
b1
a1
c2 b2
a2
x
, y
a1 b1
a1
a2
b2
a2
b1  x   c1 
      Mr  k
b2  y   c2 
c1
c2
.
b1
b2
- Denominator: determinant of the matrix with coefficients in the left side (M)
- Numerator: for x, replace x part in M with right side part, and take the determinant.
for y, replace y part in M with right side part, and take the determinant.
- We can use this method when you get the solution for the n linear equations (in
case that D  0).
- Rank of a Matrix
To find the rank of a matrix, we look at all the square submatrices and find their
determinants. The order of the largest nonzero determinant is the rank of the matrix.
cf. submatrix: a matrix remaining if we remove some rows and/or columns.
Example 6.
 1 1 2 3


  2 2 1 0
 4  4 5 6


The submatrices of the original matrix are four (123, 124, 134, and 234). Because
the first and the second columns are absolutely the same, the determinants of 123
and 123 should be zero. The absolute values of 134 and 234 should be the same.
For this reason, we need to check only 134.
2 3
 1  1 2 3
 1

 submatrix 

  2  1 0   0
  2 2  1 0   
 4  4 5 6
 4

5
6




“The rank should be less than 3.”
Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Chapter 3 Linear Algebra
Lecture 8 Vector
4. Vector (벡터)
- Notation

A  A  Ax , Ay 
- Magnitude
A  A  Ax2  Ay2  Az2
- Addition of vectors
ABBA
: commutativ
e law for addition
A  B  C  A  B  C
: associative law for addtion
- Multiplication by a constant & subtraction
- Unit vector
A
A
- Vectors in terms of components
A  iAx  jAy  kAz
A  B  i  Ax  Bx   jAy  By   k  Az  Bz 
- Multiplication of vectors 1: scalar product
A  B  A B cos
AB  BA
A  B  A times projection of B on A
 B times projection of A on B
A  A  A2
A  B  C  A  B  A  C
A  B  Ax Bx  Ay By  Az Bz


- Angles between two vectors using scalar product
AB
cos 
AB
example) Find the angles between these two vectors
A  3,6,9,
B   2,3,1
A  B  3   2   6  3  9  1  21
A  32  6 2  9 2  3 14,
cos 
B  14
AB
21
1

 ,
A B 3 14  14 2
  60
- Perpendicular and Parallel vectors
AB
A // B


AB  0
A  cB
 
Ax Ay Az


Bx B y Bz
- Multiplication of vectors 2 : vector product
A  B : vector product
A  B  A B sin  ,
A  B  A, B
A  B  0  A // B
AA  0
A  B  B  A
i  i  j j  k  k  0
i  j  k, j  k  i, k  i  j,
j  i  k.
i
A  B  iAx  jAy  kAz  iBx  jBy  kBz   Ax
j
k
Ay
Az
Bx
By
Bz
example 4.
A  2,1,1,
i
j
B  1,3,2
k
A  B  2 1  1  i  3 j  5k.
1 3 2
5. Lines and planes (직선과 평면)
- A great deal of analytic geometry can be simplified by the use of vector notation.
Such things as equations of lines and planes, and distances between points or
between lines and planes often occur in physics, and it is very useful to be able to
find them quickly. Vector notation will help you do these more easily.
‘points  vectors’
- Straight Lines
To determine a specific line, we need one point and a slope (= two points).
cf. Slope A  x2  x1 , y2  y1 , z2  z1 
 a, b, c 
Given point r0  x1 , y1 , z1 
Variable r  x, y, z 
 x  x0  at,

r  r0  At , or r  r0  At  y  y0  bt,

 z  z0  ct,
(paramet ric equat ions of a st raight line)
x  x0 y  y0 z  z0
 t , (symmet ricequat ion of a st raight line, a, b, c  0)


a
b
c
x  x0 y  y0

, z  z 0 , c  0.
a
b
- Planes
To determine a specific plane, we need a point and a normal vector.
N  r  r0   0
ax  x0   b y  y0   cz  z0   0, or ax  by  cz  d
(equation of a plane)
Example 1. Find the equation of the plane through the three points A(-1,1,1),
B(2,3,0), C(0,1,-2).
- Because points are given, what we have to do is to find the normal vector.
C
N
AC
B
A
AB
Example 1. Find the equation of the plane through the three points A(-1,1,1),
B(2,3,0), C(0,1,-2).
AB  2,3,0  (1,1,0)  (3,2,1),
AC  1,0,3
i j k
N  AB  AC  3 2  1  6i  8 j  2k.
  
1 0
3
Equation of the plane:  6x  2  8 y  3  2z  0  0
Example 2 Find the equation of a line through (1,0,-2) and perpendicular to the
plane of Example 1
A   6,8,2
 equation of the line :
x 1 y z  2
x 1
y
z2
 
(or


)
6 8
2
3
4
1
Example 3 Find the distance from the point P(1,-2,3) to the plane 3x-2y+z+1=0.
(Use the dot product.)
PR  PQcos (T he projectionis related to the dot product.)
P
Q
R
PR: distance we want to know
Q : any point on the plane
PR // N
PR  N  PQ N  n  PQ , for n  N N .
- Choose Q in the easiest way, e.g., Q=(0,0.-1) or (1,2,0) (as in the text)
PQ  0,0,1  1,2,3   1,2,4, N  3,2,1, N  14
PR   1,2,4  3,2,1 / 14  11/ 14.
Example 4 Find the distance from P(1,2,-1) to the line joining P1(0,0,0) and
P2(-1,0,2).
(Use the cross product.)
PR  PQ sin   PQ  a , where a  A / A .
A  P1P2   1,0,2,
PQ  PP1   1,2,1
PR  a  PP1  (1,0,2)   1,2,1 / 5  21 5.
Example 5 Find the distance between the lines, r=i-2j+(i-k)t, r=2j-k+(j-i)t.
P
A
n  A  B  / A  B 
n
Q
B
P  (1,2,0),
A  1,0,1
Q  0,2,1,
B   1,1,0
PQ   1,4,1, A  B  1,1,1 
PQ  n  2
3
PQ  n
Example 6. Find the direction of the lines of intersection of the planes
Note) the intersection lines are perpendicular to both the planes.
Example 7. Find the cosine of the angle between two planes
Note) The angle between the planes is the same as the angle between the
normal vectors to the planes
Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Chapter 3 Linear Algebra
Lecture 9 Matrix operation
6. Matrix operations (행렬계산)
- Matrix equations
5 
 x r u 2 1

  

 y s v   3  7i 1  i 
x  2, y  3, r  1, s  7i, u  5, v  1  i.
- Multiplication of a matrix by a number
 2
A  2i  3j  A   
3
a c
k 
b d
or AT  2 3
e   ka kc ke 
  

f   kb kd kf 
- Addition of Matrices
Note) Addition (or subtraction) can be done with the same type of matrices
 1 3  2  2  1 4  1  2 3  1  2  4  3 2 2 

  
  
  

 4 7 1   3  7  2   4  3 7  7 1  2   7 0  1
 1 3  2   2  1
  
  ?
cf. 
4 7 1  3 5 
- Multiplication of matrices
The element in row i and column j of the product matrix AB is equal to row
i of A times column j of B. [(# of row of A) = (# of column j)]
In index notation,
 ABij   Aik Bkj
k
ex.
 a c  e

AB  
 b d  g
Example 1
 4 2
,
A  
  3 1
f   ae  cg af  ch 
  
  C
h   be  dg bf  dh
1 5 3 

B  
 2 7  4
 4 2  1 5 3   4  1  2  2 4  5  2  7 4  3  2  4 

  

AB  
  3 1  2 7  4    3  1  1  2  3  5  1  7  3  3  1  4 
Example 2. Find AB and BA
 3  1
,
A  
 4 2 
 5 2

B  
  7 3
3 
 22
,
AB  
  34  2 
 1
 7

BA  
  33 13 
“not commutative”
- Zero matrix
: zero or null matrix means one with all its elements equal to zero.
cf.
 2  4
 0 0
2
  M  

M  
 1  2
 0 0
- Identity matrix or Unit matrix
1 0 0


I   0 1 0 ,
0 0 1


IA  AI  A
- Operation with Determinants
det AB  det BA  det A  det B
- Applications of matrix multiplication
0  1 x   5 
 1

  

0  y    1 
 2 3
 1  3 2  z    10

  

 xz   5 

 

Method 1   2 x  3 y    1 
 x  3 y  2 z    10

 

Method 2
0  1 x   5 
 1

  

0  y    1   Mr  k , T hen r  M 1k
 2 3
 1  3 2  z    10

  

- Inverse of a Matrix
MM 1  M 1M  I
M 1 
1
CT
det M
where Cij  cofactor of mij ,  1
i j
 M ij
Example 3.
0  1
 1


M   2 3
0 ,
 1 3 2 


det M  3
Finding the cofactor
1st row :
3
0
3 2
 6,

0 1
2nd row : 
 3,
3 2
3rd row :
0 1
 3,
3 0

2 0
1
2
 4,
1 1
 3,
1 2
1 1
 2,
2 0
2
3
1
3
 3,
1 0

 3,
1 3
1 0
 3.
2 3
 6 4 3
 6 3 3



1
1
-1
T
C   3 3 3  so M 
C   4 3 2
det M
3
 3 2 3



 3 3 3
- Rotational matrices
 cos  sin   cos


 sin  cos  sin 
 sin    cos     sin    
  

cos   sin     cos    
Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Chapter 3 Linear Algebra
Lecture 10 Linear transformation
7. Linear combinations, linear functions, linear operators
(선형결합, 선형함수, 선형연산자)
- A function of a vector, say f(r), is called linear if
f r1  r2   f r1   f r2 ,
Ex.1 A  (2,3,1),
and
f ar   af r 
f r   A  r
f r1  r2   A  r1  r2   A  r1  A  r2  f r1   f r2 .
linear
f ar   A  ar  aA  r  af r .
Ex. 2 f r   r
f r1  r2   r1  r2  r1  r2  f r1   f r2 
not linear
- F(r) is a linear vector function if
Fr1  r2   Fr1   Fr2 ,
and Far   aFr 
- Linear operator
OA  B  OA  OB
and OkA  kOA
- Matrix operators, Linear transformations
i) One set coordinate & r  R
 X  ax  by,

Y  cx  dy,
or
 X   a b  x 
   
 , or R  Mr
Y 
   c d  y 
Moving a point to some other point
 mapping or transformation
M : transformation matrix (linear operator)
ii) two sets of coordinates (x,y)  (x’,y’) & one vector r = r’
 x  ax  by,

 y  cx  dy,
or
 x   a b  x 
   
 , or r  Mr
 y 
   c d  y 
- Orthogonal transformation
: linear transformation preserves the length of a vector.
Orthogonal transformation: x2  y2  x 2  y 2
Orthogonalmaxtrix: M 1  MT
prove)
2
2
x2  y2  ax  b   cx  d   a 2  c 2 x 2  2ab  cd xy  b 2  d 2 y 2  x 2  y 2
a 2  c 2  1, b 2  d 2  1, ab  cd  0
 a c  a b   a 2  c 2

  
M M  M M  
 b d  c d   ab  cd
1
T
ab  cd   1 0 
  

2
2
b  d  0 1

det MT M  det I  det MT det M  det M   1  det MT  det M
2
 det M  1
cf. det M = 1 : rotation, det M = -1 : reflection

Example 3. Find what transformation correspond to each of these matrices
1   1
A 
2  3
1 0 
3
, B  
, C  AB, D  BA.

 1
 0  1
det A  1, rotation
det B  1, det C  det A  det B  1, det D  det B  det A
reflection
i) A
1   1
2   3
3 1 
  
 1  0 
1   1 
or


2  3
i  
ii) B : y  -y, reflection through the x axis.

1
ij 3
2
  120 rotation
iii) C
1  1
C  AB  
2  3
3  1 0 

 

 1  0  1
1   1  3 
,


2  3
1 
- We want to find the reflection line. The vector lying on the reflection line is
not changed by the reflection.
1   1  3  x   x 
 




2  3
1  y   y 
 y   x 3, say, i  j 3
1   1  3  1   1 

  



3

3
2   3
1 
 

 3  1 
1   1  3  3    3 

  











2  3
1  1    1 
 1   3 
iv) D
1 0  1  1
  
D  BA  
 0  1 2   3
3

 1 
1   1
2  3
Analyzing the D in the same way,
1   1
2  3
3  x   x 
    
1  y   y 
 y  x 3, say, i  j 3
3

1 
- Rotation in 2 Dimensions
 X   cos
   
Y 
   sin 
 sin   x 
 
cos  y 
vector rotated
(active)
 x   cos
   
 y   sin 
  
sin   x 
 
cos  y 
axes rotated
(passive)
change of basis
i, j, k   i, j, k
- Rotations and Reflections in 3 Dimensions
 cos

A   sin 
 0

 sin 
 cos

B   sin 
 0

 sin 
 cos

F 0
  sin 

cos
0
cos
0
0

0
1 
0

0
 1
0 sin  

1
0 
0 cos 
rotating along the z-axis
rotating along the z-axis
+ reflection through xy plane
rotation along the y axis
8. Linear dependence and independence
(선형종속과 선형독립)
Three vectors A=i+j, B=i+k, C=2i+j+k are linearly dependent,
because A+B-C=0.
1 1 0
Equivalently,
1 0 1 0
2 1 1
- Linear independent of functions
k1 f1 x  k2 f 2 x  k3 f3 x    kn f n x  0
In this case, these functions are linearly dependent.
ex. 1)
sin 2 x, 1  cos2 x
ex. 2)
sin x, cos x
linearly dependent
linearly independent
If f1 x , f 2 x ,, f n x  have derivative of order n  1, and if
f1 x 
f 2 x 
f1 x 
W wronskian  f1 x 
n 1
f1
f 2 ' x 
f3 x  

f3 x 
 0,
x 
fn
then, the functions are linearly independent.
Example 1.
1, x, sin x
1
x
f n x 
sin x
W  0 1 cos x   sin x  0
0 0  sin x
n 1
x 
x, sin x, 2 x  3sin x
Example 2.
x
sin x
W  1 cos x
0  sin x
2 x  3 sin x
2  3 cos x  0
0  3 sin x
- Homogeneous equations (right sides are zero)
x  y  0
1 0 0

 

0
1
0


x  y  0
x=y=0, (rank 2) = (# of unknowns)
(trivial solution)
x  y  0
1 1 0




 2 2 0
2 x  2 y  0
all points on x+y=0, (rank 1) < (# of unknowns)
(nontrivial solution)
Example 4. For what values of  does the set of eqs. have nontrivial solutions?
1   x  2 y  0

2 x  4    y  0
1 
2
2
 0,   0, 5
4
9. Special matrices and formulas
(특별한 행렬과 공식들)
- Summary
“Please take a look at (9.1) and (9.2)
- Index notation
 ABij   Aik Bkj
k
- Kronecker 
0, if i  j
 ij  
1, if i  j
1 0 0




I   ij   0 1 0 
0 0 1


 , if m  n
 cosnxcosm x dx 0, if m  n



  cosnxcosm x dx   

ij
- More useful theorems
ABC DT  DT CT BT AT
ABC D1  D1C1B 1A 1
Mathematical methods in the physical sciences 3rd edition Mary L. Boas
Chapter 3 Linear Algebra
Lecture 11 Diagonalizing matrices
10. Linear vector space (선형 벡터공간)
(Please read this section individually.)
11. Eigenvalues and eigenvectors; diagonalizing matrices
(고유값과 고유벡터 ; 행렬의 대각화)
- One set coordinate & r  R
 X  ax  by,

Y  cx  dy,
or
 X   a b  x 
   
 , or R  Mr
Y 
   c d  y 
Moving a point to some other point
 mapping, transformation, or deformation
M : transformation matrix (linear operator)
- Definition of eigenvalue and eigenvector
Some vectors are not changed in direction by transformation.
R  Mr  R  r where   const.
r : eigenvectors (characteristic vector)
: eigenvalues
- Eigenvalues (고유값)
According to the definition,
 X   5  2  x 
   
 .
 Y    2 2  y 
 X   5  2  x 
 x   x 
   
        
 Y    2 2  y 
 y   y 
Eigenvector
condition: R  r 
 5 x-2 y  μx,
 (5   ) x-2 y  0,


or 
  2 x  2 y  y,
  2 x  (2   ) y  0.
Conditionfor a solut ionot her thanx  y  0,
5
2
2
0
2
characterist ic equation of matrixM .
(5   )(2   )  4   2  7   6  0,
  1 or   6. eigenvalues 
- Eigenvectors (고유벡터)
- From the above results,  2 x  y  0 for   1
 x  2 y  0 for   6.

 R  r on 2 x  y  0

 R  6r on x  2 y  0

 5  2
.
through 
 2 2 
Any points in two straight lines can be eigenvectors.
ex.
 5  2 1   5 - 4  1 
    for   1

   
  
  2 2  2   - 2  4   2 
2 
 5  2  2  10  2  12 







   
    6  for   6

  2 2  - 1  - 4 - 2    6 
  1
4) Diagonalizing a Matrix (행렬 대각화)
5x1  2 y1  x1
 2 x1  2 y1  y1
for   1,
5x2  2 y2  6 x2
 2 x2  2 y2  6 y2
for   6.
Representing with the matrix operations,
 5 - 2  x1


 - 2 2  y1
x2   x1
  
y2   y1
x2  1 0 

  MC  CD
y2  0 6 


where, C  



1
5
2
5

2 

5  (unit vectors)
1 

5 
MC  CD, C 1MC  C 1CD  D
C 1MC  D.
 6 0
 1 0
, D  
 : no difference.
cf. D  
 0 1
 0 6
C 1MC  D
- Matrix D has elements different from zero only down the main diagonal,
diagonal matrix.
- D is called similar to M.
- When we obtain D given M, we have diagonalized M by a similarity
transformation.
5) Meaning of C and D (C와 D의 의미)
(x’, y’) rotated through  from (x, y)
 x  x' cos  y ' sin 

 y  x' sin   y ' cos
 cos
r  Cr ' whereC  
 sin 
 sin  
 specifically.
cos 
R  CR ' where R  ( X , Y ) and R'  ( X ' , Y ' ).
R  Mr,
CR  MCr   R'  C 1MCr   Dr .
- D=C-1MC is the matrix which describe in the (x’, y’) system the same
deformation (or transformation) that M describes in the (x, y) system.
- If C is chosen to make D=C-1MC diagonal, the new axes are along the
directions of the eigenvectors of M.
- The matrix C which diagonalizes M is the rotation matrix when the (x’, y’)
axes are along the directions of the eigenvectors of M.
12. Applications of diagonalization (대각화의 응용)
- Central conic section (ellipse or hyperbola) with center at the origin
Ax2  2 Hxy  By2  K ,
x
A
y 
H
 x   cos
   
 y   sin 
 x
y   x'
H  x 
   K or
B  y 
 x
y M    K
 y
 sin   x' 
 x' 
   C  
cos  y ' 
 y' 
 cos
y '
  sin 
 x
y M    K   x
 y
 x' 
  x' y 'D   K .
 y' 
x
x
sin  
 or
cos 
x
y   x'
 x
y CC 1MCC 1    K   x'
 y
y 'C T   x'
y 'C 1
 x' 
y 'C 1MC   K
 y' 
If C is the matrix which diagonalizes M, the above is the equation of the conic
relative to its principal axes.
Example 1.
5x 2  4xy  2 y 2  30
5 x  4 xy  2 y  30  x
2
2
 5  2  x 
   30.
y 
  2 2  y 
 5  2
, eigenvalues   1, 6. (from previous section)
M  

2
2


1 0
1
.
C MC  D  
0 6
1
2


 rotation matrix
 1 0  x' 
2
2
5
5

   x' 6 y '  30.
x' y '
cf.C  2
  
  arccos 15 .
1


 0 6  y ' 
5 
 5
This idea can be applied to three (or more) dimensions.
Example 2.
x
y
x 2  6xy  2 y 2  2 yz  z 2  24
0  x 
1 3

 
z  3  2  1 y   24.
 0  1 1  z 

 
Characteristic equation of thismatrix
1 
3
0
3
0
 2   1  0
1
1 
 1    
2
1
1
1 
 3
3
1
0 1 
 0
   3  13  12  (   1)(  4)(  3)
  1,   4,   3.
3 2
0
1
x'
 1 0 0  x' 

 
y ' z ' 0  4 0  y '   24 or x'2 4 y '2 3 z '2  24.
 0 0 3  z ' 

 
Let’s find C.
0  x   x 
1 3

   
 3  2  1 y    y 
 0  1 1  z   z 

   
Applying  1,   4,   3 to theaboveequation,
0  x 
1 3
 x

 
 
 3  2  1 y   1 y ,
 0  1 1  z 
z

 
 
0  x 
1 3
 x

 
 
 3  2  1 y    4 y ,
 0  1 1  z 
z

 
 
0  x 
1 3
 x

 
 
 3  2  1 y   3 y 
 0  1 1  z 
z

 
 

1
10
,0,
3
10
 1
 10
C  0
 3
 10
 when   1; 
-3
35
5
35
1
35
-3
14
-2
14
1
14


.


-3
35
,
5
35
,
1
35
 when   4; 
-3
14
,
-2
14
,
1
14
 when   3.
Example 3. Find the characteristic vibration frequencies


T  12 m x 2  y 2 ,
V  12 kx2  12 ky 2  12 k ( x  y ) 2  12 k (2 x 2  2 y 2  2 xy).
T  mx
1
2
 1 0  x 
y 
 ,
0
1

 y 
V  k x
1
2
 2  1 x 
 .
y 

1
2

 y 
(continued)
T o find theeigenvalues,
2
1
1
2
  2  4   3    1  3  0,
T o make the variablechange themat rixin V ,
x
 x' 
   C  ,
 y
 y' 
 
 

V  12 k x'2 3 y '2
 2  1
1 0
C  
,
C 1 
1 2 
 0 3

T o find a new variablein T ,
 x 
 x ' 
   C  , C 1UC  C 1C  U .
 y 
 y ' 
 
 


T  12 m x '2  y '2 .
  1,   3
(continued)
Lagrange’s equation



L  T  V  12 m x '2  y '2  12 k x'2 3 y '2
d  L  L
  
0
dt  y '  y '
d  L  L
 0,
 
dt  x '  x'
mx'  kx' ,
my'  3ky'.
x'  A sin 1t   ,
1 

y '  B sin 2t    dependinginitialconditions.
k
3k
, 2 
m
m
Finding the orthogonal transformation matrix C,
 2  1 x 
 x

   1  for   1,
  1 2  y 
 y
 1

 2


C 



1 
 for   1,
2
1
2
1
2
 1

 2
(continued)
 2  1 x 
 x

   3  for   3
  1 2  y 
 y

1 
 for   3
2
1 

2 
1 


2
 x
 x 
1
   C    x 
x  y, y  1 x  y.
2
2
 y
 y 
x  A sin 1t   ,
y  B sin 2t   ,
For B  0, y '  0. x  y 
For A  0, x'  0. x   y 
x'
A

sin 1t   .
2
2
y'
B

sin 2t   .
2
2
()(). in phase
()(). out of phase
‘Characteristic (or normal) modes of vibration’
‘Characteristic (or normal) frequencies of the system’