Quadratic Forms - Computer Science Engineering Video Lecture - Computer Science Engineering (CSE)

welcome viewers today we'll be
discussing quadratic forms this lecture
includes quadratic forms and positive
definite quadratic forms the
prerequisite to this lecture is
eigenvalues and eigenvectors together
with the organization i suggest that
before going through this lecture the
user must go through the eigenvalues and
eigenvectors together with
diagonalization we start with quadratic
forms we are familiar with this
quadratic form X square plus 2b XY plus
ey square is equal to D we call it a
quadratic form because this has X square
term XY term and Val square terms in
this expression and all these terms are
quadratic so we say this form is a
quadratic form provided a b c and d are
real numbers and for these real numbers
this expression represents a pollack
section in two dimensional plane we call
it as an exception because it is
obtained by cutting holes by place these
forms are used in many applications and
that is why we are discussing them at
this in this lecture this was a
particular form the more general term
may be x squared plus 2 bx y plus ey
square plus lea terms DX plus ey plus f
equal to 0 this more general form than
this it has all the terms which are
quadratic but this may have linear terms
also now general form is very difficult
to recognize
so we consider
with forms as a standard form in which
we do not have any cross product terms
and linear terms like X Y X Y x squared
or Y squared and ay and thumbs like DX
plus ey etc so at all are an expression
which is free of these linear terms as
well as product terms is called a
standard form its Center is a region of
that dynamic order system that is why
this is preferred like we we have X
square plus y square is equal to 1 we
know it's a circle they will have any
cross term said doesn't have any linear
terms it's a standard form of a circle
having centralized origin and radius one
similarly we have X square by a square
plus y square by B Square is equal to
one represents an ellipse again this is
free of cross terms and dunya terms and
therefore may be x squared over a square
minus y square of B square is equal to
one it represents hyperbola and we may
have a form by square is equal to four X
or X square is equal to 4a y it
represents parabola this is a circle we
all are aware of this now this
represents an ellipse the here these are
equal but in case of ellipse the major
axis and minor axis these components are
different now this represents a
hyperbola the yellow lines yellow curves
these are the axis and in fact these are
the asymptotes of the hyperbola in case
of parabola this is a parabola or
parabola may be like this now if
possible terms are present then we can
we'll arrange into a standard form by
suitably rotating the axis so if
problems are present then rotation of
axis is required and we may get an
expression which is in standard form for
example X Y is equal to 1 is this firm
so we have the tenure axes and these are
and the course representing X Y is equal
to 1 but if you rotate this axis then
the curve will look like as this and
this X Y is equal to 1 will transform to
Y square minus X square is equal to 1 by
suitable transformation and here we
don't have any cross term present over
here but this curve is the same as this
only thing is you have to rotate this
axis so after rotation this is this was
the previous axis before rotation and
when you rotate it it will be the new
axis so this term and this curves they
are same only differences that this
curve is rotated so after rotation cross
comes can be eliminated and one can very
easily recognize this to be a hyperbola
similarly this the prevents an ellipse
ax square plus BX bar plus ey square is
equal to 1 for suitable values of a B
and C and then after rotation of axis
this axis then we'll have this ellipse
and the equation will be transformed to
2 X square plus y square is equal to 1
now this is the original axis and after
rotation this is the new axis so on the
new axis it is 2 X square plus y square
is equal to 1 which is standard form and
no cost terms are present over here now
what will the effect of translation
transmission is required then many
attempts are present to bring the system
will be for a given quadratic form into
standard form for example if you have
been given this ellipse and the
expression
for this will require linear terms also
but if you translate this axis that
means to ship this origin to this place
then it will be a normal form so you are
shifting this origin to the center of
the limits and this will now be a
standard form so whenever new atoms are
present one has to shift one has to
perform translation to get the normal to
the best standard form of the quadratic
now the identification upon exception
requires rotation and translation of
coordinate axis so that we can easily
recognize by its standard form so this
rotation and translation has required to
bring the given quadratic into the
standard form and from there one can
easily recognize the quadratic now these
method can be better appreciated with
matrices and their eigenvalues and
eigenvectors see what I require to
explain here is the case of two
dimensions but if you are having three
or more dimensional quadratic forms then
things will be difficult to visualize
but with the help of matrices and new
eigenvalues and eigenvectors one can
very easily identify the quadratic forms
so in general let us say we have a three
dimensional surface we call it a
quadratic surface which is X square plus
two B X y plus 2 C X with plus ey square
plus 2 FY with plus cz square is equal
to aji so this is a quadric for
quadratic form in three dimensions for
ABCDEF all the coefficients being real
now if not centered at origin the form
of equation will be more complex now
there are no linear terms present in
this so it is centered at the origin but
if these terms are also there then the
surface will not be centered at
and form will be more difficult in
general form the quadratic is written as
X transpose ax plus BX is equal to K
this is a matrix equation where a is a
square matrix and X is a column vector
now we come to the quadratic forms some
examples if we consider this quadratic
form 2 X square plus 3 XY plus y square
in two variables then we can write this
quadratic form in the matrix form as X
transpose X where X is a column vector x
and y how we get the matrix a play the
square matrix of order 2 by 2 so we
write we write this 2 and this one in
the diagonals and this quadratic cross
term is written as 3 here and 0 here and
that makes the matrix a as 2 3 0 1 and
let us see that this actually represents
this quadratic so let us multiply this
so we have 2 X and this multiplied by
this gives me 3x plus y when x XY this
gives me 2 X square plus 3 XY plus y
square so this given quadratic is
equivalent to this matrix representation
where a is this 2 by 2 matrix
but this representation is not unique
you can check that we can write down the
same quadratic in this form right we
have a as 2 2 1 and 1 what we have done
is that we have broken down that 3 the
cross term in two parts so 2 is here and
one is here and accordingly when we
multiply it is 2
X plus y and this x this is 2x plus y
multiplied by this column vector will
give the same expression 2x square plus
YX plus 2xy plus y squared equal to 2 X
square plus 3y X plus y square so if you
break down this term as 2 & 1 we'll be
getting this
you can break it as 1 & 2 again one can
check that we'll be getting the same
expression for the quadratic and this is
not unique you can break down in any way
matter phase and you can have
corresponding matrix a so this QX y can
be written in many different ways as X
transpose ax and you may have different
values however there is only one way in
which the matrix a is symmetric matrix
so if you want to write down a matrix
which is symmetric then we can have only
one possible way to represent this
quadratic as this matrix so let us see
how we do this for symmetric matrix we
divide the product term into two equal
parts that is if we have this quadratic
then three XY is divided into two parts
three XY by 2 here and 3 XY by 2 here
and that makes the matrix a as 2 and 1/2
is this square term this one is this
square term and 3 by 2 corresponding to
this and 3 by 2 corresponding to this so
this matrix is the Associated matrix a
with this given quadratic and when my
notice that this is a symmetric matrix
in the next example I am considering a 3
dimensional surface 2 X square plus 3 XY
plus y square plus 2 XZ plus y Z plus Z
square so I need Q X Y Z is equal to XYZ
and a three-dimensional 3x3 square
matrix a multiplied by the column matrix
X Y Z so this 3 by 3 matrix is obtained
as first term
to corresponding to this term then
corresponding to this I have one here
from here I will be having one here this
3xy is divided into two equal parts so 3
by 2 here and 3 by 2 here only x and y
part but 2x which will contribute here
and here since it is 2 so half of it one
is here and half of this is appearing
here and visored will appear in this
part this was it so half of Buzzard is
coming here and another half is here so
that is how I obtained this matrix a in
general if I have a quadratic expression
in n variables then this can be written
in this form solution bij X I XJ I and J
varies from 1 to M so how to get bi J's
so I write bij as B IJ plus b GI by 2
multiplied by X I X C so I am writing
this expression in this form and from
here I can get the matrix a I J so I'll
do the typical element of the matrix a
and this is equal to BI I when I is
equal to J and when I is not equal to J
that means corresponding to cross terms
I will write down this as bij plus BJ by
2 and if this is what I am doing here so
bij and B a B GI is 3 here so divided by
2 that is a IJ and if I define my matrix
a IJ in this manner then one can easily
observe that a IJ is equal to a GI
because if I is equal to J they are
equal and when I and J are not equal
when a IJ is equal to AJ I so for the
matrix a I have a IJ is equal to AJ and
that simply means that the matrix is
symmetric so with this definition I can
always represent a given quadratic in
the matrix form where the matrix is a
symmetric matter now I formally define
quadratic form we say that if a is
symmetric matrix then the real valued
function G from RM to R 1 defined as GX
is equal to X transpose X where X is a
column vector x1 x2 xn transpose it is
called a real quadratic form in the
variables X 1 X 2 X n so if you can
define such a function GX represented as
X transpose ax then this GX is a
quadratic form in variables X 1 X 2 X n
the matrix a is a matrix of quadratic
form for example if I have a
2-dimensional vector X so we have X Y as
a column vector a is a symmetric matrix
a B in the diagonals and C C in the and
non diagonal terms then X transpose ax
defines a quadratic form one can check
it easily in the second example I have a
3 dimensional column vector X and a is
this symmetric form D and D here C and C
here F and F here so it is a quadratic
form by suitably changing the variable X
is equal to py where P is and all
singular matrix we can write down the
quadratic expression px X is a column
vector is equal to py transpose that
what I am writing for X transpose a x py
one can simplify
it is y transpose P transpose so P
transpose into a and P multiplied by Y
because multiplication is associative so
you can regroup the terms and what we
have is void transform of our transpose
B Y now we can say this Q X is now
transformed to another quadratic in why
we call it Q 1 Y it is y transpose B Y
so a quadratic can be transformed to
another system and this transformation
will be X is equal to py so we have
given initially Q X and then we have
transformed it into Q 1 Y now we define
that if a and B are square matrices of
order n then B is congruent to a
provided B is equal to P transpose ap
for a given non singular matrix P so the
two matrices a and B are congruent if
they are related by this expression so B
is equal to P transpose ap for some more
singular matrix P then B and E are
congruent and we can also say that a is
congruent to B if this condition is
satisfied and maybe we can in short say
that a and B are congruent
in fact converse is more general concept
than similarity of symmetric matrices by
an orthogonal matrix P since we have
already gone through this lecture on
similar matrices we have we say that B
is equal to P inverse ap then B is
similar to a and when the P happens to
be an orthogonal matrix then P inverse
and P transpose they are same and in
that case we can say that this is
nothing but B is similar to a where P
happens to be an orthogonal matrix but
congruence is more generalized
concept then simply symmetric matrix of
submit a small general concept then
similarity of symmetric matrices by an
orthogonal matrix P hence here for this
concept we require P has to be a non
singular which is not the case in this
relationship we further prove that
Congress is an equivalence relation by
equalization I mean to say that a is
congruent to a and if a is congruent to
B then B is also congruent to a that is
symmetric relation and finally
transitive that is if a is congruent to
b and b is congruent to C then a is
congruent to C so in this these things
can be very easily proved and we
conclude that congruence is a
equivalence relation the quadratic forms
q x and q 1y are equivalent if the
matrices a and b are congruent so this
is what we derive let us stick in the
example if Q X is 2x square plus 2xy
plus 2 y square then Q X can be written
as X transpose ax where a is the
corresponding matrix it is 2 or 2 and
this term is 1 and 1 so this is the
representation for Q X let us write down
changed system UV is equal to P times
the original system X Y where the matrix
P is let us take this matrix P is 1 by
root 2 1 1 minus 1 1 how we arrive at
this matrix that I am NOT discussing at
the moment but if we use this
transformation then Q 1 comes out to be
UV P transpose ap x UV and that means if
you perform this multiplication then Q 1
comes out to be we are
these values here q1 comes out to be u
square plus 3v square the idea is I have
started with this quadratic in which
some cross terms are present but after
this transformation which I have applied
here the final form the congruent form
is free of cross product um so it is now
written as u square plus 3v square that
means given a quadratic we can transform
into a form where cross term will not be
present and we know that handling such
things are simpler as compared to this
so Q X can be transformed to a more
suitable equivalent form as u square
plus 3 v square it is free of product
term and this can be done for this is an
example only this can be done in general
and for that we'll have principal axis
theorem according to this the quadratic
form QX is equal to X transpose ax is
equivalent to this form q1 Y is equal to
Y transpose B Y which is lambda 1 by 1
square plus lambda 2 y 2 square plus
lambda n by n square where lambda 1
happens to be the eigen values of the
matrix associated with this quadratic a
so B is a diagonal matrix and these
terms are the eigen values of the matrix
a so this is called principal axis
theorem so for a symmetric matrix this
is possible so let us try to prove this
here is the proof we consider Q X is a
quadratic represented as X transpose ax
is the Associated matrix and it is a
real symmetric matrix now a can be
diagonalized by an orthogonal matrix P
such that D is equal to P inverse ap
this is a result which we
have already discussed in my earlier
lectures so if a real symmetric matrix
then it can always be diagonalized by an
orthogonal matrix P which is the matrix
of eigenvectors of this matrix a so team
tease the matrix of eigenvectors then
penis is in penis invertible subpoenas
exist in that case G is equal to penis
ap will be a diagonal matrix and since
it is this penis I can write it as P
transpose because it's an orthogonal
matrix so P inverse is equal to P
transpose so B will be P transpose ap or
we can say if we can write down D in
this form then by the definition of
congruent we can say that a and D
matrices are congruent now this is
possible because P is orthogonal that is
all I have already explained and the
diagonals of D are eigen values of a so
all this requires the earlier lectures
we're on diagonalization and with this
this QX is represented as X transpose ax
equal to X transpose P transpose ap X
and then we can use this substitution Y
is equal to px and then we can easily
check that QX will transform to q1 Y so
QX is written as P Y transpose ap 1 this
simplifies to Y transpose P transpose a
py we can re group the terms and it is
VY transpose multiplied by P transpose
ap multiplied by Y and I write down P
transpose ap as B so this expression
becomes y transpose B Y so this Q
is now transformed to q1 by where we
this is a again a quadratic form but the
associated matrix is B so a and B are
congruent matrices so the false Q and Q
X and QN X are equivalent and
application of October granulation does
not require the matrix P or I ghen
letters need not be computed only eigen
values of a are requires that's the
advantage of this form you don't have to
require the matrix P only the
eigenvalues are required this is
expected in the example so let us
consider the quadratic form for x X 1 X
2 plus X 2 X 3 plus X 1 X 3 is a
quadratic in three dimensions so
determinates equivalent quadratic form Q
1 X Y Z bitches in which the Associated
matrix is a diagonal matrix or the this
form is free of cross terms so let us
consider the matrix corresponding to
this since all the value come all these
square terms are absent so diagonal will
be 0 and it is 4 times so it is 2 here
and 2 here corresponding to X 1 X 2 and
these this 2 and 2 corresponding to X 1
X 3 and this 2 and this 2 corresponding
to 4 times X 2 X 3 so this is the
Associated matrix with this quadratic
form
now to find eigenvalues consider this
first equation it is determinant of
lambda I minus a so this is lambda minus
lambda will appear in the diagonal and
this is rest of it is a so it is a minus
lambda equal to 0 and you simplify it
the characteristic equation will come
out to be this is a simplification and
this gives me this expression which will
be further
amplified to give lambda is equal to
minus 2 and minus 2 and 1 is 1 eigen
value is 4 so 2 roots are repeated and 1
is 4 we have 3 eigen values for this
given matrix a so the equivalent
quadratic form is simply minus 2y 1
square minus 2x2 square plus 4 by 3
square we don't have to calculate and we
don't have to compute P and P transpose
and perform the multiplication to arrive
at this quadratic form only eigen values
are needed so there are two eigenvalues
minus 2 minus 2 so we have 2 terms minus
2y we'll square minus 2 y 2 square
corresponding to them and third term is
4 because the eigen value is 4 so we
have here 4 by 3 square the americaland
form may be this here again here again i
am having two eigenvalues minus 2 minus
2 and one value for but what I am doing
here is an associating these two
negative values with by 2 and y3 and y4
is associated with y1 in this case it
for was associated with y3 so these are
alternative forms and they are
equivalent now we say this theorem as a
principal axis theorem because y1 y2 y3
are actually the principal axis of the
quadratic surface and these axes lie
along y 1 y 2 y 3 X's in the new system
and that's the reason we say this that
this form is it this form is a and
standard form and the theorem changes
changes of a coordinate axis to
corresponding principal axis and that's
why we say the theorem as principal axis
theorem
now these are some of the results which
we have obtained earlier and actually
these forms a basis for finding out q1
why we know that eigenvalues of
symmetric matrices are real so these
eigen these values are real so we always
have to we can always transform our
given quadratic in this form where all
these coefficients are real because
eigen values are real so this is always
possible if I ghen values of a matrix
are complex then this would not have
been possible but since the matrix is
symmetric matrix so the eigen values are
we will are always real and we can
always find out this form and existence
of Q one wise guaranteed due to the real
eigenvalues so if lambda 1 lambda 2
lambda n are the eigen values of a then
the Q X which is given as X transpose X
can be transformed to the real eigen
values lambda 1 lambda 2 lambda n are
arranged in a manner so that all
positive eigen values appear first and
then all negative values and finally the
zeros eigen values so if lambda 1 lambda
2 lambda P are real positive values and
lambda 1 lambda 2 lambda are are real
values then lambda P plus 1 lambda P
plus 2 lambda n are 0 values so are
happens to be the rank of the matrix P
and out of these are P are positive and
remaining will be negative
while lambda P plus 1 2 lambda N or
eigen values are 0 now we consider an
another diagonal matrix H with diagonal
elements as 1 over lambda 1 1 over
lambda P 1 over under root lambda P plus
1 and then
the negative terms are written with
minus sign under root so lambda lambda R
is negative so we write it as minus
lambda R it's under root and then with I
don't 1 1 1 corresponding to these zero
values so I form a matrix H with these
eigen values now let us consider this
matrix D 1 as H transpose D H now HD and
H are all diagonal matrices and H is
having positive terms as 1 over under
root lambda and H transpose is also
having 1 over under root lambda while d
is having under root lambda so when you
multiply this every term in the diagonal
of D 1 will be either having 1 or will
be having minus 1 depending whether the
corresponding eigenvalue is positive or
negative and then the values
corresponding to 0 will be having 0 over
here so d1 will either be having 1 or
minus 1 or 0 so this way we can write
down the diagonal matrix D 1 and once we
can express D 1 as this matrix we can
say that D and D 1 are congruent and
that means a given quadratic QX is
equivalent to a quadratic form QP y
which is equal to Y 1 square plus y 2
square plus YP square positive terms
minus YP plus 1 square minus YP plus 2
square up to Y R square with negative
terms they are the rank of this
associated matrix was R so we'll have at
the most our terms some of them will be
positive some of them will be negative
the difference between Q then Y and Q 2
why is that in the coefficients in Q 1 y
the coefficients are eigenvalues but in
Q 2 y the coefficients are plus 1 or
minus 1 depending upon the sign of
corresponding eigenvalue so for a given
Patrick's QX we can always transform it
into a form where Q 2 y is y 1 square
plus y 2 square plus YP square minus YP
plus 1 square up to minus y our square
so this particular form is called
canonical form so when P is fixed Q 2 is
unique so this is the result which we
are having and a given expect given
polynomial can always be transformed to
the corresponding canonical form and so
unique form now I introduce a term
inertia which is a trial consisting of
three values this first value represents
how many eigen values are positive the
second term represents the number of
negative eigenvalues and zero is the
number of zero eigen values so inertia
associated with the given given
quadratic is a try indicating the number
of positive eigen values negative eigen
values and 0 eigen values and from there
one can find out the canonical canonical
form now the difference between the
number of positive eigen values and
negative eigen values is called the
signature of quadrature quadratic form
so if I am having T number of eigen
values which are positive and are being
the ring then R minus P will be the
negative eigen values remaining will be
0 so P minus R minus P is equal to 2 P
minus R so we say that 2 3 minus R is
the signature of the quadratic form we
say if Q 1 and Q 2 are equivalent
quadratic forms then they have same rank
and same signature these are easy to
prove this result is easy to prove the
next result is that Q 1 and Q 2 have
same rank and signature
then they are equivalent this is a just
converse of this let us just say this
with an example I have been given a
quadratic 4 by 1 square minus 2y 2
square minus 2 by 3 square its
eigenvalues are 4 minus 2 minus 2 and D
is 4 minus 2 minus 2 actually this
quadratic has been taken from earlier
example and we have already computed the
eigenvalues of that matrix as minus 2
and minus 2 so for this quadratic i ghen
values are 4 minus 2 minus 2 and
correspondingly we have the diagonal
matrix as 4 minus 2 minus 2 in off
diagonal terms are all zeros then the
eigenvalues of H will be 1 by 2 it is
underneath as 1 by lambda so it is 1 by
2 and then the next eigen value will be
1 by minus of the eigen value so that is
why it is coming out to be 1 by root 2
and another eigen value of H will be 1
by root 2 now let us now compute D 1 as
X transpose D H so I am writing H
transpose now H is a value matrix so
it's transpose will be the same as H and
the values in the Dardanelle will be 1
by 2 which is under root of 1 by 4 then
1 over under root 2 it is unbilled of
minus of minus 2 so it is 1 by root 2
and we have 1 by root 2 then this the
matrix the diagonal elements are not
shown here but they are 0 elements and
will have this diagonal matrix H and you
multiplied this multiplied by this and
this multiplied by this is 1 this this
and this multiplication gives me minus 1
this this and this is minus 1 and
correspondingly we have d 1 is equal to
a diagonal matrix with 1 minus 1 and
minus 1
only cuter why comes out to be y 1
square minus y 2 square minus y 3 square
so these are two negative eigenvalues so
it is minus y 2 square minus y 3 square
and since it has one positive eigenvalue
and two negative eigenvalues so we can
say P is equal to 1 rank of this matrix
is 3 so number of negative values will
be R minus P that is 2 and from here in
a choice is positive values one negative
values to 0 values is 0 so number of 0
values is 0 here so unless R is this
triad and signature of this quadratic
form is s is equal to minus 1 because we
are having no eigenvalues compared to
the more negative eigenvalues compared
to positive eigenvalues so signature
comes out to be negative now we start
with the general form of the quadratic
equation X transpose X plus BX plus C
equal to 0 of course we have till now
concentrated on this part x transpose ax
so let us consider we have a general
form X transpose ax plus BX plus C equal
to 0 then if a is not diagonal matrix
then we rotate it by this transformation
X equal to py and it will become
diagonal matrix and if B is not 0 then
we translate the axis so that we can get
rid of this and we finally have X
transpose B X or we can say Y transpose
B y plus D equal to 0 so this can be
easily converted into a standard form
and the classification of this standard
form actually depends upon this
eigenvalue problem ax is equal to lambda
X and that is how we identify identify
or classify 2d surfaces
this eigenvalue problem if it has two
eigenvalues which are positive then the
corresponding standard form will be an
ellipse it's you know share will be two
because number of eigenvalues are tools
two dimensions so negative values will
be 0 and no eigen value is 0 in this
case it is hyperbola
when both a eigenvalues are negative
when when the product lambda 1 and
lambda 2 is negative so one eigenvalue
is positive and one eivin value is
negative in this case so one for number
of positive values one negative
eigenvalues no value which is 0 and it
is parabola and one eigenvalue is 0 so
either lambda 1 is equal to 0 or lambda
2 is equal to 0 then it will represent a
parabola it has one positive value it
has no negative value and one 0 value so
depending upon the inertia corresponding
to given matrix in the given
representation we can classify the
quadratic as ellipse hyperbola or
parabola and corresponding inertia is
given as this now for 3d surfaces we can
similarly classify a quadratic surfaces
we say if you know she is 3 0 0 then it
is ellipsoid that means all the three
eigenvalues are positive
no eigen value is zero and nothing is 0
no eigen value is 0 in this case then we
have ellipsoid and then in a phase 2 0 1
then we call it elliptic paraboloid so
two eigenvalues are positive one value
is 0
no eigen value is negative we'll have
elliptic paraboloid then in a shell
still 1 0 that makes 2 positive
eigenvalues 1 negative eigen value then
we say it is height of the light of one
sheet then 1 to 0 is an iShare we have
one positive value two negative eigen
then we'll have hyperboloid of two
sheets and whenever sells 1 1 1 that
means we love one positive one negative
and one zero eigenvalue then we call
that surfaces hyperbolic paraboloid 1 1
2 represents a parabolic cylinder so
that is how we classify different
surfaces now in this example you have to
identify the conic section so the conic
section is given as X square plus y
square plus Z square minus 4 XY is equal
to 4 it's a three dimensional surface so
the identify what it represent we have
to calculate its inertia for that we
have to first write down the
corresponding matrix a and terms in the
diagonal are 1 1 1 and this minus 4
corresponds to minus 2 and minus 2 in
this positioning the matrix a then we
calculate the eigenvalues for this we
consider the characteristic equation as
this and this simplifies to this
equation characteristic equation lambda
minus 1 multiplied by lambda minus 1
square minus 4 and further
simplification will give me lambda minus
1 minus 2 and lambda minus 1 plus 2 so
the eigenvalues are given from this
expression and they are 1 minus 1 and 3
so we will have 2 positive eigenvalues
namely 1 and 3 and one negative value
minus 1 so the inertia for this given
matrix is 2 1 0 and according to the
qualification according to the
classification this surface is height of
the light of one sheet now we will
discuss positive definite quadratic
forms many times we are interested to
know the sign of this expression ax
square plus 2 bx y plus ey square it's a
quadratic expression let us say equal to
d and in that expression we write it as
x transpose ax so we want to know
whether this expression is positive or
negative
so we want to know under what condition
this expression is positive under what
condition may be negative and under what
conditions it is always positive so
those conditions there for two positive
definite quadratic form so I will define
the quadric the quadratic form is
positive definite if x transpose ax is
always positive for positive x and it is
0 only when X is equal to 0 so an
expression X transpose ax is bit is
always positive then that expression is
or that quadratic form is positive
definite the quadratic form is positive
semi definite when B can be positive for
some nonzero X it may be positive for
other values but there may be some value
of x which is not 0 but D is positive
the quadratic form is negative semi
definite wendy can be negative for some
nonzero X also so symmetric matrix a and
the quadratic form X transpose ax are
positive definite if X transpose X is
positive for X not 0 we call the form as
negative definite when x transpose ax is
negative for some for X not 0 and
positive semi definite if X transpose X
will be positive already 0 for all X now
the tip semi definite then X transpose X
is negative for all X negative or 0 for
all X and we say the form is indefinite
if we cannot determine the sign of X
transpose X if it is positive always we
call it positive definite if it is
negative always then we call it negative
definite
for nonzero values of x and a positive
semi-definite or semi indefinite and the
people selling a definite when will have
this expressions and if the sign cannot
be determined then it may be indefinite
the quadratic form is indefinite wendy
can be both positive and negative can
take any value so in this example we try
to decide the positive definiteness of
the given quadratic form so in the first
we have associated matrix as 1 1 1 5 in
the circuit will have the associated
matrix 1 3 3 1 and in the third will
have this matrix will go one by one this
matrix is associated with this quadratic
expression x squared plus 4y square plus
2xy we can write down this form as X
plus y whole square I am rewriting this
expression plus 4y square see X square X
plus y square is equal to X square plus
1 square plus 2xy so 2 X Y is coming
here and Basco article from this so what
remains here is 4y square now this
expression is 0 when x is equal to 0 and
Y is equal to 0 and for all values of x
whatever positive or negative this
expression will always be 0 or we can
say that this expression is positive
definite or this quadratic form is
positive definite quadratic form it's
positive definite in the second example
I have this matrix and corresponding to
this matrix all of this form X square
plus 6xy plus y squared and this form
I'll write it as X plus 3 y whole square
minus a12 square and one can notice that
4x is equal to minus 2 and y is equal to
1 X is equal to minus 2 and Y's
- one then this term will be more than
this term and that makes QX negative so
this is not positive definite because
it's sign will be QX is negative for a
value X is equal to -2 and Y is equal to
1 so this is not positive definite in
the other example the 3 by 3 matrix it
is a diagonal matrix it comes out to be
X square plus 2 y square minus Ceaser to
square one can notice that if X is equal
to zero Y is equal to zero then if that
is whatever be the value of Z then this
is negative so it is again not positive
definite so 0 0 1 QX is negative and it
is not positive definite now in this
theorem will like to state the
conditions under which we can ensure
that given quadratic is always positive
definite so the theorem states that the
three conditions the following three
conditions are equivalent one condition
is if QX is positive definite second is
all the eigen values of a are positive
third is all the sub matrices have
positive determinants so I will start
with a given quadratic x I transpose X I
is positive
I'll first prove that they implies B
means if QX is positive definite then it
implies all the eigen values of a are
positive so with this given positive
definite form let us consider an organ
value problem ax I is equal to lambda
ixi now in this form is a symmetric
matrix so for this form we can multiply
this matrix by X all transpose from left
as well as on the right so XL transpose
SCI is equal to lambda X I'll transpose
X I and we know Excel transpose X is
nothing but the norm of X I which is
always positive so X transpose ax I is
always positive
provided lambda I is positive so if a
given metric given form is quadratic
definite then all the eigen values of a
should be positive so I started with a
positive definite form and I have proved
that it has all eigen values which are
positive so a implies B now to consider
do you implies a I will start with all
positive eigen values and I will show
that because following matrix is
positive definite so assuming all
lambdas are positive and the matrix is
symmetric because this matrix is related
with the quadratic then the eigen values
from set of all independent eigen
vectors this is a result which I am
deriving from my other lectures on eigen
values and eigen vectors so they form n
independent eigenvectors so we can write
down X any vector X as a linear
combination of these n independent
eigenvectors so we'll have X is equal to
C 1 X 1 plus C 2 X 2 plus CN xn I
multiplied that a so ax is equal to C 1
X 1 plus C 2 X 2 plus C n Axl then I
multiplied by X transpose pre multiply
pre multiplication so it is X transpose
ax is equal to C 1 X 1 plus C 2 X 2 plus
CL X and this is X transpose X transpose
is X is nothing but this so this is X
transpose multiplied by C value X 1 plus
C 2 X 2 plus C n ax n whatever I have
obtained here and then if you perform
this multiplication it is calling this
it is transpose
so if you multiply this they'll have
civil square lambda 1 plus C 2 square
lambda 2 plus CL square lambda n square
so C's are arbitrary so X transpose X is
positive and that means eigenvalues are
positive because Argan values are
positive we'll have X transpose ax is
positive and that means we'll have
positive definite quadratic form so
they've started with positive
eigenvalues and we could prove that we
have a positive definite quadratic form
so do you guys see is fruit now let us
try a fly see so given that QX is a
positive definite this that's the
meaning of a the V of 0 V are heavy
eigenvalues are positive because PS is
positive definite so beam is eigenvalues
are positive so we have this then what
is determinant of a determinant a is
productive eigenvectors now we consider
X transpose X as X K 0 what is XK
transpose is a vector which is having
nonzero values and rest of them are
zeros so I reduce this matrix a into K
by K matrix that is a K and rest of them
are these elements in E so I can write
it as X K transpose and rest of the
elements are 0 so I'm only those values
which are corresponding to a K they are
nonzero and rest of them are 0 so this
particular X transpose is chosen so
accordingly we'll have this column
vector so when you perform this
multiplication
I'll have XK transpose a K X transpose
but X transpose X is positive so XK
transpose a k x xk transpose is positive
provided determinant
is positive this means the sub-matrix a
K has positive determinant but what is a
KA things are some matrix of a and this
can be any order matrix it can be order
1 it order to order 3 and so on
in fact this result should be true for
all values of K and that means if we
have a positive definite net positive
definite form then all sub matrices will
have positive determinant now that
proves C that means starting from a we
can prove the see the reverse of this
that is if I if we have a matrix having
all sub matrices determinant 0 then we
can prove that it is a positive definite
form that is C implies a I did I leave
this an exercise for the viewer in this
example we are using the theorem which
has just now stated and with the help of
that theorem would like to see whether a
given quadratic form is positive
definite or not so if in the first
example if I have a 2 by 2 matrix then
one can notice that its determinant is 1
into 5 minus 1 is positive and since it
has a positive determinant and all sub
matrices have positive determinant so it
has all positive eigenvalues and it is a
positive definite form so first is it is
positive definite form but in the second
case first sub matrix is having
determinant positive but if you consider
this sub matrix then its determinant is
1 minus 9 is negative so it is not
positive definite and in the third case
again the one of the eigen value is
negative so again this is also not
positive definite form so in this
example will have only one positive
definite form in the second example
determinant is negative so we'll have
will not have a positive definite form
in this exam
one eigenvalue is negative then it's
this is also not positive definite form
solely one is positive for them if a is
positive definite matrix then a square
is also positive definite I leave this
as an exercise for the viewer to prove
this further if arias is also positive
definite if a is positive definite so
these are two exercises user can try
easily that gives that comes with this
we come to the end of the lecture today
we have covered quadratic surfaces we
have discussed then under what
conditions a given quadratic form is
positive definite quadratic form that's
all thank you