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1. Determinants and Hyper Sphere Equations
Three points in the plane not all on the same straight line determine a circle. A simple geometric construction of the circle
proceeds as follows. Construct two line segments between different pairs of the three points. Each of these segments is a
chord of the circle. Construct and extend the perpendicular bisectors of each chord. These meet at the center of the circle.
The segment from this point of intersection to any of three original points is a radius of the circle.
This construction can be extended to three points in three dimensions. Given P1, P2 and P3 in three dimensional space
construct the segment from P1 to P2. Through the mid point of this segment construct the plane P perpendicular to the
segment. Construct the segment from P3 to P2. Construct the perpendicular bisector of this segment in the plane of P1, P2
and P3. This bisector intersects the perpendicular bisector of the segment P1 to P2 in the plane of P1, P2 and P3 at a point
in plane P. This point is the center of the circle through P1, P2 and P3. Let N1 be the normal to the plane of P1, P2 and P3
through the circle’s center. By the Pythagorean Theorem every point on N1 is equidistant to P1, P2 and P3. N1 must be
perpendicular to a diameter parallel to the segment P1 to P2. Hence, N1 must lie in the plane P. Now consider a fourth
point P4 not coplanar with P1, P2 and P3. Performing an analogous construction using P1, P2 and P4 shows that the
normal N2 through the center of the circle through P1, P2 and P4 also lies in plane P. Hence, the normal lines N1 and N2
must intersect at a point O in plane P. Point O is equidistant to P1, P2, P3 and P4 and is thus the center of a sphere that
contains the points P1, P2, P3 and P4.
This construction is illustrated below.
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Despite the ease and simplicity of the geometric construction, at times (such as in computer graphics or surveying) it would
be beneficial to have an analytic representation of the circle determined by the three points. This can proceed according to
the same steps as in the geometric construction. However, the algebra, while not difficult is messy and seems “to hide” the
solution. A compact form for the equation of the circle would seem to be in order. Let C be the circle with center at (a, b)
and radius R that passes through the three points  x1 , y1  ,  x2 , y2  and  x3 , y3  . If the points are not collinear the system of
three equations in three unknowns,
2ax1  2by1  R 2  a 2  b 2  x12  y12
2ax2  2by2  R 2  a 2  b 2  x22  y22
2ax3  2by3  R 2  a 2  b 2  x32  y32
has a unique solution. If   2a ,   2b , and   R 2  a 2  b 2 , this can be represented by the linear system
2
 2
y1 1    x1  y1 
y2 1      x22  y22  .


y3 1     x 2  y 2 
3
 3
Solving this system corresponds to transforming the following augmented matrix to reduced row echelon form.
2
2
x
1 y1 1 x1  y1


 x2 y2 1 x22  y22 


 x3 y3 1 x32  y32 


Suppose now that  x, y  is another point in the plane. Unless  x, y  lies on C, the system represented by the augmented
matrix A has no solutions.
x
y 1 x2  y 2 


 x1 y1 1 x12  y12 
A

 x2 y2 1 x22  y22 


2
2
 x3 y3 1 x3  y3 
 x1
x
 2
 x3
If  x, y  lies on C, A must have a row of zeroes when reduced to row echelon form. In this situation the rows of A would be
linearly dependent and det  A  0 . Performing a cyclic permutation on the columns of A, the equation of the circle that
passes through the three points  x1 , y1  ,  x2 , y2  and  x3 , y3  is given concisely by equation (1).
x2  y 2
x
y
x12
x22
x32
x1
y1 1
x2
y2 1
x3
y3 1



y12
y22
y32
1
0
(1)
For completeness, suppose that the points are collinear. Initially assume that all of the points, assumed distinct, are on the
same vertical line, i.e. the three points are  C , y1  ,  C , y2  ,  C , y3  with y1  y2 , y1  y3 y2  y3 . Then the determinant can
be expanded as follows
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x2  y 2
2
C 
2
C 
2
C 
y12
y22
y32
C
Now C
C
WisMATYC
x
C
C
C
y
1
C
y1 1
2
2
 x y C
y2 1
C
y3 1


C 2  y12
y1 1
y2 1  x C 2  y22
y3 1
C 2  y32
C 2  y12
y1 1
y2 1  0 , C 2  y22
y3 1
C 2  y32
y1 1 0
y2 1  0
y3 1 0
y2 1  y C 2  y22
C 2  y32
y3 1
C 2  y12
C
y1
C 1  C 2  y22
C
y2
C 2  y32
C
y3
C 1
C 1
0 1
C 1  C 2  y22
0 1  0 , and the remaining two determinants are related
C 2  y32
C 1
C 2  y12
y1 1
C 2  y12
C 1
September 25, 2010
0 1
to nonzero three by three Vandermonde determinants.
C 2  y12
y1 1
y12
C 2  y22
y2 1  y22
C 2  y32
y3 1
C 2  y12
C
y1
y12
C
y1
C 2  y22
C
y2  y22
C
y2  C y22 1 y2  C 1 y2
C 2  y32
C
y3
y32
C
y3
y1 1
y32
1
y12
y1
y2 1   1 y2
y22    y2  y1   y3  y1  y3  y2 
y3 1
y32
1 y3
y12 1
y1
y32 1 y3
1
y1
1 y3
y12
y22  C  y2  y1   y3  y1  y3  y2  .
y32
Thus equation (1) reduces to x  C  0 . For the case y  Mx  B , with x1  x2 , x1  x3 , x3  x2 equation (1) becomes
x2  y 2
x12
x22
x32



y12
y22
y32
x
x1
x2
x3
y
1
x1
y1 1
2
2
 x  y x2
y2 1
x3
y3 1


Mx1  B 1
Mx2  B 1  x
Mx3  B 1
x12  y12
Mx1  B 1
x22  y22
Mx2  B 1  y x22  y22
x32  y32
Mx3  B 1
x1
Now, subtracting B times the last column from the middle column in x2
x3
Mx1  B 1 x1
Mx2  B 1  x2
Mx3  B 1 x3
x1
x2
x3
Mx1 1
Mx2 1  M
Mx3 1

x1
x2
x3

x12  y12
x32  y32
x12  y12
x1
Mx1  B
x2 1  x22  y22
x2
Mx2  B
x32  y32
x3
Mx3  B
x1 1
x3 1
Mx1  B 1
Mx2  B 1 gives the result that
Mx3  B 1
x1 1
0 x1 1
2
2
x2 1  M 0 x2 1  0 . So the coefficient on x  y vanishes.
x3 1
0 x3 1
Since x12  y12  x12   Mx1  B   1  M 2 x12  2MBx1  B 2 , the coefficient on  x can be simplified.
2
1  M 2  x12  2MBx1  B2
1  1  M 2  x22  2MBx2  B 2
1
1  M 2  x32  2MBx3  B2
x12  y12
Mx1  B 1
x22  y22
Mx2  B
x32
Mx3  B

y32
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1  M 2  x12  2MBx1
1  1  M 2  x22  2MBx2
1 1  M 2  x32  2 MBx3
Mx1  B 1
Mx1 1
Mx2  B
Mx2 1
Mx3  B
Mx3 1
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1  M 2  x12
 1  M 2  x22
1  M 2  x32
WisMATYC
Mx1 1


x12
1 x1
Mx2 1   M 1  M 2 1 x2
  x2  x1   x3  x1  x3  x2   0
x32
1 x3
Mx3 1

x22   M 1  M 2
September 25, 2010
The coefficient on y can be expressed as follows.
1  M 2  x12  2MBx1  B2
1  M 2  x22  2MBx2  B2
1  M 2  x32  2MBx3  B2
1  M 2  x12  2MBx1
1  1  M 2  x22  2MBx2
1 1  M 2  x32  2MBx3
x1 1
x1 1
x2
x2 1   1  M 2
x3

  x2  x1   x3  x1  x3  x2 
x3 1
Finally, the last determinant takes the form given below.
1  M 2  x12  2MBx1  B2
1  M 2  x22  2MBx2  B2
1  M 2  x32  2MBx3  B2
So equation (1) becomes
x1
x2
x3
1  M 2  x12  2MBx1  B2
Mx2  B  1  M 2  x22  2MBx2  B 2
Mx3  B 1  M 2  x32  2MBx3  B 2
Mx1  B
x2  y 2
x
y
x12
x22
x32
x1
y1 1
x2
y2 1
x3
y3 1



y12
y22
y32
x1
B
x2
B  B 1  M 2
x3
B

  x2  x1   x3  x1  x3  x2 
1

 1 M 2
  x2  x1   x3  x1  x3  x2  xM  y  B  0 . This is equivalent
to y  Mx  B . Thus even when the points are collinear, but distinct, the determinant generates the equation that describes
the correct geometric relation between the points.
This entire analysis can be generalized to n dimensions where n + 1 points not all in the same n-dimensional hyper-plane lie
on the surface of an n-dimensional hyper-sphere. Let the n + 1 points be designated as  x11, x12 , x13 , , x1n  ,
 x21, x22 , x23 ,


, x2n  ,… and xn1,1, xn1,2 , xn1,3 ,
expression using an n + 2 by n + 2 determinant.
x12  x22  x32 
, xn1,n . The equation for the hyper-sphere is given by the following
xn2
x1
x2
xn
1
2
2
2
x11
 x12
 x13

x12n
x11
x12
x1n
1
2
2
2
x21
 x22
 x23

x22n
x21
x22
x2n
1
xn21  xn22  xn23 
2
xnn
xn1
xn 2
xnn
1
xn1,1
xn1,2
xn21,1  xn21,2
 xn21,3

xn21,n
0.
xn1,n 1
As an example consider the following four points in 3D:  2,6,11 ,  4,5, 3 ,  3, 2, 8  and  7, 4,1 . The following
determinant generates the equation of the sphere that contains these points.
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x2  y 2  z 2
161
50
186
77
x
y
2 6
4 5
7 4
3 2
September 25, 2010
z
11
3
11
8
1
1
1 0
1
1
Wolfram Alpha provides a ready and easy way to simplify this determinant.
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 7005 9497 7801 
Thus the sphere’s center is at  
,
,
 and the radius is R 
 2554 2554 2554 
Winplot can generate the sphere.
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September 25, 2010
1031486499
. From this information
 51081277 
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September 25, 2010
2. Rayleigh-Ritz Variational Calculations of Extreme Eigenvalues of 2 by 2 Matrices
Suppose A is an n x n real symmetric matrix. In many applications such as bound states in Quantum Mechanics and
economic optimizations the extreme eigenvalues max and min are of particular interest. Consider an n x 1 vector  called
a “trial function”. The eigenvectors of A are linearly independent and an orthonormal basis for R n consisting of
 
eigenvectors of A is always possible. Designate this orthonormal basis by  j
and max  1  2
n
j1
n


with A j   j  j ,  j ,  k   jk ,


 n  min . The trial function can be expanded as    ,  j  j .
j1

n
n

n
j 1
k 1

j 1 k 1
 A,     A  ,  j   j ,   ,  k   k     ,  j   ,  k   A j ,  k 

n
   j   ,  j    ,  k    j ,  k     j   ,  j    ,  k   jk    j   ,  j 
n
n
n
j 1 k 1
n
n
j 1 k 1
j 1
  ,  
A,   


For   0 , the Rayleigh Quotient is defined as
n
j 1
 ,  
 min  ,  j 
n
j 1
 ,  
j
2
j
 ,  
. It can be bracketed as follows:
  ,  
A,   



n
 ,  
j 1
max
 ,  
n
2
j 1
j
j 1
 ,  
2
j
  ,    A,  
  ,  


n
min
2
2
2
j
 ,  
 ,  
 A,    
min 
 ,   max
max
Any choice of   0 gives a lower bound to max and an upper bound to min . By allowing  to vary over all non zero
vectors of R n we have the following characterizations.
 A,  
R , 0   ,  
 A,  
max  max
R , 0   ,  
min 
min
n
n
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If  varies over just a subspace W of R n then min 
 A,  
or if  2 W
W , 0   ,  
max  max
 A,  
W ,  0   ,  
max
and
is to 1 and  2 .Let  
then min 
 A,  
W ,  0   ,  
min

 ,  
, then
September 25, 2010
 A,     max  A,  
and max
. If 1 W
W ,  0   ,  
W , 0   ,  
min
then
 A,  
. How well the variational calculations
W ,  0   ,  
min
approximate the extreme eigenvalues depends on “how close” the subspace W
 A,    A, 

 with  ,    1 . So in practice variational calculations are
 ,  
done using normalized trial functions, i.e.  ,    1. This is equivalent to varying the trial function over a subspace of
the unit n-sphere.
a c 
 . The eigenvalues are solutions of the characteristic
 c b
Consider the 2 x 2 real symmetric matrix A  
polynomial  2    a  b   ab  c2  0 .
max  1 
min  2 
ab
a  b
2
 4c 2
2
ab
a  b
2
 4c 2
2
ab
 a b 
2
 
 c
2
 2 
2

ab
 a b 
2

 
 c
2
2


2
cos   
 A,   and
 with 0    2 spans the unit circle (the unit 2-sphere). Thus, 1  0max
 2
 sin   
The trial function   
2  min  A,   . This can be verified as follows.
0 2
 a cos    c sin   cos   
2
2
f     A,    
,
  a cos    2c cos   sin    b sin  
 c cos    b sin   sin   
Using the trigonometric identities
cos 2   
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1  cos  2 
2
; sin 2   
1  cos  2 
2
; sin   cos   
sin  2 
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a  b a  b

cos  2   c sin  2  ; cos   
2
2
f   
September 25, 2010
a b
 a b 
2
2 
 c
2


2
c
; sin   
 a b 
2

 c
2


2
ab
ab
 a b 
 a b 
2
2
 
 
  c cos   cos  2   sin   sin  2   
  c cos  2   
2
2
 2 
 2 
2

2
 
 ab
 a b 
2
 
So the maximum of f   at   is
  c  1 and the minimum of f   at    is
2 4
2
2
 2 
2
ab
 a b 
2
 
  c  2 . Let 1 and  2 be normalized eigenvectors of A with A1  11 , A 2  2 2 , and
2
2


2
 1, 1    2 , 2   1 . Since A is a symmetric matrix,  1, 2   0 and the expanded inner product  A,   is
 A C 
1
1
 C2  2  , C11  C2  2   C12 1  1 , 1   C22 2   2 ,  2   2C1C2  1  2  1 ,  2   C121  C222
This is subject to the constraint that  ,    C12  1 , 1   C22  2 , 2   2C1C2  1 , 2   C12  C22  1 . The inner
product thus determines a paraboloid in the 2D space of  C1 , C2  and the maximum (minimum) of the Rayleigh Quotient is
the maximum value on the surface subject to the circular constraint C12  C22  1 . When the two eigenvalues are of the
same sign the surface is an elliptic paraboloid, while when the eigenvalues differ in sign it is a hyperbolic paraboloid. This
is illustrated in the following set of figures.
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WisMATYC
September 25, 2010
c  di 
 a
. The extremes of Rayleigh Quotient need to be computed as an
b 
c  di
For a 2 x 2 Hermitian matrix A  
Hermitian product over C 2 , so that the maximum of f   is a lower bound to max and the minimum of f   is an upper
bound to min . This is seen explicitly by solving the characteristic polynomial  2    a  b   ab  c2  d 2  0 .
max  1 
min  2 
a  b
ab
2




 4 c2  d 2
2
 a  b
ab
2
 4 c2  d 2
2

ab
 a b 
2
2
 
 c d
2
 2 

ab
 a b 
2
2
 
 c d
2
 2 
2
2
ab
 a b 
2
 
The maximum of f   ,
  c  1 with equality if and only if d = 0. Similarly, the minimum of f   ,
2
 2 
2
ab
 a b 
2
 
  c  2 with equality if and only if d = 0.
2
 2 
2
a d 
2
 the eigenvalues solve     a  b   ab  dc  0 , so that .
c
b


For a general real 2 x 2 matrix A  
max  1 
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ab
a  b
2
2
 4dc
ab
 a b 

 
  dc
2
 2 
2
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min  2 
ab
a  b
2
 4dc
2
cos   
 on the unit circle and cos   
sin





September 25, 2010
ab
 a b 

 
  dc
2
 2 
2
a b
For   
 a  b  c  d 
2
2
cd
, sin   
 a  b  c  d 
2
2
,
 a cos    d sin   cos   
2
2
f     A,    
,
  a cos     c  d  cos   sin    b sin  
 c cos    b sin   sin   
a
b
cd 
 1  cos  2    1  cos  2    
 sin  2 
2
2
 2 
ab
ab
 a b   c  d 
 a b   c  d 
 
 
 
 cos  2  cos    sin  2  sin    
 
 cos  2   
2
2
 2   2 
 2   2 
2

2
2
2
ab
 a b   c  d 
Hence, the maximum of the Rayleigh Quotient over the unit circle is
 
 
 while the minimum is
2
 2   2 
2
2
ab
 a b 
 a b   c  d 
 
  dc  0 the eigenvalues of A are complex so that the real extremes of
 
 . If 
2
 2 
 2   2 
2
2
2
 a b 
f   , are not very relevant. So assume that the eigenvalues of A are real, i.e. 
  dc  0 .
 2 
2
 c2  d 2 
ab
ab
 a b   c  d 
 a b 

 




cd


  1 with equality if and only if A is a
 



2
2
 2   2 
 2 
 4 
2
f max
2
2
 c2  d 2 
ab
a b
 a b   c  d 
 a b 
 




cd


  2 . Thus, when
 



2
2
 2   2 
 2 
 4 
2
diagonal matrix. Similarly, f min 
2
2
A has real eigenvalues the interval 2 , 1    f min , f max  . To look at this same problem another way consider that if
2  1 , then the real eigenvectors of A, assumed normalized, form a basis for R 2 . It might be interesting to see what form
the extremes of the Rayleigh Quotient have in terms of these eigenvectors. One can express   C11  C2 2 , but now
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Page 13
WisMATYC
September 25, 2010
 1 , 2   0. The Rayleigh Quotient can then be written as
 A,     1C11  2C22 , C11  C22   1C12  2C22  C1C2  1  2 1 , 2 
subject to the normalization constraint that  ,    C12  C22  2C1C2  1 , 2   1. From the Cauchy–Schwarz inequality
 1, 2    1, 1  2 , 2   1 . Thus the descriminant of the conic section described by the normalization equation
is 4  1 ,  2   4  0 and the normalization means that C1 and C2 are constrained to lie on an ellipse.
2

  
Making the orthogonal transformation to new coordinates  1 and  2 ,  1   
 2  



C
  
transformation  1   
 C2  


1
2
1
2
1 
2   C1 

and using the inverse
1  C2 
2 
1 
2  1 
   , the equation of the ellipse becomes  12 1   1 ,  2     22 1   1 ,  2    1 .
1   2 
2 
The Rayleigh Quotient becomes  A,   
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1
2
1
2
1
2

2
1

   2 1 2 
2
2
2
2

2
1

   2 1 2 
2
2
 12   22
2
Madison Area Technical College
 1  2  1 ,  2  .
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WisMATYC
September 25, 2010
From the constraint equation one can obtain an expression for  2 in terms of  1 .
2  
1   12 1   1 ,  2 
1   1 ,  2 
So that
 A,   
1  2

2
1

1   1 ,  2     22 1   1 ,  2     1  2   1 2
2
 
 1 2   1  2   1 2
2

1  2
2
1   12 1   1 ,  2  
  1  2   1
1   1 ,  2  
Setting the derivative of this expression with respect to  1 equal to zero yields
 1  2  
1   1 ,  2  


0
1   1   1 ,  2  
2
1   1 1   1 ,  2  

 12 1   1 ,  2 
2
1
From which it follows that extreme values of  A,   require that 1   12 1   1 ,  2     12 1   1 ,  2   or
1  
1
2 1   1 ,  2  
, 2  
1
2 1   1 ,  2  
with the minimum value of  A,   occurring at the two positions
on the ellipse where  1 and  2 are of opposite sign and the maximum value of  A,   occurring at the two positions on the
ellipse where  1 and  2 are of the same sign. The maximum value of  A,   is given by
 A,  max 
1  2
  1  2   1 2
2
 
1
 1 2   1  2 
2
2 1   1 ,  2   2 1   1 ,  2 
 
1  2
   
 1 2
 1 2  1 2  1
2
2
2
2
2 1  , 

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1
2

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Page 15
The corresponding minimum is  A ,  max 
WisMATYC
1  2
2

September 25, 2010
1  2
2 1   1 ,  2 
2

1  2
2

1  2
2
 2 . The constrained
optimization is illustrated in the following 3 D plots.
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