Diagonalization (Part - 2) - Mathematics, Engineering Video Lecture - Engineering Mathematics

welcome viewers today I will be talking
on diagonalization this lecture is in
continuation to my earlier lecture on
diagonalization then I had discussed the
realization of real matrices today I
will discuss the analyzation of complex
matrices this lecture constitutes two
parts the unitarily diagonalizable and
application to the concepts applied to
solution of system of ordinary
differential equations I will first
start with unitary matrices so unitary
matrix is defined as a matrix with
complex entries satisfying the condition
that LS is equal to a star here a star
is transverse conjugate and that means
we first take the transpose and then we
take the conjugate or we first take the
conjugate and then take the transpose
and this the matrix is the same as a
inverse so first thing is that a has to
be invertible only then we can say that
a inverse is equal to a star and if the
matrix happens to be a real matrix then
a star is nothing but a transpose
because conjugator for real matrix is
the matrix itself so a star is nothing
but a transpose and in that case we have
a unitary matrix as the same as
orthogonal matrix in other words we say
that a real unitary matrix is orthogonal
matrix because those will have a inverse
is equal to a transpose similarly where
we define normal matrices a square
matrix a with complex entries is called
normal if
a star is equal to a star a so this
product a with a star that is transpose
conjugate and a star multiplied by a
they should be equal the difference here
is that if a a star is equal to a star a
is equal to identity then we can write
down a inverse is equal to a star but
the product may not be identity but
still these two products are equal and
then we say the matrix a is a normal
matrix so every permission matrix is a
normal matrix because a a star is equal
to a star a if it is a normal matrix but
we know that a star is equal to a for a
hermitian matrix what is hermitian
matrix hermitian matrix is a matrix with
complex entries so that a star is equal
to a so when a star equal to a then this
product as well as this product equal to
a square which satisfies the condition
of normal matrices and that's why we say
that every hermitian matrix is a normal
matrix and every literal matrix is
normal it's just an extension of what we
have discussed just now you can check
that a a star is equal to a star a is
equal to I so this is I but both of them
are equal this is what is required for a
normal matrix and that is why we say
that every military matrix is a normal
matrix so let us elaborate this with
this example a where I have a matrix of
2 by 2 order here some of the elements
are complex we have to check whether it
is normal or not so for this purpose I
have to first constitute a star
so first I'll take the conjugate and
then transpose the conjugate of this
matrix means the a particular element
here will be transposed here so one plus
I transpose 1 plus ie its conjugate is 1
minus I conjugate means this imaginary
part will be negated and then 1 minus I
will become 1 plus I so this is a star
and when we take the transpose Rho
becomes columns and - 1 - I first row
will become first column and second row
will become second column so that is the
transpose and if you compare this matrix
with the given matrix we can notice that
this is nothing but a so a star is equal
to a so it is a this matrix a star is
equal to a so it is a hermitian matrix
and we have already proved that position
matrix is a normal matrix and in this
sense the given matrix a is normal
matrix so now I state some simple
results and I leave them to the viewers
they can prove them simple first simple
result is if a is unitary then row
vectors form an orthonormal set
and if row vectors form an orthonormal
set then the matrix is unitary we write
this as a unitary if and only row
vectors form an orthonormal set this
result is not only true for row vectors
this is applied to column vectors also
and the results read like this is
unitary if and only if column vectors
form an orthonormal set this is first
result to be proved by our viewers the
second is the unitary matrix have eigen
values such that the magnitude of an
eigen value is 1 so all the
that eivin vectors all the eigenvalues
immediately met unitary matrix have
magnitude one so this is another result
which is to be proved by the viewers now
in this example a matrix is given you
have to show that it is a unitary matrix
so 2 by 2 matrix it has complex entries
and what we do is we consider the row
vectors of a they are 1 plus I by 2 and
1 plus I by 2 and the second row vector
is 1 minus I by 2 and minus 1 plus I by
2 so these are 2 row vectors we
calculate their magnitude so norm of r1
bar is equal to square of first
component plus square of second
component and model 1 plus I square is
equal to real part square plus imaginary
part square divided by 2 so it is 1 plus
1 2 divided by square of this 4 so this
number is equal to 1 by 2 similarly this
is also 1 by 2 so we have 1 so norm of
row vector is 1 then on the same lines
one can find the norm of second vector
and it is 1 minus I by 2 magnitude of
this plus minus 1 plus I by 2 this is
the second component and it's square and
when you calculate this it is 1 by 2 and
this is also 1 by 2 so some of this is
equal to 1 now R 1 dot R 2 that is in
the part of R 1 and R 2 it is 1 plus I
by 2 multiplied by 1 minus I by 2 and
it's cons you
8 plus 1 plus I by 2 into minus 1 plus I
by 2 it's conjugate and when you perform
this multiplication this comes out to be
1 plus I by 2 and conjugate of this is 1
plus I by 2 plus this remains as such
but its conjugate will be minus 1 minus
I by 2 so this is one can take this
minus common and it is 1 plus I by 2 and
1 plus I by 2 with negative sign so this
cancels out with this and what we have
is that in effect of R 1 and R 2 that is
R 1 dot R 2 is 0 so what we have is we
have two rows R 1 and R 2 there dot
product is 0 and their magnitude is
equal to 1 and that means the matrix
which is given to us has row vectors
which are orthonormal they are normal
because of this they are orthonormal
because their magnitudes are 1 so in
this way we say that a star is equal to
a inverse
it is a unitary matrix so this we are
not actually evaluating a inverse and
then saying a inverse is equal to a star
we are saying that this matrix has row
vectors which are orthonormal and that
is how we prove a inverse is equal to a
star actually we are applying the result
which we have stated in my earlier
slides this means that the matrix a is
unitary matrix now before we proceed
further we'll verify this now if a LS is
equal to a star then the product of the
two matrices should be identity so we
verify this and we calculate this a into
a star is equal to this is what I have
the matrix a and this is a star and when
we multiplied this multiplied by this
this multiplied by this this is 1 this
row multiplied by
this row multiplied by this column is
zero this row multiplied by this column
is zero and this multiplied by this is
one so a a star is equal to identity
matrix now we define unitarily
diagonalizable matrices we say that a
square matrix a with complex entries is
unitarily diagonalizable if there exists
a unitary matrix P such that P inverse
ap is diagonal so basic differences the
matrix is diagonalizable we should have
a matrix P which is invertible such that
P inverse ap is diagonal but the
difference here is the matrix P is not
only invertible but it is unitary matrix
so if we can find such a matrix P then
we say the matrix a is unitarily
diagonalizable now to prove this we say
that the following are equivalent first
a is unitarily diagonalizable and the
second is a has an orthonormal set of n
eigenvectors and third statement is is
normal that means a is unitarily
diagonalizable if it is normal and it is
normal matrix then eigenvectors from
different eigen spaces of a are
orthogonal
not only orthogonal but they are
orthonormal this particular result forms
the basis for a procedure for u iterally
die analyzing a given matrix so let us
discuss a procedure for unitarily daniel
eisenman
is find the basis for each eigen space
is of a
in the step two we apply Graham shallot
process to Auto normalize the basis for
each eigen space of a and finally mr.
step three the we form the matrix P
whose columns are the orthonormal basis
now in the next example we illustrate
this procedure for unitarily
diagonalizable matrix a so in this
example our matrix a is given as a two
by two matrix we are considered two by
two matrices because they are simple so
you have to find I am NOT unitary matrix
P which will diagonalize this given
matrix a to find matrix P we have to
form we have to first obtain the eigen
values and then corresponding eigen
vectors for this we first write down the
characteristic equation for the given
matrix a so the characteristic equation
for this is the determinant lambda minus
2 minus 1 plus I minus 1 plus I and
lambda minus 3 one can simplify it it is
lambda minus 2 into lambda minus 3 plus
1 plus I into minus 1 plus I perform
this multiplication and it is lambda
square minus 5 lambda plus 6 this part
and from here we'll get minus 1 minus 1
and that gives us lambda square minus 5
lambda plus 4 equal to 0 and when you
factorize it it is lambda minus 1 into
lambda minus 4 equal to 0 and that gives
us the eigen values as lambda is equal
to 1 and lambda is equal to 4 to get the
eigen vectors corresponding to these
eigen values we have to solve this
equation a minus lambda I multiplied by
X equal to 0 so let us
a minus lambda X equal to zero so it is
2 minus lambda 1 plus I 1 minus I 3
minus lambda X equal to 0 and 4 lambda
is equal to 1 we have to solve this
equation will get eigenvectors
corresponding to lambda is equal to 1
and when we solve this equation for
lambda is equal to 4 will get
eigenvector corresponding to lambda is
equal to 4 so first for lambda is equal
to 1 this equation become 1 1 plus I 0 1
minus I 2 0 this is an Augmented matrix
corresponding to this system so this
system can be transformed I am writing
only the this part the coefficient
matrix part here I have applied
elementary transformation and this will
only the coefficient matrix is written
here because this will not be affected
so we transform this matrix to this
matrix and from here one can notice that
this Augmented matrix will finally
become this matrix and this matrix has
one I then one row completely zero and
that means that means we can have either
on vector of given matrix as minus 1
plus I K into K the K is an arbitrary
value this is an eigen vector
corresponding to lambda is equal to 1
and when we write K is equal to 1 then
let us say the eigen vector U becomes
minus 1 plus I and 1 this indicates that
it is a column vector so it's a
transpose here so U is a column vector
which is actually an eigen vector
corresponding to lambda is equal to 1 on
the same lines we can
get Argan values I then we can find
eigen vector corresponding to lambda is
equal to four so we consider this
augmented matrix apply elementary
transformations number of elementary
transformations are required and finally
this matrix will reduce to this matrix
with one row completely reduced to zero
and from here one can easily write down
the eigen vector as one plus I times K
into two K because this will satisfy
this equation and that means the
eigenvector is the eigenvector is one
plus I in comma two when we write K is
equal to 1 so we have another
eigenvector
so in this cells u1 is equal to minus 1
plus I by root 3 comma 1 by root 3 this
is what we have obtained after
normalizing the given vector u U and
this is a row vector because I am NOT
writing transpose here so it's au vector
so u 1 is equal to minus 1 plus I by
root 3 comma 1 by root 3 is a
orthonormal vector it's magnitude is 1
and V 1 is 1 plus I by under root 6 and
2 over under root 6 because the
magnitude of V 1 is 1 by root 6 so V 1
is a row vector with this as the first
component and this is the second
component and it is again a row vector
now u 1 MV 1 from a set of orthonormal
vectors their dot product is 0 and each
of them is of magnitude 1 and that means
we have a matrix P which is having the
two vectors as column vectors this is
first column vector this is second
column vector and these column vectors
are having magnitude 1 and their
doctorate is 0
so they are orthonormal and this
characteristic makes the matrix PS a
unitary matrix and then penis AP is
equal to a diagonal matrix so we can be
here we have in this particular step we
have found a matrix P which is given a
matrix for the given matrix so we say
that P inverse ap is a diagonal matrix
basically we are applying the result
which we have already stated the proof
has not been given here and the proof of
that theorem can again be left to the
readers so previous AP is a diagonal
matrix and the diagonal elements here
are the eigenvalues of the matrix a they
are 1 and 4 so this way we say that a is
unitarily diagonalizable
now here we discuss applications to
eigenvalue problem to the system of
differential equations consider system
of differential equations in two
variables x and y x and y are two
dependent variables depending upon an
independent variable T and we write the
first equation is the first order
differential equation relating the
derivative DX by DT with x and y so we
say DX by DT is equal to ax plus B Y a
and B be constants in the second
equation also relates derivative in
respect to Y with the variables x and y
so dy by DT is equal to CX plus dy so
here we have two equations two
differential equations which are of
first order they are coupled because
first equation involves X as well as Y
and the second or second equation in y
also involves the variable X so this is
coupled with X and this is coupled with
Y so we have a system of coupled first
order
differential equations now this system
can be written in this form in this
matrix form here D by DT of this matrix
X Y so we have a column vector on the
left hand side and we have taken its
derivative so deep left hand side is d
by DT of X Y is equal to a b c d this is
the coefficient matrix on the right hand
side multiplied by the column vector XY
and this is 2x2 matrix this is a column
vector this is a column vector and we
write the dynamical system we call this
as a double dynamical system and we
write it as DX by DT is equal to ax I
denote this matrix by a this column as X
so this is nothing but DX by DT and the
whole system is DX by DT is equal to a X
so X is a column vector in r2 because
dimensional X is 2 so X belongs to r2
now together with this differential
equation it's not a single equations
actually a system of equations and this
system up to a differential equation can
be written in a compact form in the in
this matrix differential equation and
this has to be solved under specific
initial conditions so we specify the
initial conditions as X at T is equal to
0 is equal to X naught actually this
condition simply means that X at T is
equal to 0 is X naught and Y at T is
equal to 0 is y naught so that initial
condition is also written in a compact
form and this system is called a
dynamical system we call it dynamical
system because it is changing with time
so we are considering how the dynamics
of the vector X is changing with time so
that is the meaning of a dynamical
system
now we generalize this concept and we
write down DX by DT is equal to ax where
X belongs to RN so earlier I have
written a matrix equation in two
variables but X need not be belonging to
R - it may be a column vector consisting
of M components in that case we say X
belongs to RM and this equation DX by DT
is equal to X means that a is a square
matrix of order n now one represents a
homogeneous system of n linear first
order differential equations in any
differentiable variables because X is a
column vector so this left hand side is
actually n terms and the right hand side
we have this product and X is a column
vector so actually this system
represents n differential equations
because each equation is having D by D
by DT involved in it and since we are
considering derivatives so we should
have the functions as differentiable
functions differentiable vary this all
these variables should be differentiable
we call this as a homogeneous system
because all the terms on the right hand
side they are involving X or Y there is
no term which is free of the variable X
and vice so there is no constant matrix
added here and in this sense we say this
represents a homogeneous system of M
differential equations each equation is
linear in its own right and that is why
it is a linear system of differential
equations the vector X 1 which is having
any components is a column vector we say
it's a solution of the given system 1 if
it satisfies the system of ordinary
differential equations or in short we
say it oh de so any function which
satisfies this differential equation
is called the solution of this system so
X we denote this solution vector as X 1
all solutions of 1 will form a subspace
of the vector space of differentiable
real valued and vector functions if you
look at this this is a solution what is
X 1 X 1 has to be a function of T it has
to be a differentiable function X 2 also
has to be a differentiable function
because all this is they are to be
substituted here if any of them is not
differentiable then you cannot say that
it is DX by DT when you cannot form DX
by dt there is no question of satisfying
this differential equation
so first thing is that this has to be
differentiable function and when you
substitute it here this should satisfy
this differential equation so this each
solution vector is a vector in M
dimension so if we consider all vectors
which having which is having n
components each of them is a
differentiable function then it will
form a vector space but if we consider
all the solution space that is the space
constituting the solutions of this
differential equation then we say it
will form a subspace of this vector
space of different n differentiable
vector functions the next if X 1 X 2 xn
are solutions of a system 1 then for
arbitrary constants the linear
combination X is also a solution so XT
is written as C 1 X 1 plus C 2 X 2 plus
C and X n if we can write down the
solution as XT consisting of this
then we say that X 1 X 2 X n is a
fundamental system for one the initial
condition X 0 is equal to X 1 0 if we
denote the initial condition in this
manner then when we substitute this
initial condition here then all these
constants c1 c2 CN which can be obtained
so in this state at this state these are
arbitrary constants so x1 x2 xn they are
n solutions the linear combination is
also solution but if you will do is
initial can this initial condition then
all these c1 c2 CN will be determined
and what we have is a particular
solution so a particular solution is the
general solution obtained using initial
conditions which will be free of
arbitrary constants so these n
conditions
use with this will eliminate these
arbitrary constants and what we have is
a solution free of arbitrary constants
and that such a solution is called a
particular solution the problem of
determining the solution of system one
with given initial condition is known as
initial value problem so let us consider
DX by DT is equal to a 1 a 2 n and n by
n diagonal matrix multiplied by the
column vector X 1 X 2 xn which is
nothing but X together with the initial
condition X 0 is equal to X naught so
this constitute an eigen value problem
but it is a special case in the sense
that the matrix a happens to be a
diagonal matrix that means only diagonal
entries are nonzero rest of the entries
are zeros I am considering this because
this is easy to solve so let us solve
this system here you can reduce this
system as DX 1 by DT is equal to a 1 X 1
so
this equation is no more coupled with
rest of the equations the X to the by DT
is equal to a2 times x2 it is again be
coupled with other equations so if the
equations are decoupled especially when
we have a diagonal matrix here then each
equation can be solved independent of
other equations so we first solve the
first equation in the system that is DX
1 by DT is equal to a 1 into X 1 this is
a first-order differential equation and
it's very easy to solve
and if solution will be X 1 into X 1 is
equal to X 1 0 e times E or exponential
a T here what is X 1 0 is the arbitrary
constant and we can we can and Cal
calculate this using this initial
conditions similarly we can calculate
the second solution DX 2 by DT is equal
to a 2 into X 2 and so on so in general
the solution of this system is written
as X I is equal to X I 0 e a i t i will
take values from 1 to n i is equal to 1
will have solution of first equation I
is equal to 2 will have solution of
second equation and so on so this is a
general solution once we have obtained
this then X is equal to X 1 comma X 2
comma X n and then the column vector is
a solution of system 1 this system 1 so
this represents a solution of this
equation and if you substitute these
values the way we have computed here
this is the solution vector of this
differential equation and with this the
general solution is written in this form
where I have written this first term as
X 1 0 1 0 0 the column vector e a 1 T
and the circuit corresponding to this I
have written X 2 0 a column vector x e a
2 T
and so on so if you sum it up the it
will be first component will be x10 ea1
t only this contribution will be coming
for the second time contribution will be
coming from this term only and for the
NS term it is contributing contribution
is coming from this term so one can
write this general solution in this form
so XT is a linear combination of these
solutions we will estate this with in
this example so if I have given a 3 by 3
system DX 1 by D DT the D by DT of X 2
DT and D by DT of X 3 is equal to this
diagonal matrix multiplied by the column
vector X 1 X 2 X 3 so let us solve this
system
this has diagonals 1 minus 2 and minus
and 1 minus 2 & 3 and that makes X 1 is
C 1 et X 2 s C 2 e minus 2 T and X 3 or
C 3 e 3 T and that means XT is equal to
C 1 times the column vector 1 0 0 ET
plus C 2 times the column vector 0 1 0 e
minus 2 T plus C 3 as 0 0 1 e 3 T so we
have obtained the general solution now
if the matrix is given if the matrix a
is given in the diagonal form then
things are simple but if a is the
diagonal then system can be transformed
to a diagonal system and then we can
solve it and here comes the importance
of diagonalization what we have
discussed in these lectures so if the
matrix a can be diagonalized as this d
is equal to P inverse ap then we can
solve the dynamical system DX by DT is
equal to ax X belonging to RN on the
same lines as I have Illustrated just
now so if this is a
system given to us then we can change
the variable X to P you so here is a new
variable and we write down X as P you
with this transformation P the
invertible constant matrix then we can
pre multiply the given equation by P
inverse so before we apply this
transformation we have to make sure that
p is invertible and it is a constant
matrix so if you do that then previous
DX by DT is equal to P inverse ax since
penis is a constant matrix so we can
write down this left hand side as d by
DT of P inverse X and that is why I've
assumed that it is invertible and
constant matrix so left hand side is
this and right hand side is P inverse a
and a this X I am replacing by PU and
this simply means da by DT is equal to P
inverse ap into u so that gives me D u
by DT is equal to D u so with this
transformation this matrix D is diagonal
and I have obtained a system of
equations in which the matrix D is
diagonal and then we can very easily
solve the system of equations and we
know from our earlier discussions that u
T can be expressed as C 1 an arbitrary
constant multiplied by this column
vector multiplied by E into lambda 1 T
what is lambda 1 lambda 1 is the
diagonal first diagonal element in the
matrix D then we have at the second term
here C 2 arbitrary constant multiplied
by this column vector where the second
term is 1 and rest the terms are 0 and
this will be multiplied by e lambda 2 T
lambda 2 being the second term in the
diagur
matrix D and so on the last term will be
CM then at arbitrary constant and that
component in this vector will be one
rest of them are 0 and multiplied by
exponential of lambda and T so we have
obtained solution UT but of us where our
equation was in X so we have to
transform this solution to its original
coordinate system so that means we apply
P on both the sides so P of U P is equal
to c1 P of this vector ax lambda 1 T
plus c2 P of the vector multiplied by e
lambda 2 T and so on and that makes P UT
as X of T the function X is a function
of T is equal to c1 and I write them C
into 1 0 0 if you multiply there if you
perform this multiplication this simply
means the first row first column a the
matrix P so this P 1 e lambda 1 T plus
c2 and this multiplication is P times 0
1 0 0 is nothing but P 2 so it is e
lambda 2 T and so on and the final term
will be the last value P n so this p
multiplied but this is nothing but P n
so we write down XT as C 1 P 1 e lambda
1 T plus C 2 P 2 e lambda 2 T and so on
so are what I am writing is that the
solution vector XT is this linear
combination in which P 1 is the eigen
vector corresponding to eigen value
lambda 1 of the matrix D which is the
diagonal matrix similar to the given
matrix a so solution can be easily
obtained from the matrix D so basically
what we are doing is we are
diagonalizing the matrix a
and then we are finding the solution in
this particular manner so if DX by DT is
equal to ax and lambda 1 lambda 2 lambda
and our eigen values of a and P 1 P 2 PN
are eigen vectors of a then the solution
of this differential equation is
expressed as this in terms of eigen
values and eigen vectors this gives us a
procedure for solving system of
equations according to this the step one
constitutes the computation of eigen
values lambda 1 lambda 2 lambda n of the
given matrix a in the step two we
compute the eigen vectors P 1 P 2 P n of
a corresponding to the eigen values
lambda 1 lambda 2 lambda n and the step
three is we compute the general solution
of one as XT is equal to C 1 P 1 e
lambda 1 T plus C 2 P 2 e lambda 2 T and
lastly C n pn e lambda and T we restate
this procedure in this example so a 2 by
2 matrix being given to us first
according to first step we have to first
form the eigen values for this the
characteristic equation for a is the
determinant of this 2 by 2 matrix it is
simplified to lambda minus 4 multiplied
by lambda minus 1 plus 2 equal to 0 we
further simplified it is lambda square
minus 5 lambda plus 6 equal to 0 and
vectorizing it it is lambda minus 2 into
lambda minus 3 equal to 0 and that gives
us i can value Czar's
2 n 3 and from here lambda is equal to 2
we calculate the eigen vector by solving
this equation 2 minus 2 1 minus 1 x
equal to 0 and one can easily see that
one row will be
one row can be transformed to zero and
the eigenvector will be 1 comma 1
transpose for lambda is equal to 3 we
have to solve this system 1 minus 2 1
minus 2x equal to 0
again these two rows are identical and
this gives us the eigenvector s 1 comma
2 and once we have obtained eigenvalues
and corresponding eigenvectors then XT
is equal to C 1 multiplied by the first
vector e 2 T and then plus C 2 the
second vector 1 2 into e raised to power
3 T so this is the solution of given
system of differential equations in this
example I have seen where I have two
distinct roots of the characteristic
equation and correspondingly I have got
two independent eigen vectors but there
may be a situation where we have
repeated roots matrix cannot be
diagonalizable then things will be more
involved I am NOT discussing all these
things at the moment so once we got this
solution then in the component form we
can write down X 1 as C 1 e 2 t plus C 2
e 3 T that's the first component and the
second component of this vector X 2 is
written as C 1 e 2 T plus 2 times C 2 e
3 T so we can get this as a solution of
given differential equation let the
initial conditions be specified as X 1 0
is equal to 1 X 2 0 is equal to 0 see in
this equation we have two arbitrary
constants involved in it so we in fact
will have infinitely many solutions of
the given system of equations because
these coefficients can take infinitely
many values so to fix the ideas we can
specify the initial conditions and for
the given
initial conditions they are two in
number so we can substitute them here
two equations two unknowns and one can
solve for c1 and c2 and what we get is
the particular solution
so if we substitute this here t is equal
to 0 x1 is 1 so left hand side is 1 and
right hand side t is equal to 0 it is C
1 this is 1 for T is equal to 0 this
factor is also 1 so it's C 1 plus C 2
from the right hand side and similarly
substituting this in the second equation
will have 0 on the left hand side and on
the right hand side will have C 1 plus 2
C 2 these two equations can be solved
easily and the solution will be C 1 is
equal to 2 and C 2 is equal to minus 1
when we substitute this here we get the
solution X 1 as 2 e 2 t minus e 3 T and
X 2 as 2 e 2 t minus 2 C 2 e 3 T so in
this example we have been given a system
of differential equations together with
initial conditions or we say we have
been given an eigen value problem and
this eigen value problem is solved using
the concept of eigen values and eigen
vectors so this concept is useful in
solving system of first-order linear
homogeneous differential equations now
this can also be use in performing
qualitative analysis of dynamical system
but qualitative analysis of a dynamical
system I mean to say that a given system
DX by DT is equal to ax we may like to
answer questions like existence of
stationary solutions that means does
this system have stationary solutions or
does there exist
Saoirse solutions for this system or we
may like to know are these solutions
stable
so if this is if this stationary
solution exists are they stable or what
is the behavior of solutions as T tends
to infinity and if any possibility that
this system has a periodic solutions so
when we discuss qualitative behavior of
dynamical systems we actually trying to
answer these questions and these
questions can very easily be answered
with the help of eigenvalues and
eigenvectors related to this
differential system so let us illustrate
this for this we require concepts like
equilibrium point sometimes we call it
stationary point we define a stationary
point as X naught Y naught for the given
system such that DX by DT is equal to 0
by this I mean that 0 0 is equilibrium
point because this system means DX by DT
is equal to ax is equal to 0 now ax is a
homogeneous system and this homogeneous
system will always have zero zero at the
solution so 0 0 is equilibrium point
then the other equilibrium point
possible for this dynamical system for
the this equilibrium point or more
equilibrium points we may like to
perform the phase plane analysis but
phase plane analysis I mean to say that
if x and y are state variables of phase
space analysis then what are the
solution trajectories on XY plane and
these to decrease when they are drawn on
XY plane they constitute phase portrait
and this analysis is called phase plane
analysis so analyzing these trajectories
on the phase plane comes under phase
plane analysis it's the two donation
then it is phase plane analysis if the
analysis if the system has more than two
variables then what we have is a phase
space analysis
so we try to see how this trajectories
look like on XY plane
and then the behavior of solution
Mirek Librium point characterize the
type of equilibrium point so what are
the equilibrium points what are their
characteristics we illustrate with the
help of eigenvalues and eigenvectors let
us see that we have the solution of a
system as XT is equal to C 1 P 1 e
lambda 1 T plus C 2 P 2 e lambda 2 T for
a two-dimensional system
now since this solution depends upon
icon values lambda 1 and lambda 2 so the
behavior of the solution depends upon
the eigenvalues lambda 1 and lambda 2 so
we consider different cases the case 1
is then I when the eigenvalues are real
and in this case I subdivide into case 1
as eigen values are positive eigen
values may be negative eigen values may
be one positive one negative so I first
discuss the case when eigen values are
positive when the eigen values are
positive and the street then solution
goes away from 0 0 as T increases this
can be observed from this solution when
lambda 1 and lambda 2 are positive this
and this component will increase as time
increases and will be going away and
away from 0 0 and such in such a
situation we say the equilibrium point 0
0 is unstable by in the second case when
the eigen values are negative then these
negative values means that this term as
well as this term will become smaller
and smaller as T increases that means
will be going closer and closer to the
equilibrium point 0 0 and this means
that the equilibrium point is stable in
case 3 when one eigen value is positive
other is negative we say that there are
some trajectories where the solution
goes away
from zero zero as T increases because
corresponding e lambda 1 T increases but
on some other trajectories the solution
goes closer to zero zero as T increases
because lambda 2 is negative means Imai
a lambda 2 T tends to 0 as T tends to
infinity for negative lambda 2 so there
are some trajectories which will go
close to 0 but there are some other
trajectories where solution goes away
from 0 0 in such a situation we say that
the equilibrium point is a saddle point
in the set next case we consider that
the eigenvalues are complex when the
eigenvalues are complex we can write an
eigen value lambda 1 as a plus IB and
since I am the complex eigen roots of a
characteristic equation always occurring
third so the second lambda 2 will be a
minus IB it is the complex conjugate of
lambda 1 and if we have this lambda 1
and lambda 2 then e lambda 1 T is
written as e raised to power a T plus IB
T and I beauty is written as cos BT plus
I sine BT similarly corresponding to
lambda 2 we can write down e lambda 2 T
as e 8e x cos BT minus I sine B T now in
this case solution will depend what is
the sine of a if is positive then this
value will remain bounded because there
are sine cosine functions so they the
values cannot increase beyond 1 so 1 or
minus 1 so this value this remains
bounded so as T tends to infinity this
may go to infinity or may tend to 0
depending upon whether a is positive or
negative so in the case when a is
positive then we always go away
zero-zero because of this positive
exponent and in that case we say the
solution is unstable however in the case
to when a is negative then this term
will always grow closer and closer to
zero and this remains bounded so the
total term will be going closer to zero
and we say that the solution is unstable
in the third case when a is equal to
zero then we'll have a periodic solution
what we can do is that when a is equal
to zero then first part will be cos BT
plus I sine BT the second part will be
cos BT minus I sine BT and the solution
will be linear combination of these two
and these this linear combination can be
solved and one can find out the
trajectory which is periodic so when a
is equal to zero we'll have a periodic
solution and what we are doing here is
that this is the phase space in fact the
phase plane because it's two dimension
this is x-axis and let us say this is y
axis and when both at lambdas are
positive they the solutions will tend to
zero from all the sides whatever be
whatever be the value of lambda one will
always tend to zero if they are negative
so we say the solution is stable but
then both the lambdas are negative then
we'll go away and away from the solution
and we say solution is unstable so this
is the behavior of this solution whether
we are going towards the equilibrium
point or we are going away from the
equilibrium point if you are going
closer to the equilibrium point we say
the solution is stable when we are going
away from it it is unstable so these are
the cases when lambda 1 and lambda 2 are
real and negative this is the case when
lambda 1 lambda 2 are real and positive
in all these cases I am concern
during that they have to values of
lambda 1 and lambda 2 are distinct but
if it is complex value and the real part
is 0 then we will always have a closed
trajectory and close to victory means
the solution is periodic we start from
this point and we always come back to
that point and this describes a periodic
nature of the solution and whether will
be moving on this curve or will be
moving on this curve that depends upon
the initial distribution or initial
condition specified with the dynamical
system this is the case when the
solution is unstable and this is the
case when we have complex eigenvalues
and if the real part is positive then if
we start from this equilibrium this
position then will go away and away from
the 0 equilibrium point and we say the
solution is unstable so when lambda 1
and lambda 2 are complex and real part
is positive then this is the direction
in which will be moving on the phase
space and we say the solution is
unstable but if we are having a the real
part of the complex root as negative
then if you start from this initial
condition then will move towards the 0 0
equilibrium point this in this manner
and we say the solution is stable so
from wherever we start we ultimately
tend to this 0 0 equilibrium point and
such a point is called stable
equilibrium point so that is how on the
that is how we describe the behavior of
the solution on the basis of eigenvalues
and eigenvectors of the given dynamical
system so we illustrate this for some
simple problems simple physical problems
for this purpose I start with mass
spring damper system so we have a simple
system consisting of a
in which is attached to a spring the
spring constant K and the damper is
present in the system in which the
damping is there which is proportional
to the velocity DX by DT and C the
damping coefficient so we'll have em D
to X DT square plus C DX by DT plus KX
is equal to zero so this simple second
order differential equation represents
mass spring damper system similar type
of second-order equation can be obtained
in RLC circuits where L is the
inductance R is the resistance and C is
the capacitance and E is the external
EMF applied in the circuit so we write
down the difference differential
equation as LD to I DT square plus R di
by DT plus 1 over C I is equal to de by
DT as I being the current in the circuit
now if you compile these two equations
then one can easily see that these two
systems are similar when I is related to
X X is the displacement in this and I is
current in this so the two systems are
identical I guess
ie is the same as X the inductance is
identified with M in this system are the
resistance in this RLC circuit is
identified by C in this math simple mass
spring damper system and one over C C
being kinda C being the capacitance and
this is identified as K in mass spring
damper system in this I want to say that
whatever we may affect of K in mass
spring damper system the sale effect is
observed by 1 over C in RLC circuit
whatever effects thing is having here
same effect R is having in the RLC
circuit to solve this system we first
introduce DX by DT is equal to Y
and this substitution will may will
reduce this second-order equation into
two first order differential equations
so we write down the X by DT is equal to
Y here and D 2 X by DT square will come
out to be minus C by minus KX and that
makes our system as d by DT XY is equal
to 0 1 minus K minus C multiplied by X Y
so a circular system can be reduced to
two first order differential equations
when there is no damping present that
means C is equal to 0 then we have D by
DT into D by DT of XY is equal to 0 1
minus K 0 X n y if you have to solve
this system we say the characteristic
equation is lambda square plus K is
equal to 0 and from here one can find
out that eigen values are purely
imaginary and the on the basis of our
discussion we know that this solution is
periodic solution as the values are
purely imaginary so we can solve we can
get the we can solve and we can get the
solution of differential equation with
the help of eigen values and eigen
vectors so this is the end of this
lecture and in this lecture we have
started with and some diagonalization of
complex matrices then we have discussed
how we can solve the ordinary
differential equations how we can solve
system of ordinary differential
equations what is the role of
eigenvalues and eigenvectors in
predicting the behavior of solution of
such systems thank you