<a name="docA001">
2009-06-08-19-10 start
This file http://freeman2.com/tute0008.htm
is Freeman's reading notes. Although Freeman
always keep correct view point. But Freeman's
capability is limited, plus no one proofread
this file. Then you can still find wrong view
point. When read, please put question mark as
often as possible. If you suspect any view 
point wrong, please ask a math expert near by.

<a name="docA002">,<a name="textbook">
This file is a note for reading inequality
book written by Professor J. Michael Steele
The Cauchy-Schwarz Master Class  　★★★★★
Below use 'textbook' as abbreviation.
Freeman also read web pages online, and will 
indicate the source URL at discussion point.

<a name="docA003">
Freeman study mechanical engineering. 
Engineering mathematics do not teach 
inequality. Above book is first inequality
book. First time read, it was very hard.
Although high school time learned
  a*a + b*b >= 2*a*b
But this little knowledge do not help.

<a name="docA004">
This file follow textbook chapter section
order, but not continuous, Freeman skip 
those uncertain sections/problems.

This file first function is to learn 
inequality. Second function is to learn
how to use html code to write math 
equations.

<a name="docA005">　Index begin　Index this file
On 2009-01-27-10-08 Freeman accessed
the next page
http://www.sftw.umac.mo/~fstitl/10mmo/inequality.html
save as sftw.umac.mo_text_math_eqn_good.htm
Above page is the main reference for html
math equation.

In order to let reader to build  html math
equation, previous file tute0007.htm page 
end has math symbol and internal code.

This file third function is to display how
to draw curves in web page. The main engine
is XYGraph v2.3 - Technical Figures

Thank you for read Freeman's inequality page.
2009-06-08-19-47 stop

<a name="docA006">
2009-08-23-15-00 start
□　Exercise 1.14 solution

Exercise 1.14 problem (a) is solved by hint
Exercise 1.14 problem (b) is out of LiuHH's
reach. What is "cardinality of a set B⊂Z3" ?
Without clear understand the definition, 
LiuHH is unable to solve problem (b)
LiuHH major in mechanical engineering.
LiuHH is mathematics admirer and outsider.

<a name="docA007">
To solve problems in
The Cauchy-Schwarz Master Class
LiuHH's goal is to solve 50% of the exercise
problems. In reality, solve 40% is good enough.
Please do not be surprise that LiuHH skip some
text section, skip some problems.
2009-08-23-15-12 stop

<a name="ch04d001">　Index begin　Index this file


2009-11-15-19-41 start 
■　Exercise 4.7 problem statement
　　textbook page 70
（Representations of Complex Inner Product）

<a name="ch04d002">
(a) if 〈.,.〉 is a complex inner product
and if α∈C and αN=1 but α2≠1, then show 
that one has the representation

where as usual ∥w∥=〈w,w〉1/2 .

<a name="ch04d003">
(b) Similarly show that for any complex 
number inner product one has

<a name="ch04d004">
One benefit of identities such as these
is that they may help us convert facts
for ∥.∥ into facts for 〈.,.〉 or vice-
versa. One can say that these are "just"
variants of polarization identity, but
there are times when they are just the
variant one needs.
2009-11-15-20-06 here




<a name="ch04d005">　Index begin　Index this file
2009-11-15-20-12
■　Exercise 4.7 hint
　　textbook page 241

To prove the identity (4.26), expand the
inner product squares and use
 1+α+...+αN-1=(1-αN)/(1-α)=0 ---eqn.AM001

For the second identity, just expand and
integrate. This exercise is based on
D'Angelo (2002, pp.53-55) where one 
finds related material.
2009-11-15-20-17 stop






<a name="ch04d006">
2009-11-15-20-19
■　Exercise 4.7 solution


Expand eqn.4.26 as following

<a name="ch04d009">
2009-11-15-20-48 here
if x≠y, then 〈x,y〉 is a complex 
number. In eqn.4.26 right side must 
be the same complex number.
eqn.4.26 right side 〈x+αny , x+αny〉 is
a real number. αn is a complex number.
x and y at right side contribute to 
real number only. 
When x≠y, how can right side αn carry
left side 〈x,y〉 signal?! There must be
a good reason. LiuHH skip Exercise 4.7 
before find out this reason.
2009-11-15-21-02 stop



<a name="ch04d010">　Index begin　Index this file
2009-11-16-09-45 start
■　Exercise 4.8 problem statement
　　textbook page 70
（A Concrete Model of an Abstract Space）

If x1,x2,...,xn are linearly independent
elements of the (real or complex) inner
product space (V,〈.,.〉) , we define a
new sequence e1,e2,...,en by setting
  e1=x1/∥x1∥ ---eqn.AM002
and by applying the two part recursion

<a name="ch04d012">
2009-11-16-10-17 here
for k=2,3,...,n. This algorithm is known
as the Gram-Schmidt process, and it
provides a systematic tool for reducing
questions in an inner product space to
questions for real or complex numbers.
In this exercise we develop the most 
basic properties of this process, and
in the next four exercises we show how
these properties are used in practice.

<a name="ch04d013">
(a) Show that {ek, 1≦k≦n} is an
    orthonormal sequence in the sense 
    that for all 1≦j,k≦n one has

<a name="ch04d014">
2009-11-16-10-33 here
(b) Show that {xk, 1≦k≦n} and
    {ek, 1≦k≦n} satisfy the triangular
    system of linear relations
    x1=〈x1,e1〉e1 ---eqn.AM004
    x2=〈x2,e1〉e1+〈x2,e2〉e2 ---eqn.AM005
    .....
    xn=〈xn,e1〉e1+〈xn,e2〉e2+...+〈xn,en〉en ---eqn.AM006
2009-11-16-10-41 stop







<a name="ch04d015">　Index begin　Index this file
2009-11-16-12-55 start
■　Exercise 4.8 hint
　　textbook page 241
The first part of the recursion (4.28)
gives us //zk is defined in (4.28)
   〈zk,ej〉=0 for all 1≦j＜k ---eqn.AM007
and gives us //1≦j＜k done, k≦j≦n to go
   〈ek,ej〉=0 for all 1≦j＜k ---eqn.AM008
<a name="ch04d016">
The normalization
   〈ek,ek〉=1 for all 1≦k≦n ---eqn.AM009
is immediately from the second part 
of the recursion (4.28), and the
triangular spanning relations just 
rewrite the first part of the 
resursion (4.28)
2009-11-16-13-05 stop




<a name="ch04d017">
2009-11-16-13-20 start
■　Gram-Schmidt Process computer program
In 2009-Feb Liu,Hsinhan read
The Cauchy-Schwarz Master Class and
wrote a program to convert a set of 
input vectors to a set of orthonormal 
base vectors. 
English version URL is
http://freeman2.com/gramsch2.htm local
Chinese version URL is
http://freeman2.com/gramsch1.htm local

<a name="ch04d018">
First select "Space dimension  2 to 102"
for example, select 12, then page display
12*12 boxes for 12 input vectors. Because
it is 12 dimension space, each vector has
12 components, need 12 vectors to span
the whole space. 
<a name="ch04d019">
For test purpose, you can
click "Random number", next click 
[Run GS Process] button.
For twelve dimension test, number range 
from zero to eleven, not from 1 to 12 !!

After click [Run GS Process] button
Box 1, verify input data
Box 2, Output ortho-normailzed vectors

<a name="ch04d020">
Important point is to check Box 2 second
half output. It contains 
ortho-normailzed vectors norm value
      This value must be one.
ortho-normailzed vectors scalar product
      value. This value must be zero.
If all one and all zero, output is correct.

<a name="ch04d021">
LiuHH tested gramsch2.htm, program output
should be reasonable.
You are welcome to test run.
http://freeman2.com/gramsch2.htm local

Many other programs
JavaScript program list
http://freeman2.com/jsindex2.htm local
2009-11-16-13-47 stop

<a name="readlocal">
2009-11-19-10-33 start
"local" link is effective only if you
save the related web page into your
computer in same directory as this
tute0019.htm file. AND you view page
from local tute0019.htm
2009-11-19-10-35 stop

<a name="ch04d022">　Index begin　Index this file
■　Exercise 4.8 discussion
2009-11-16-16-20 start
Gram-Schmidt process use two equations
they are eqn.4.28
Reproduce in simple format below
zk＝xk－∑[j=1,k-1]{〈xk,ej〉ej} ---eqn.AM010
and
ek＝zk/∥zk∥ ---eqn.AM011
k=2,3,...,n; k-1 is not ∑ final bound,
n-1 is final.

<a name="ch04d023">
Let x1,x2,...,xn be input sequence
Let z1,z2,...,zn be working sequence
Let e1,e2,...,en be output sequence

input x-sequence is not length-one
and is not orthogonal to each other

<a name="ch04d024">
working z-sequence is not length-one
but is orthogonal to already done
base e-vectors in progress.
(vector and sequence are same thing)

output e-sequence is length-one
and is orthogonal to each other.
Satisfy our requirement.

<a name="ch04d025">
Suppose we have five dimension problem,
we have five sequences in hand and each
sequence has five elements. Then n=5. 
Suppose we work to k=3, eqn.AM010 say:
to build z3 use the next equation
z3＝x3－∑[j=1,3-1]{〈x3,ej〉ej} ---eqn.AM012
Summation is compact, expanded equation
may be better understood
z3＝x3－{〈x3,e1〉e1 + 〈x3,e2〉e2}
or
z3＝x3－〈x3,e1〉e1 - 〈x3,e2〉e2 ---eqn.AM013
<a name="ch04d026">
In eqn.AM013, z3,x3,e1,e2 are vectors.
〈x3,e1〉 and 〈x3,e2〉 are pure number
(non-vector). e1,e2 are already build
orthonormal vectors. (e1,e2 are our 
goal and done)

<a name="ch04d027">
eqn.AM013 say in x3 sequence, 
delete x3 component along e1, 
that is －〈x3,e1〉e1 part

eqn.AM013 also say in x3 sequence, 
delete x3 component along e2, 
that is －〈x3,e2〉e2 part
After delete, x3 become z3.

<a name="ch04d028">　Index begin　Index this file
then z3 has no component along e1
 and z3 has no component along e2
This fact can be verified by taking
dot product between z3 and e1
or e2 as below
  〈z3,e1〉=eqn.AM013 dot with e1

<a name="ch04d029">
[x3－〈x3,e1〉e1 - 〈x3,e2〉e2] dot e1 ---eqn.AM014
In eqn.AM014, blue terms are non-vector,
In eqn.AM014,  red terms are vector,
only vector participate dot product.
blue terms are just coefficients.
Dot result is next line
〈x3,e1〉－〈x3,e1〉〈e1,e1〉 - 〈x3,e2〉〈e2,e1〉 ---eqn.AM015
eqn.AM014 is a vector equation.
eqn.AM015 is a scalar (non-vector) equation.
<a name="ch04d030">
eqn.AM015 is a pure number expression,
not a vector any more.
eqn.AM015 red term are new created dot
result (pure number, not vector)
eqn.AM015 blue term are old created dot
result (pure number, not vector)
<a name="ch04d031">
Because e1 and e2 are orthonormal,
〈e1,e1〉 = 1 ---eqn.AM016
〈e2,e1〉 = 0 ---eqn.AM017
then eqn.AM015 become
〈x3,e1〉－〈x3,e1〉*1 - 〈x3,e2〉*0 ---eqn.AM018
<a name="ch04d032">
that is
〈z3,e1〉=〈x3,e1〉－〈x3,e1〉 = 0---eqn.AM019
eqn.AM019 tell us: vector z3 component
along e1 direction is zero, that is
z3 perpendicular to e1
similar reason for
z3 perpendicular to e2

<a name="ch04d033">　Index begin　Index this file
z3 length is not one, we carry
out eqn.AM011, change z3 to e3.
One done and shorter way to go.

When we process to higher order and work
back with already normalized bases 
e1, e2 ... same reason applies.

<a name="ch04d034">
How about initial condition e1 ?
first vector x1 no previous vector
we can not carry out eqn.AM010.
for first vector, we do only eqn.AM011
normalization.
2009-11-16-17-50 here

<a name="ch04d035">
Please goto
http://freeman2.com/gramsch2.htm#graph01
see simplest 2D figure illustration.

Click "GS 1" button
we see initial condition two vectors OA
and OB. For 2-D problem, we need only 
two vectors.

<a name="ch04d036">
Click "GS 2" button
normalize OA vector to OC. Both OA and 
OC are on same line, but |OC|=1.

Click "GS 3" button
Project OB on to OC
OB has component OD along OC.

<a name="ch04d037">
Click "GS 4" button
remove OD from OC,
that is OB + (-OD)
Since vector BE = -vector OD
then  vector OB +  vector BE
achieve our goal. Result is vector OE
Now vector OE do not have component 
along OC.
Last step, normalize OE to OF.
Red line OC and OF are our answer.
2009-11-16-18-00 here




<a name="ch04d038">
■　Exercise 4.8 solution

2009-11-16-18-01 start
LiuHH need goto kitchen to cook.
Can you help me to fill out 
Exercise 4.8 solution part?
Thank you very much.
2009-11-16-18-02 stop



<a name="ch04d039">　Index begin　Index this file

2009-11-16-19-07 start
■　Exercise 4.9 problem statement
　　textbook page 71
（Gram-Schmidt Implies Cauchy-Schwarz）

Apply the Gram-Schmidt process to the
two term sequence {x,y} and show that
<a name="ch04d040">
it reduce to the inequality
  |〈x,y〉|≦√[〈x,x〉〈y,y〉] ---eqn.AM020
to a bound that is obvious. Thus the
Gram-Schmidt process gives us an
automatic proof of the Cauchy-Schwarz
inequality.





<a name="ch04d041">　Index begin　Index this file
2009-11-16-19-16 here
■　Exercise 4.9 hint
　　textbook page 241
Without loss of generality may we 
assume that  ∥x∥=1 ---eqn.AM021
The Gram-Schmidt relations are then 
given by     x=e1 ---eqn.AM022
and
   y=μ1e1+μ2e2 ---eqn.AM023
<a name="ch04d042">
Orthogonality gives us 
  〈x,y〉=μ1 ---eqn.AM024
and
  〈y,y〉=|μ1|2+|μ2|2 ---eqn.AM025
and the bound
  |μ1|≦(|μ1|2+|μ2|2)1/2 ---eqn.AM026
is obvious.
<a name="ch04d043">
But this says
  |〈x,y〉|≦〈y,y〉1/2 ---eqn.AM027
which is Cauchy-Schwarz inequality when
 ∥x∥=1 
2009-11-16-19-29 here



<a name="ch04d044">　Index begin　Index this file
2009-11-16-19-37
■　Exercise 4.9 solution


Let e1 and e2 be two dimensional
base vectors. Let
  x=λ*e1+0*e2 and λ=1 meter ---eqn.AM028 
and  
  y=μ1e1+μ2e2 ---eqn.AM023
Here λ=1 and not change.
<a name="ch04d045">
eqn.AM028 is same as eqn.AM022
eqn.AM028 show up the value one λ
Because λ=1 carry physics dimension, 
for example meter. Unit vector e1 
carry direction, not carry physics
unit. If write x=e1, drop physics
length λ, that may cause confuse 
if one check for physics dimension 
consistency in a equation.

<a name="ch04d046">
Orthogonality gives us 
  〈x,y〉=λ*μ1 ---eqn.AM029
If x,y are observed length data, 
then λ*μ1 is length square.
and
  〈y,y〉=|μ1|2+|μ2|2 ---eqn.AM025
and the bound
  |λ*μ1+0*μ2|≦(|μ1|2+|μ2|2)1/2 * (λ2+0)1/2 ---eqn.AM030
are both length square.

<a name="ch04d047">
■　Assume ∥x∥=1 with caution
Remember λ=1 (meter). 
eqn.AM030 simplify to
  |μ1|≦(|μ1|2+|μ2|2)1/2 ---eqn.AM026
eqn.AM026 magnitude is obvious.
However we need keep in mind that
physics dimension appearance changed.
Area look like length. 
Set a parameter to be one, this is a
common practice. For example in
relativity theory, set light speed c
to one and neglect c completely, then
equation look like  time=length
Simplification is common, but do 
not forget the whole equation.

<a name="ch04d048">
eqn.AM026 says
  |〈x,y〉|≦〈y,y〉1/2〈x,x〉1/2 ---eqn.AM031
which is Cauchy-Schwarz inequality when
 ∥x∥=1 

Exercise 4.9 problem statement is correct
Exercise 4.9 hint is correct. 
When we simplify our work and set
  x=e1 ---eqn.AM022
Advanced reader has no trouble to recover
  x=λ*e1+0*e2 and λ=1 meter ---eqn.AM028 
But for freshman level reader, it is not
easy to figure out what is going on.

Although Genie is not visible, but
remember Genie is around somewhere.
2009-11-16-20-09 stop



<a name="ch04d049">　Index begin　Index this file
2009-11-16-20-19 start
■　Exercise 4.10 problem statement
　　textbook page 71
（Gram-Schmidt Implies Bessel）

If {yk: 1≦k≦n} is an orthonormal
sequence from a (real or complex) 
inner product space (V,〈.,.〉), then
Bessel's inequality assert that

<a name="ch04d051">
2009-11-16-20-32 here
Show that the Gram-Schmidt process 
yields a semi-automatic proof of 
Bessel's inequality. Incidentally,
one should also note that the case
n=1 of Bessel's inequality is
equivalent to the Cauchy-Schwarz 
inequality.
2009-11-16-20-36 here





<a name="ch04d052">
2009-11-16-20-38 here
■　Exercise 4.10 hint
　　textbook page 241
From the Gram-Schmidt process applied 
to {y1,y2,...,yn,x} one finds 
e1=y1, e2=y2, ... , en=yn ---eqn.AM032
and  en+1=z/∥z∥ ---eqn.AM033
where 
  z = x - (〈x,e1〉e1+〈x,e2〉e2
    + ... +〈x,en〉en) ---eqn.AM034
provided that z≠0 ---eqn.AM035
Taking inner products and using
orthonomality then gives us

<a name="ch04d054">
and since |〈x,en+1〉|2 gives us 
Bessel's inequality when z≠0.
When z=0 one finds that Bessel's 
inequality is in fact an identity.
2009-11-16-20-58 stop





<a name="ch04d055">　Index begin　Index this file
2009-11-17-10-09 start
■　Exercise 4.10 discussion
2008-11-03-20-15 click "purchase" button
2008-11-06-13-11 LiuHH received textbook
"The Cauchy-Schwarz Master Class"
Start from 2008-11-06-17-53 until now
(2009-11-17) read textbook many times.

<a name="ch04d056">
2009-01-27-07-56 marked at Exercise 4.10
book margin
"Equation 4.29 has question, if x=length
 if y=length, equation left side is
 length to fourth power, right side is
 length to second power, 
 length fourth power can not compare
 with length second power"

<a name="ch04d057">
LiuHH read book several times, second 
book margin comment to equation 4.29 is
"2009-11-16-18-33 yk is i,j,k and not
 have physics dimension. x has length unit"
2009-11-17-10-29 here

<a name="ch04d058">　Index begin　Index this file
■　Vector physics dimension and
　　direction carrier
Assume 3-D space distance vector
 x=[x1,x2,x3] ---eqn.AM037
Distance physics dimension is meter.

Assume 3-D space Cartesian coordinates
(ground coordinates) unit vectors are
 i,j,k 
Assume 3-D space rotated coordinates
unit vectors are 
 e=[e1,e2,e3] ---eqn.AM038

<a name="ch04d059">
We write vector equation for distance
vector x as
  x=x1i+x2j+x3k ---eqn.AM039
x is a distance vector, x has length
  ∥x∥=√[x1*x1+x2*x2+x3*x3] ---eqn.AM040
x has direction eqn.AM039

<a name="ch04d060">
We can define a rotated coordinate 
system, let unit vector point to x
direction (see eqn.AM039). Assign this
new unit vector as e1, the
expression for new unit vector e1 is
  e1=x/∥x∥ ---eqn.AM041
(eqn.AM041 is same as eqn.AM002)
<a name="ch04d060a">
eqn.AM041 tell us that
  e1=eqn.AM039/eqn.AM040
It is a long expression, e1 is a short 
symbol. All of e1,e2,e3 eventually
express with ground unit vector i,j,k.
e2's i,j,k expression is complicate.
e3's i,j,k expression is even more 
complicate. e3 come from z3, please 
see eqn.AM013 for z3.

<a name="ch04d061">
x vector eqn.AM039 has direction, 
x vector has physics dimension "meter"
  for length. Components x1, x2, x3
  carry physics dimension.

<a name="ch04d062">
Like x vector, unit vector e1 also carry
direction information. How about physics
dimension ? Please review eqn.AM041
unit vector e1=x/∥x∥ ---eqn.AM041
  numerator x has direction and
  physics dimension meter.
  denominator ∥x∥ has only physics
  dimension, ∥x∥ has no direction
  information.
<a name="ch04d063">
The result is that unit vector e1
carry only direction information,
e1 not carry physics dimension
information, because division in 
eqn.AM041 cancelled physics 
dimension factor !
more related comment

<a name="ch04d064">　Index begin　Index this file
■　Physics dimension and direction summary
unit vector [i,j,k] or [e1,e2,e3] carry
  only direction information.
Vector component [x1,x2,x3] carry only
  physics dimension information.
Distance vector  x=x1i+x2j+x3k ---eqn.AM039
  carry both direction and physics
  dimension information.
<a name="ch04d065">
Physics dimension example:
Length, meter; time, second; mass, kg;
Acceleration, meter/second/second;
Force, kilo-gram*meter/second/second;
Energy, kg*meter*meter/second/second;
Density, kilo-gram/cubic-meter  etc.
Red color are vector, black are scalar.
Summary end here.

<a name="ch04d066">
Now back to eqn.4.29
We can not compare length fourth power
with length second power. If x is length 
vector, eqn.4.29 tell us that 
  length 2nd power ≦ length 2nd power
In this case, yk must play the role of
unit vector!
2009-11-17-11-08 here

<a name="ch04d067">
Exercise 4.10 problem statement first 
word is
[[
If {yk: 1≦k≦n} is an orthonormal
sequence .....
]]
Unit vector [i,j,k] or [e1,e2,e3] has
orthonormal property. Then first word
strongly suggest that {yk: 1≦k≦n} 
is unit vector set.
2009-11-17-11-20 stop

<a name="ch04d068">
2009-11-17-15-39 start
The following use "vector" and "sequence"
equivalently.
Two vectors x, y dot product is elements
multiplication between two vectors.

Vector x and vector y can be of same
physics quantity. For example length
multiply by length get area.

<a name="ch04d069">
Vector x and vector y can be of different
physics quantity too. For example length
multiply by force get energy (work, heat).

Within one vector, all elements should
be of same physics quantity.
For example
Vector x all elements are length.
Vector y all elements are force.
<a name="ch04d070">
[[
There are exception, but must handle very 
carefully. For example quaternion has
four elements, all be pure number. First
element relate to rotation angle. Other
three elements relate to rotation axis
vector components.
]] (2009-11-19-20-42 modify)

<a name="ch04d071">
eqn.4.29 use two different dot product.
〈x,yk〉 use length x vector multiply 
with no-physics-unit element yk (or ek)
One multiplication 〈x,yk〉 get length unit
and |〈x,yk〉|2 get length square unit.

Second dot product in eqn.4.29 〈x,x〉 is
length multiply by length. get length 
square directly.
2009-11-17-16-04 here




<a name="ch04d072">　Index begin　Index this file
2009-11-17-17-28 start
■　Exercise 4.10 solution


Exercise 4.10 problem statement say
[[
{yk: 1≦k≦n} is an orthonormal
sequence from a (real or complex) 
inner product space (V,〈.,.〉)
]]
<a name="ch04d073">
this indicate that 
{yk: 1≦k≦n} is a set of n vectors
like
y1 is a vector [1, 0, 0, ... 0]
y2 is a vector [0, 1, 0, ... 0]
.....
yn is a vector [0, 0, 0, ... 1]
and y sequence can not be one vector like
y = [y1, y2, ... , yn]
For one vector, "orthonormal" is undefined.
(I am the only one, orthogonal to who?!)

<a name="ch04d074">
Second point about {yk: 1≦k≦n} 
is that like unit vector ek,
 yk carry direction information only. 
 yk do not carry physics dimension
    information.
with this understanding, we can explain
that if x sequence is observed length 
quantity, eqn.4.29 left side is length 
square and right side is length square
too.
(length2≦length2 OK
 length4≦length2 NO)

<a name="ch04d075">
Exercise 4.10 problem statement given
[[
{yk: 1≦k≦n} is an orthonormal
sequence from a (real or complex) 
inner product space (V,〈.,.〉)
]]
Gram-Schmidt process for yk is trivial!
Because {yk: 1≦k≦n} is already
orthonormal.
<a name="ch04d076">　Index begin　Index this file
■　Gram-Schmidt process key point
Gram-Schmidt process convert a set of
not orthogonal and not normalized
vectors to a new set of vectors which
are orthogonal to each other and each
vector is normalized (length=1)
Input sequences order strongly influence
answer. First vector direction is used
other vector direction changed.

<a name="ch04d077">
Exercise 4.10 hint suggest
[[
From the Gram-Schmidt process applied 
to {y1,y2,...,yn,x} one finds 
e1=y1, e2=y2, ... , en=yn ---eqn.AM032
and  en+1=z/∥z∥ ---eqn.AM033
where 
  z = x - (〈x,e1〉e1+〈x,e2〉e2
    + ... +〈x,en〉en) ---eqn.AM034
provided that z≠0 ---eqn.AM035
]]
<a name="ch04d078">
We consider n+1 vectors {y1,y2,...,yn,x}
each vector has n elements.
Gram-Schmidt process convert from
{y1,y2,...,yn,x}
to
{e1,e2,...,en,z/∥z∥}

<a name="ch04d079">
yk to ek is direct replacement
(yk already orthonormal)
eqn.AM034 is new result.
eqn.AM034 say from x vector
delete x components along e1 direction
     (that is the term "－〈x,e1〉e1")
delete x components along e2 direction
     ("－〈x,e2〉e2")
.....
delete x components along en direction
the result is called z.
This vector deletion (add negative
of component vector) is Gram-Schmidt
process.

<a name="ch04d080">　Index begin　Index this file
from eqn.AM033 we have
  z=en+1*∥z∥---eqn.AM042
from eqn.AM034 we have
  x=〈x,e1〉e1+〈x,e2〉e2
    + ... +〈x,en〉en + z ---eqn.AM043
put z in eqn.AM042 to eqn.AM043 get
  x=〈x,e1〉e1+〈x,e2〉e2 ---eqn.AM044
    + ... +〈x,en〉en + en+1*∥z∥
<a name="ch04d081">
Red terms in eqn.AM044 are vectors
Black terms in eqn.AM044 are scalar.
Use eqn.AM044 to calculate 〈x,x〉, only
red vectors participate dot product.
(eqn.AM044 is vector equation
 eqn.AM045 is scalar equation)
<a name="ch04d082">
Dot product result is
  〈x,x〉=〈x,e1〉2〈e1,e1〉
          +〈x,e2〉2〈e2,e2〉 + ... 
          +〈x,en〉2〈en,en〉
          +〈en+1,en+1〉*∥z∥2 ---eqn.AM045
In eqn.AM045, red terms are new-created
scalar. All 〈ek,ek〉 are one
because ek are all normalized.
<a name="ch04d083">
Drop one and rewrite eqn.AM045 as next
  〈x,x〉=〈x,e1〉2+〈x,e2〉2 + ... 
          +〈x,en〉2+∥z∥2 ---eqn.AM046
From eqn.AM044 we see
  〈x,en+1〉=0+0+...+〈en+1*∥z∥,en+1〉
  〈x,en+1〉=∥z∥*〈en+1,en+1〉=∥z∥*1=∥z∥ ---eqn.AM047
<a name="ch04d084">
Square eqn.AM047 get
  〈x,en+1〉2=∥z∥2 ---eqn.AM048
Put eqn.AM048 into eqn.AM046, get
  〈x,x〉=〈x,e1〉2+〈x,e2〉2 + ... 
          +〈x,en〉2+〈x,en+1〉2 ---eqn.AM049
eqn.AM049 is same as eqn.AM036
Exercise 4.10 hint reasoning continue
here. Problem solved.
2009-11-17-18-52 stop


<a name="ch04d085">　Index begin　Index this file


2009-11-17-20-10 start
■　Exercise 4.11 problem statement
　　textbook page 71
（Gram-Schmidt and Product of Linear Forms）

Use the Gram-Schmidt process for the
three term sequence {x,y,z} to show
that in a real inner product space
one has
  〈x,y〉〈x,z〉≦ ---eqn.4.30
   0.5*[〈y,z〉+∥y∥∥z∥]∥x∥2
a bound which we used earlier (page 61)
to illustrate the use of isometries and
projections.
2009-11-17-20-15 here






<a name="ch04d086">
2009-11-17-20-17
■　Exercise 4.11 hint
　　textbook page 242

Without loss of generality we can assume
that x, y and z are linearly independent
and ∥x∥=1, so the Gram-Schmidt relations 
<a name="ch04d087">
can be written as
  x=e1 ---eqn.AM050
  y=μ1e1+μ2e2 ---eqn.AM051
  z=ν1e1+ν2e2+ν3e3 ---eqn.AM052
from which we find 
  〈x,x〉=1 ---eqn.AM053
  〈x,y〉=μ1 ---eqn.AM054
  〈x,z〉=ν1 ---eqn.AM055
  〈y,z〉=μ1ν1+μ2ν2 ---eqn.AM056
<a name="ch04d088">
The bound (4.30) asserts
  μ1ν1 ≦ 0.5*[μ1ν1+μ2ν2  ---eqn.AM057
   +(μ12+μ22)1/2+(ν12+ν22+ν32)1/2]
or
  μ1ν1-μ2ν2 ≦   ---eqn.AM058
   (μ12+μ22)1/2+(ν12+ν22+ν32)1/2
which is immediately from Cauchy's
inequality.
2009-11-17-20-36 stop



<a name="ch04d089">　Index begin　Index this file
2009-11-17-20-46 start
■　Exercise 4.11 discussion
Problem given
[[
three term sequence {x,y,z} 
]]
<a name="ch04d090">
Exercise 4.11 hint suggest
[[
  x=e1 ---eqn.AM050
  y=μ1e1+μ2e2 ---eqn.AM051
  z=ν1e1+ν2e2+ν3e3 ---eqn.AM052
]]
<a name="ch04d091">
If someone ask a question:
[[
for general case consideration, 
why not start from
  x=λ1e1+λ2e2+λ3e3  ---eqn.AM050_ill
  y=μ1e1+μ2e2+μ3e3  ---eqn.AM051_ill
  z=ν1e1+ν2e2+ν3e3  ---eqn.AM052
?
]]
Can you answer this question?

<a name="ch04d092">
Gram-Schmidt process, 
step one, choose first vector 
 x=[x1,x2,x3]
normalize this vector, define the 
result vector as 
  e1=x/∥x∥ ---eqn.AM053
<a name="ch04d093">
Gram-Schmidt process, let first 
vector become
  x=e1 ---eqn.AM050
(or as x=∥x∥*e1 if length is not one)
We should not write first vector as
  x=λ1e1+λ2e2+λ3e3  ---eqn.AM050_ill
<a name="ch04d094">
because
(a) unit vector e1 is chosen along
    first vector. first vector has 
    no component along e2 and e3
(b) when work on first vector, only
    e1 is defined, but e2 and e3
    are undefined at first step.

<a name="ch04d095">
Similar reason apply to second vector
We can write second vector as
  y=μ1e1+μ2e2 ---eqn.AM051
but can not write it as eqn.AM051_ill
2009-11-17-21-09 stop



<a name="ch04d096">　Index begin　Index this file
2009-11-17-21-11 start
■　Exercise 4.11 solution


Exercise 4.11 hint apply Gram-Schmidt 
process at first step and get eqn.AM050
eqn.AM051, eqn.AM052 three relations.

<a name="ch04d097">
eqn.4.30 use 〈x,y〉, 〈x,z〉, 〈y,z〉 
and  ∥y∥, ∥z∥, ∥x∥ terms.
eqn.AM053 to eqn.AM056 find out these
term values.
eqn.4.30 is a wait-for-prove equation,
use eqn.AM050 to eqn.AM056 relations
<a name="ch04d098">
re-write wait-for-prove eqn.4.30 as
eqn.AM057 then convert to
  μ1ν1-μ2ν2 ?≦?   ---eqn.AM058
   (μ12+μ22)1/2+(ν12+ν22+ν32)1/2
"?≦?" means "wait-for-prove"

<a name="ch04d099">
By Cauchy's inequality, eqn.AM058 is
true for [μ1, -μ2, 0] and [ν1, ν2, ν3]
two sequences.
We can drop '?' from "?≦?", the 
beginning wait-for-prove eqn.4.30
is true.
2009-11-17-21-31 stop



<a name="ch04d100">　Index begin　Index this file
2009-11-18-09-46 start
■　Exercise 4.12 problem statement
　　textbook page 71
（A Gram-Schmidt Finale）

Show that if x,y,z are elements of 
a (real or complex) inner product
space V and if 
  ∥x∥=∥y∥=∥z∥=1 ---eqn.AM059
<a name="ch04d101">
then one has the inequality
  |〈x,x〉〈y,z〉-〈x,y〉〈x,z〉|2≦ ---eqn.4.31
  {〈x,x〉2-|〈x,y〉|2}{〈x,x〉2-|〈x,z〉|2}
and the inequality
  〈x,x〉2(|〈y,z〉|2+|〈y,x〉|2+|〈x,z〉|2)
  ≦ 〈x,x〉4+〈x,x〉〈z,y〉〈y,x〉〈x,z〉
   +〈x,x〉〈y,z〉〈x,y〉〈z,x〉 ---eqn.4.32
<a name="ch04d102">
At first glance, these bounds may seem
intimidating, but after one uses the
Gram-Schmidt process to strip away the
inner products, they are just like the
kind of bounds we have met many times
before.
2009-11-18-10-03 stop






<a name="ch04d103">
2009-11-18-11-20 start
■　Exercise 4.12 hint
　　textbook page 242

With the normalization and notation
used in the solution of Exercise 4.11,
the left side L of the bound (4.31)
can be written as
  |〈x,x〉〈y,z〉-〈x,y〉〈x,z〉|=
  |{(μ1ν1j+μ2ν2j)-μ1ν1j}|2
  =|μ2ν2j|2 ---eqn.AM060
(LiuHH note: if μ=1+2i, then μj=1-2i 
       somethingj=its complex conjugate)

<a name="ch04d104">
and the right side R can be written as
  {〈x,x〉2-|〈x,y〉|2}{〈x,x〉2-|〈x,z〉|2}
  =(1-|μ1|2)(1-|ν1|2)
  =|μ2|2(|ν2|2+|ν3|2) ---eqn.AM061
Since we have
  1=∥y∥=|μ1|2+|μ2|2 ---eqn.AM062
and
  1=∥z∥=|ν1|2+|ν2|2+|ν3|2 ---eqn.AM063
These formulas for L and R make it
evident that L≦R.

<a name="ch04d105">
Now, to prove the bound (4.32) it 
similarly reduces to showing
  |μ1ν1j+μ2ν2j|2+|μ1|2+|ν1|2
  ≦ 1+(μ1jν1+μ2jν2)μ1ν1j
     +(μ1ν1j+μ2ν2j)μ1jν1 ---eqn.AM064
and by expansion, this is the same as
  |μ1|2+|ν1|2+|μ1ν1|2 ---eqn.AM065
   +|μ2ν2|2+2Re{μ1ν1jμ2jν2}
  ≦ 1+2|μ1ν1|2+2Re{μ1ν1jμ2jν2}
<a name="ch04d106">
After cancelling terms, we see it
suffice for us to show
  L≡|μ1|2+|ν1|2+|μ2ν2|2
   ≦ 1+|μ1ν1|2 ---eqn.AM066
but the substitution 
  |μ2ν2|2=(1-μ12)(1-ν12-ν32) ---eqn.AM067
<a name="ch04d107">
gives us
  L=1+|μ1ν1|2+|ν3|2(|μ1|2-1)
   ≦ 1+|μ1ν1|2 ---eqn.AM068
since
   |μ1|2 ≦ 1 ---eqn.AM069
This exercise is based on Problems 
16.50 and 16.51 of Hewitt and Stomberg
(1969, p.254)
2009-11-18-12-15 stop





<a name="ch04d108">　Index begin　Index this file
2009-11-18-14-26 start
■　Exercise 4.12 solution


<a name="ch04d109">
From Exercise 4.11 hint
[[
we can assume
that x, y and z are linearly independent
and ∥x∥=1, so the Gram-Schmidt relations 
can be written as
  x=e1 ---eqn.AM050
  y=μ1e1+μ2e2 ---eqn.AM051
  z=ν1e1+ν2e2+ν3e3 ---eqn.AM052
<a name="ch04d110">
from which we find 
  〈x,x〉=1 ---eqn.AM053
  〈x,y〉=μ1 ---eqn.AM054
  〈x,z〉=ν1 ---eqn.AM055
  〈y,z〉=μ1ν1j+μ2ν2j ---eqn.AM056a
]]
<a name="ch04d111">
Above is Exercise 4.11 hint
Below is Exercise 4.12 hint
eqn.4.31 mark left side L and right side R
  L=|〈x,x〉〈y,z〉-〈x,y〉〈x,z〉|2 ---eqn.AM070
   ≦ 
  R={〈x,x〉2-|〈x,y〉|2}{〈x,x〉2-|〈x,z〉|2} ---eqn.AM071

<a name="ch04d112">
Here, be careful, 
Exercise 4.11 is 
   "in a real inner product space"
Exercise 4.12 is
   "a real or complex inner product space"
Because complex include real, but real
not include complex.
The following use complex rule apply to 
both Exercise 4.11 and Exercise 4.12
formulas, that is use complex conjugate
for an inner product.

<a name="ch04d113">　Index begin　Index this file
Do substitution and calculation as below
  L=|〈x,x〉〈y,z〉-〈x,y〉〈x,z〉|2
  L=|1*(μ1ν1j+μ2ν2j) - μ1*ν1j|2
Here 〈x,x〉=1 use eqn.AM059 or eqn.AM053
     〈y,z〉=μ1ν1j+μ2ν2j use eqn.AM056a
eqn.AM056a is complex version of eqn.AM056
     〈x,y〉=μ1 use eqn.AM054
<a name="ch04d114">
The red term
     〈x,z〉=ν1j use eqn.AM055
〈x,y〉 and 〈x,z〉 are on equal foot,
why change 〈x,z〉 to complex conjugate?
not change 〈x,y〉 to complex conjugate?
Look like we have freedom to choose the
one we want to change. LiuHH is still 
learning, just follow textbook hint,
and cancel μ1ν1j from L. get
  L=|μ2ν2j|2 ---eqn.AM072
2009-11-18-15-18 here

<a name="ch04d115">
Next goto eqn.4.31 right hand term R
  R={〈x,x〉2-|〈x,y〉|2}{〈x,x〉2-|〈x,z〉|2} ---eqn.AM071
  R={    1   -  |μ1|2  }{   1  -   |ν1|2 }  ---eqn.AM073
Above two lines use eqn.AM053 to eqn.AM055
How to solve eqn.AM073 ?
Problem given condition is
 ∥x∥=∥y∥=∥z∥=1 ---eqn.AM059
<a name="ch04d116">
Exercise 4.11 hint suggest
[[
  x=e1 ---eqn.AM050
  y=μ1e1+μ2e2 ---eqn.AM051
  z=ν1e1+ν2e2+ν3e3 ---eqn.AM052
]]
(Above three equations are application
 of Gram-Schmidt process. The heart of
 this exercise.)
<a name="ch04d117">
then 
  1=∥y∥2=|μ1|2+|μ2|2  ---eqn.AM074
and
  1=∥z∥2=|ν1|2+|ν2|2+|ν3|2  ---eqn.AM075
re-write as
  1-|μ1|2= |μ2|2  ---eqn.AM076
and
  1-|ν1|2= |ν2|2+|ν3|2  ---eqn.AM077
<a name="ch04d118">　Index begin　Index this file
Now put eqn.AM076 and eqn.AM077 into
eqn.AM073, we find
  R=|μ2|2*(|ν2|2+|ν3|2)
that is
  R=|μ2|2*|ν2|2 + |μ2|2*|ν3|2  ---eqn.AM078
compare with
  L=|μ2ν2j|2 ---eqn.AM072
that is
  L=|μ2|2*|ν2|2 ---eqn.AM072a
<a name="ch04d119">
We find
  R=L + |μ2|2*|ν3|2  ---eqn.AM079
R-L is a square of real number,
R-L ≧ 0 is ensured, that is R ≧ L or
L ≦ R for eqn.4.31
2009-11-18-15-47 here

<a name="ch04d120">
Above solved eqn.4.31
Below solve  eqn.4.32

From eqn.4.32, define L and R as following
  L=〈x,x〉2(|〈y,z〉|2+|〈y,x〉|2
           +|〈x,z〉|2) ---eqn.AM080
  ≦ // from eqn.4.32
  R=〈x,x〉4 ---eqn.AM081
     +〈x,x〉〈z,y〉〈y,x〉〈x,z〉
     +〈x,x〉〈y,z〉〈x,y〉〈z,x〉

<a name="ch04d121">
Apply eqn.AM050 to eqn.AM056 to L and R
that is 
"Gram-Schmidt process to strip away the
inner products"

<a name="ch04d122">
Start from L
  L=〈x,x〉2(|〈y,z〉|2+|〈y,x〉|2
           +|〈x,z〉|2) ---eqn.AM080
  L= 12(|μ1ν1j+μ2ν2j|2+|μ1|2+|ν1|2)
Drop one, get
  L= |μ1ν1j+μ2ν2j|2+|μ1|2+|ν1|2 ---eqn.AM082
<a name="ch04d123">　Index begin　Index this file
Expand square term get
  L= |μ1ν1j|2+|μ2ν2j|2+2Re{μ1ν1j*μ2jν2}
    +|μ1|2+|ν1|2 ---eqn.AM083
2009-11-18-16-18 here

<a name="ch04d124">
square term |μ1ν1j+μ2ν2j|2
why choose 2Re{μ1ν1j*μ2jν2}
why  not   2Re{μ1ν1j*μ2ν2j}
LiuHH is not math major, so LiuHH
can puzzle for a while.
2009-11-18-16-29 here

<a name="ch04d125">
Complex conjugate change direction, but
not change absolute value, when consider
absolute value, drop complex conjugate
sign. L become
  L= |μ1ν1|2+|μ2ν2|2+2Re{μ1ν1j*μ2jν2}
    +|μ1|2+|ν1|2 ---eqn.AM084

<a name="ch04d126">
Next, goto eqn.4.32 right side
because
  〈y,z〉=μ1ν1j+μ2ν2j ---eqn.AM056a
then
  〈z,y〉=μ1jν1+μ2jν2 ---eqn.AM056b
<a name="ch04d127">
similarly
  〈x,y〉=μ1 ---eqn.AM054
  〈x,z〉=ν1 ---eqn.AM055
become
  〈y,x〉=μ1j ---eqn.AM054a
  〈z,x〉=ν1j ---eqn.AM055a
<a name="ch04d128">　Index begin　Index this file
eqn.4.32 rignt side R is
  R=〈x,x〉4 ---eqn.AM081
     +〈x,x〉〈z,y〉〈y,x〉〈x,z〉
     +〈x,x〉〈y,z〉〈x,y〉〈z,x〉
  R= 14 // 1 come from eqn.AM053
     +1*(μ1jν1+μ2jν2)*μ1j*ν1
     +1*(μ1ν1j+μ2ν2j)*μ1*ν1j ---eqn.AM085
<a name="ch04d129">
But textbook page 242 line -8 use
  R= 14
     +1*(μ1jν1+μ2jν2)*μ1*ν1j
     +1*(μ1ν1j+μ2ν2j)*μ1j*ν1 ---eqn.AM086
LiuHH do not know why.
Suggest reader solve problem first,
Your solution could be better and correct.

<a name="ch04d130">
Follow textbook page 242 line -8
next step is expand, 
μ multiply μ_conjugate
ν multiply ν_conjugate
happy canceling, get
  R=1+2|μ1ν1|2+2Re{μ1ν1j*μ2jν2} ---eqn.AM087
<a name="ch04d131">
Now combine L=eqn.AM084 and R=eqn.AM087
get
  L= |μ1ν1|2+|μ2ν2|2+2Re{μ1ν1j*μ2jν2}
    +|μ1|2+|ν1|2 ---eqn.AM084
 ?≦?
  R=1+2|μ1ν1|2+2Re{μ1ν1j*μ2jν2} ---eqn.AM087
<a name="ch04d132">
cancel again, get
  L= |μ2ν2|2+|μ1|2+|ν1|2
 ?≦? //  ---eqn.AM088
  R=1+|μ1ν1|2
In eqn.AM088 continue next term
  |μ2ν2|2=|μ2|2*|ν2|2 ---eqn.AM089
from eqn.AM074
  |μ2|2= 1-|μ1|2 ---eqn.AM090
and
  |ν2|2= 1-|ν1|2-|ν3|2 ---eqn.AM091
<a name="ch04d133">　Index begin　Index this file
Put eqn.AM090 and eqn.AM091 into eqn.AM088
eqn.AM088 become
  L= (1-|μ1|2)*(1-|ν1|2-|ν3|2)
     +|μ1|2+|ν1|2
 ?≦? //  ---eqn.AM092
  R=1+|μ1ν1|2
<a name="ch04d134">
Expand eqn.AM092 get
  L= 1-|ν1|2-|ν3|2
      -|μ1|2*1+|μ1|2*|ν1|2+|μ1|2*|ν3|2
     +|μ1|2+|ν1|2
 ?≦? //  ---eqn.AM093
  R=1+|μ1ν1|2
<a name="ch04d135">
cancel 1, cancel |μ1|2, cancel |ν1|2
left
  L=|μ1|2*|ν1|2+|μ1|2*|ν3|2-|ν3|2
 ?≦? //  ---eqn.AM094
  R=|μ1|2*|ν1|2
<a name="ch04d136">
final focus is
  +|μ1|2*|ν3|2-|ν3|2  ---eqn.AM095
rewrite as
  +(|μ1|2-1)|ν3|2  ---eqn.AM096
but
  1=∥y∥=|μ1|2+|μ2|2 ---eqn.AM062
tell us eqn.AM096 can be written as
  +(|μ1|2-|μ1|2-|μ2|2)|ν3|2  ---eqn.AM097
|μ1|2 cancel out.
<a name="ch04d137">
eqn.AM094 is in fact
  L=|μ1|2*|ν1|2-|ν3|2*|μ2|2
 ?≦? //  ---eqn.AM098
  R=|μ1|2*|ν1|2
the term -|ν3|2*|μ2|2 is non-positive.
eqn.AM098 is true, then the starting
equation eqn.4.32 is true.
2009-11-18-17-34 stop





<a name="ch04d138">　Index begin　Index this file
2009-11-18-18-23 start
■　Exercise 4.13 problem statement
　　textbook page 72
（Equivalence of Isometry and Orthonormality）

This exercise shows how an important
algebraic identity can be proved with
help from the condition for equality 
in the Cauchy-Schwarz bound. 
<a name="ch04d139">
The task
is to show that if the nxn matrix A 
preserve the Euclidean length of each
v in Rn then its columns are orthonormal.
In the useful shorthand of matrix 
algrbra, one needs to show
  ∥Av∥=∥v∥ for all v∈Rd ↔ ATA=I ---eqn.AM099
where I is the identity matrix, AT 
is the transpose of A, and ∥v∥ is 
the Euclidean length of v.

<a name="ch04d140">
As a hint, one might first show that
  ∥Av∥≦∥v∥ ---eqn.AM100
that is , one might show that the 
transpose AT does not increase 
length. One can then argue that 
if Cauchy-Schwarz inequality is 
applied to the inner product
  〈v,ATAv〉 ---eqn.AM101
then equality actually holds.
2009-11-18-18-36 stop







<a name="ch04d141">　Index begin　Index this file
2009-11-18-18-39 start
■　Exercise 4.13 hint
　　textbook page 243

Following the hint, we first note
  ∥ATv∥2=〈ATv,ATv〉
  = 〈v,AATv〉 ≦ // ---eqn.AM102
  ∥v∥*∥AATv∥ = ∥v∥*∥ATv∥
so by division
  ∥ATv∥≦∥v∥ ---eqn.AM103
<a name="ch04d142">
Next by Cauchy-Schwarz inequality
and the property of A and AT we have
the chain
  ∥v∥2=〈Av,Av〉=〈v,ATAv〉
  ≦∥v∥∥ATAv∥≦∥v∥∥Av∥
  =∥v∥2 ---eqn.AM104
<a name="ch04d143">
so we actually have equality where 
the first inequality is written.
This tell us that there is a λ 
(which possibly depend on v) for
which we have 
  λv=ATAv ---eqn.AM105
<a name="ch04d144">
This relation in turn gives us
  λ〈v,v〉=〈v,ATAv〉
        =〈Av,Av〉=〈v,v〉 ---eqn.AM106
so in fact λ=1 (and hence it does
not actually depend on v). We 
<a name="ch04d145">
therefore find that 
  v=ATAv ---eqn.AM107
for all v, so
  ATA=I ---eqn.AM108
as claimed. This argument follows 
Sigillito (1968)
2009-11-18-18-56 stop


<a name="ch04d146">　Index begin　Index this file
2009-11-18-20-49 start
■　Exercise 4.13 discussion

 〈v,AATv〉 ≦ ∥v∥*∥AATv∥
left side is two vector dot product,
right side is two vector length product,
This is typical application of Cauchy
inequality.

<a name="ch04d147">
Next
  ∥v∥*∥AATv∥ = ∥v∥*∥ATv∥
this is application of
  Au = u ---eqn.AM109
here u=ATv

Conclusion is
  ∥ATv∥≦∥v∥ ---eqn.AM103

<a name="ch04d148">
In eqn.AM104
[[
  ∥v∥2=〈Av,Av〉=〈v,ATAv〉
  ≦∥v∥∥ATAv∥≦∥v∥∥Av∥
  =∥v∥2 ---eqn.AM104
]]

<a name="ch04d149">
  〈v,ATAv〉≦∥v∥∥ATAv∥
is application of Cauchy inequality.

  ∥v∥∥ATAv∥≦∥v∥∥Av∥
is application of eqn.AM103
in eqn.AM103 write v as u
  ∥ATu∥≦∥u∥ ---eqn.AM103b
and u=Av

<a name="ch04d150">　Index begin　Index this file
  ∥v∥∥Av∥=∥v∥2
is application of Av=v

eqn.AM105 and eqn.AM106 are
same application as above.
2009-11-18-21-09 here


<a name="ch04d151">
2009-11-18-21-10 here
■　Exercise 4.13 solution


Exercise 4.13 hint already solved
this problem.
2009-11-18-21-10 stop




<a name="ch04d152">
2009-11-18-22-37 start
continue from 
unit vector e1 carry only direction
information. ...
<a name="ch04d153">
unit vector e1 physics dimension 
cancellation has nothing to do with 
the magnitude of ∥x∥, it is the 
physics dimension ∥x∥ possess in
denominator cancel numerator physics 
dimension.
<a name="ch04d154">
In other words, instead of
  e1=x/∥x∥ ---eqn.AM041
if we write
  e0=x_meter_vector/1_meter ---eqn.AM041b
the result is a pure vector e0 carry
direction information only and its 
length is not one.
Because unit length vector is so
convenient, everybody like to use
  e1=x/∥x∥ ---eqn.AM041
2009-11-18-22-52 stop

2009-11-19-11-33 done proofread
2009-11-19-15-51 done spelling check



========= Chapter four end here =========




<a name="ch05a001">　Index begin　Index this file
2009-11-20-14-18
■■Chapter 05: Consequences of Order

■　Cauchy Converse defined
Determine the circumstances which 
suffice for nonnegative real numbers 
ak, bk k=1,2,...,n 
to satisfy an inequality of the type

<a name="ch05a003">
Let us review Cauchy's inequality

<a name="ch05a005">
2009-11-20-14-31 here
then eqn.AM202 is possibly true.
eqn.AM201 is Cauchy Inequality
eqn.AM202 is Cauchy Converse
eqn.AM201 is a theorem
eqn.AM202 is a conjecture at present 
time. 
"?≧?" indicate conjecture
"?≧?" indicate uncertain
conjecture=guessed, not proved.

<a name="ch05a006">　Index begin　Index this file
■　Cauchy Converse not allow negative
Cauchy Inequality use two sequences
[a1,a2,...,an] and [b1,b2,...,bn]
require both a-seq. and b-seq. all 
elements be real, less than zero is
OK. This apply to eqn.AM201.
<a name="ch05a007">
But negative elements cause trouble
in eqn.AM202, if eqn.AM202 left side
one is negative and other one is 
positive, then ajbj is negative, it 
can not be greater than the squared
positive side. For this reason, the 
Cauchy Converse require both a-seq. 
and b-seq be non-negative (positive
or zero).

<a name="ch05a008">
From Cauchy Inequality
 to  Cauchy Converse
from both a-seq. and b-seq. all elements
     be real
 to  both a-seq. and b-seq. all elements
     be non-negative
Whether one change is enough to ensure
that Cauchy Converse is a well-defined
problem?
We are not certain at this moment.
2009-11-20-14-57 atop

<a name="ch05a009">　Index begin　Index this file
■　Cauchy Converse trial example
2009-11-20-16-28 start
If two sequence, each has just one 
element we get Cauchy identity.
Two one-dimension vectors must 
collinear, for just one direction
available, when two sequence are
collinear, Cauchy inequality become
Cauchy identity. 

<a name="ch05a010">
Now try two sequence, each has two
elements, let
a-sequence=[1,a] and a>0
b-sequence=[1,b] and b>0

Simplified Cauchy inequality is
  √[(1+a*a)(1+b*b)]≧ (1+ab) ---eqn.AM203

Simplified Cauchy Converse is
  √[(1+a*a)(1+b*b)]?≦?ρ(1+ab) ---eqn.AM204

<a name="ch05a011">
Because given a>0 and b>0, remove
square root, square two equations
preserve inequality sign. We find

squared Cauchy inequality is
  (1+a*a)(1+b*b) ≧ (1+ab)2 ---eqn.AM205

<a name="ch05a012">
squared Cauchy Converse is
  (1+a*a)(1+b*b)?≦?ρ2(1+ab)2 ---eqn.AM206

Expand eqn.AM206, find
  1+b*b+a*a+a*a*b*b?≦?ρ2[1+2a*b+a*a*b*b] ---eqn.AM207
since given a>0 and b>0, 
rewrite eqn.AM207 as

<a name="ch05a014">
2009-11-20-17-07 here
If f(a,b) has an asymptote
   f(a,b)=constant
then ρ can be defined.

<a name="ch05a015">
If a*b = constant k1 ---eqn.AM209
then eqn.AM208 denominator is
 1+k1+k1+k1*k1 still a constant
but eqn.AM208 numerator is
 1+b*b+a*a+k1*k1 
this is not a constant, because a 
or b can vary to infinity, then 
eqn.AM208 do not have an asymptote
   f(a,b)=constant

<a name="ch05a016">　Index begin　Index this file
■　Cauchy Converse ratio bound
Textbook instruct consider the ratio
bound

<a name="ch05a018">
2009-11-20-17-44 here
Although wondering about eqn.AM210
(k=1 only in AM210, no k=2 etc.)
LiuHH follow textbook eqn.5.2 closely
From eqn.5.2, 
 M - ak/bk ≧ 0
and
 ak/bk - m ≧ 0
we can build next inequality

<a name="ch05a021">
2009-11-20-18-13 here
whole equation multiply by bk2
  M*akbk-M*m*bk2-ak2+m*akbk ≧ 0 ---eqn.AM212
that is
  ak2+M*m*bk2 ≦ m*akbk+M*akbk ---eqn.AM213
or textbook written as
  ak2+(mM)*bk2 ≦ (m+M)*akbk ---eqn.5.4
  for k=1,2,3,...,n
Write eqn.5.4 as a summation

<a name="ch05a023">
2009-11-20-18-32 here
We need to convert an additive bound
eqn.5.5 to a multiplicative bound. We 
have done before, please see Magic 
adding to multiplying, normalization.
Convert from variables ak, bk to akn
and bkn. Super n indicate normalized.
<a name="ch05a024">
If we do the same trick as before, 
this time we get into trouble, because
eqn.5.5 is valid for
  m ≦ ak/bk ≦ M ---eqn.5.2
  where k=1,2,3,...,n
then it is valid for
  m ≦ akn/bkn ≦ M ---eqn.5.2 aux
But constraint eqn.5.2 aux is not
compatable with normalization process.

<a name="ch05a025">　Index begin　Index this file
■　Cauchy meet AM-GM
One need a smart idea.

Cauchy Inequality need two sequences
AM-GM  Inequality need one sequence
Here, Cauchy Converse use two sequences
eqn.5.5 left side is a,b two sequences
squared sum, the genius idea is that
we create a two elements one sequence

<a name="ch05a026">
first element is eqn.5.5 left most
term ∑ak2
second element is eqn.5.5 middle
term (mM)∑bk2

YES !, it is one sequence
then we apply AM-GM Inequality to this
two elements one sequence, and find

<a name="ch05a029">　Index begin　Index this file
■　Cauchy Converse answered
2009-11-20-19-20 here
Now define
  A=(m+M)/2 ---eqn.AM216 (eqn.5.6)
and
  G = √(mM) ---eqn.AM217 (eqn.5.6)
In eqn.AM215 change m and M to A and G
then for all
  ak＞0 and bk＞0 ---eqn.AM218
with
  0＜m≦ak/bk≦M＜∞ ---eqn.AM219
we have the following bound

<a name="ch05a031">
2009-11-20-19-52 here
ρ in eqn.5.1 and in eqn.AM202 is 
A/G in eqn.5.7, then 
we answered Cauchy Converse problem
<a name="ch05a032">
for all
  ak＞0 and bk＞0 ---eqn.AM218
with
  0＜m≦ak/bk≦M＜∞ ---eqn.AM219
define
  A=(m+M)/2 ---eqn.AM216 (eqn.5.6)
and
  G = √(mM) ---eqn.AM217 (eqn.5.6)
set 
  ρ=A/G ---eqn.AM220
then Cauchy Converse problem eqn.5.1
is true.
2009-11-20-19-58 stop

2009-11-20-21-50 done proofread
2009-11-21-10-39 done spelling check

<a name=20091217>
2009-12-17-11-09 start
Update 2009-12-17 change all tute*.htm
(from tute0007.htm to tute0023.htm)
first:
Correct 'Limit' link from '#docA06'
to '#docA006'
second:
Change Javascript index to read from
jslist1e.js so that update jslist1e.js
then update ALL tute*.htm.
2009-12-17-11-23 stop

<a name="Copyright">　Index begin　Index this file
2009-06-19-10-48
If you are interested in inequality,
suggest you buy the book
The Cauchy-Schwarz Master Class
written by Prof. J. Michael Steele
The Cauchy-Schwarz Master Class is
this web page's textbook.

To buy textbook, that is to show thanks
to Prof. J.M. Steele's great work.
and it is also respect copyright law.
Thank you. Freeman 2009-06-19

The Cauchy-Schwarz Master Class
J. Michael Steele ★★★★★
ISBN 978-0-521-54677-5
2009-06-19-10-56