Now, we start with Stability Analysis in State
Space. In this first part is the Concept of
Eigenvalues and Eigenvectors.
In this Eigenvalues and Eigenvectors, we study
introduction to Eigenvalues and Eigenvectors,
Definition of Eigenvectors, Definition of
Eigenvalues, Properties of Eigenvalues and
numerical examples. Now, introduction to Eigenvalues
and Eigenvectors, the most important thing
is that this Eigenvalues and Eigenvectors
are useful in the control system.
In control system, we have discussed earlier
the control engineer has to perform modeling,
then he has to check stability, then he has
to design the controller. For all these issues,
Eigenvalues plays a very important role. If
there is no Eigenvalue, no control system
or we can, in other words, we can say Eigenvalues
and Eigenvectors in the heart of the control
engineering. And now, what is control engineering?
Control engineering is applicable to all types
of systems. It may be electrical mechanical,
civil, computer. Therefore, Eigenvalues plays
an important role in control system.
So, these Eigenvalues plays an important role
in all type of systems. For example, if you
take a rigid body and if you move up, so,
it will rotate. So, it has some axis of rotations.
So, this axis of rotations is called Eigenvector
and it is the movement of inertia is called
Eigenvalue. If you take, if we can consider
another example that is if you consider the
vibration of mechanical systems, in that case,
the natural frequency of vibrations that is
called Eigenvalue and it is particularly,
we can say the mode of this vibrations this
is called Eigenvector.
So, this Eigenvalues and Eigenvectors are
usefully everywhere, where there is the system.
Now, how to define this Eigenvalues and Eigenvectors?
Now, we will see the definition of Eigenvectors.
Suppose, A x is equal y where A is n cross
n matrix, x is n cross 1 matrix, y is n cross
1 vector. From above equation, we can say
that the n cross n matrix operator A operate
on n cross vector x we get new transform n
cross 1 vector y. That is, we have a matrix
which is of order n cross n, when it is operate
on x which is of order n cross 1, we can get
y which is of order n cross 1. That is A operated
on x, we can y for example, if I will take
the example.
Let us say A equal to 1 2 3 4. What is the
order of a matrix is 2 cross 2. And now, this
has to operate on x, let us say x is vector
that is n cross 1, that is said here 1 2.
Now, this x, so, order is 2 cross 1, two rows,
one column. Now, what output will get output
y is nothing but the rows what is rows, these
two rows, 1 and 2 operate on column this column.
So, what you get here 1 2 plus 2 4 will get
5 and here this row, operate on this will
get 3 plus 8 11. That is y will get as 5 11
Now, another definition, Eigenvectors of system
matrix A are all vectors x i not equal to
0 which under the transformation of matrix
A becomes multiples of themselves; that means,
here Eigenvectors are non-zero. They are never
0; that means, here it is mentioned that a
x I, there is an operation on this vector
which is nonzero become multiples of this
themselves; that means, we can write down
as lambda into x i. A x i equal to lambda
x i; that means, here this what is happening
here a matrix operated on x and we get the
new vector.
So, we will find that in that case, you to
see what is the directions. So, it is along
same directions.
So, here it is shown Eigenvector, that is
third definition Eigenvector can also defined
as vector x such that matrix operator, that
is A matrix transforms it to vector lambda
x where to earlier. That is A x equals to
lambda x. So, here it is mentioned the same
thing to vector lambda x this vector has the
same direction is state space as the vector
x. If you take any variables, it has a magnitude
and directions ok.
So, after transformation, if direction remains
the same, then that is called Eigenvector.
If the direction is not same then, we cannot
call as an Eigenvector. For Eigenvector this
vector should be always nonzero, it should
not be 0; that means, here I am showing, so,
here x 1, x 2. Now, here this A x and now,
let us say lambda x. So, this is not valid
that is this a x and lambda x must be in the
same directions. So, here I am showing it
again, so here, x 1 x 2.
Now, here I am showing A x and I can say it
is lambda x, then this is called Eigenvector,
this is correct.
Now, what are the definitions of Eigenvalues?
Actually, Eigenvalues is nothing but a combination
of German and English. So, because here there
are two words, Eigen and values, therefore,
here Eigenvalues is the combination of German
and English word. In this, German word is
Eigen and English word is value, English word
is value. The meaning Eigen is the characteristics
and meaning of value the characteristics of
the system, value is characteristics of the
system and whereas Eigen is nothing but the
characteristics.
Now, how to define Eigenvalues? Eigenvalues
of a matrix A are the values of lambda that
satisfy the equations A x i into lambda x
i and as I told, x i should not be equal to
0.
So, this is the definition of Eigenvalues
and here lambda I is nothing but the Eigenvalue;
that means, A operated on x, that is sketching
of the vector and it could be the lambda;
that means, sketching vectors is been sketching,
length had been increased that is basically
an Eigenvalue. So, the factor which helps
in sketching that is called Eigenvalue and
the x is the Eigenvector. Now, why the Eigenvalues
of the system matrix as defined as A minus
lambda I equal to 0 means whenever you calculate
the Eigenvalues.
Why it is said that A equals to lambda I.
So, further purpose, what we do? We write
this A x i equal to lambda i into x i. Now,
this equation, assuming an identity matrix
A x i into lambda i into I into x i, then
write this equation A minus lambda i into
x i, lambda into I equals to 0; that is a
minus lambda into x i equals to 0. We have
to replace all the elements on the one side
and it is said that A i minus lambda equals
to 0, that is the way to calculate the Eigenvalues,
but we should know, why it is so.
So, for that purpose, we not taken the same
equation and then I have solved it. I have
compared. Now, how to do this? How to move
further? So, now, what I have done this x
i equal to A minus lambda i into I inverse
into 0. Now, here we have got 0 and this is
the inverse. So, x i become zeros, but the
condition of Eigenvalues that x I should be
never 0 even earlier definition we have seen,
Eigenvector should never be 0, it should be
always some value.
But if you see the literature, it is said
in order to calculate Eigenvalues, we have
to make a minus lambda I equal to 0, but here
we have got these equations. Therefore, what
we can do here if you write this equation
a minus lambda I inverse. So, it can be written
as adjoint of a minus lambda I divide by determinant
of a minus lambda I into 0 there is x i. So,
now, here this in order to have x i to be
non-zero, what is the possibility because
this cannot be 0, if this is 0, this total
product is 0; that means, if this factor is
0, then 0 by 0 cannot be defined or determined.
And therefore, in order to check the stability
of the system or the Eigen particularly, we
can say Eigenvalues, we have to check A minus
lambda I equal to 0.
So, now, we have to see some properties of
Eigenvalues. First, for any general matrix
if the coefficients of the matrix A are real,
then it is Eigenvalues are always real or
complex conjugate pair; that means, here if
I am saying A is matrix, it has some elements
1 2 3, 4 5 6, 7 8 9 these are real elements.
So, Eigenvalues of A, if you calculated like
this lambda I minus A. So, that Eigenvalue
can lie on the real axis or it may be complex
conjugate. So, any possibility will be available
when, when you are given a matrix is real
and it is general form. Now, we coming to
real symmetric matrix. If matrix A is real
symmetric, what is real symmetric the row
and column elements are same; that is if you
say take any matrix?
If the row elements and the column elements
are the same, that in that case that matrix
is called really symmetric matrix. And when
the matrix real symmetric, their all Eigenvalues
a lying on the real axis, there will be no
complex conjugate Eigenvalues here it is written.
A matrix a is real, symmetric row and column
elements are same, then the Eigenvalues are
always real no complex conjugate Eigenvalues.
For example, if I write this for this part
here write 1 4 6 this is row. Now, it takes
elements here 4, 6, then, I can write 5 it
is 6, 8 you can take 8 here and 7; see here
1 4 6, 1 4 6, 4 5 8, 4 5 8, 6 8 7, 6 8 7.
So, this is real symmetric and in that case,
the Eigenvalues of this are lying on the real
axis only. That is this point, Eigenvalues
of a matrix A and it is transposed.
The Eigenvalues of A and it is transposed
matrix are always the same; that means, if
you take the A, let us say A matrix has Eigenvalues
lambda 1, lambda 2, lambda 3 and if you make
transpose of this, so, will we will always
get the Eigenvalues lambda 1, lambda 2, lambda
3. Eigenvalues of matrix say and it is inverse
that is A inverse.
Suppose if you matrix A is Eigenvalues lambda
i equal to 1 2 3, up to n just like this is
and if you make it A inverse, so, A inverse
is also Eigenvalues. So, in that case the
Eigenvalues are nothing but one upon lambda
i where i equal to 1 2 3, up to n that is
this part.
Eigenvalues and determinants: The product
of the Eigenvalues of a matrix equals the
determinants of the matrix. See here, if what
are the Eigenvalues are there if you multiply
it that is nothing but the determinant of
a matrix. Then trace Eigenvalues and diagonal
elements. In the previous part, we were seen
that using the trace of the matrix we can
determine the transfer function model of the
system.
So, here the same trace, the trace is nothing
but equal to sum of Eigenvalues that is what
let us say lambda 1, lambda 2, lambda 3, up
to say lambda n. Similarly, sum of diagonal
elements is also equal to the trace of the
matrix that is if you take a trace of the
diagonal elements; let us say a 11, a 22,
a 33, a nn. So, this concept, the trace equal
to sum of diagonal elements we have already
use in determining the transfer function from
a given state-space model.
Then Singular matrix and Eigenvalues; matrix
is singular, if and only it has 0 Eigenvalues
matrix is singular. Singular means, the determinants
is 0. Let us say this determinant delta and
let us say these are Eigenvalues, lambda 1,
lambda 2 up to lambda n; what is say it has
0 Eigenvalue. So, 0 Eigenvalues why, because
if singular is there, open delta equals to
0, the delta means determinants equal to 0
and this is only possible when one of the
Eigenvalue equals to 0, lambda 2 equal to
0 or lambda n equals to 0. Therefore, it is
shown that matrix is singular, if and only
it has 0 Eigenvalue.
Now, 0 Eigenvector and 0 Eigenvalue, the Eigenvalue
can be 0, but Eigenvector cannot be a 0 vector;
that means, we are in A X equal to lambda
X.
So, what is said Eigenvalue can be 0, but
Eigenvector never be 0; that means, theoretically,
it is possible that A 0 into lambda 0, but
what about the actual things because actually,
Eigenvalue can be 0, but Eigenvector never
be; for further purpose let us say if I have
taken matrix as 1 2 and here you can take
2 4. So, we can write down this as here minus
2 1 equal to. So, what will get, you will
get it as this is X, this is A. So, here A,
X is being multiplied.
So, it is minus 2 plus 2 and here we will
get minus 4 plus 4. Now, here it is 0, 0.
So, now, x we have taken here. Now, what we
can write we can write lambda X. So, we can
get lambda X equals to 0. So, now, we can
write lambda and what is X, X is minus 2,
1; that equals to 0, but this is not 0, this
x is non zero; that means, possibilities lambda
equals to 0; that means, Eigenvalue can be
0, but this is not true for Eigenvector. The
same Eigenvector cannot be associated with
different Eigenvalues.
Suppose if you taken A X equal to say lambda
1 X and then we can also take A X equals to
lambda 2 X, assume that X is the Eigenvector
for the lambda 1, X is a Eigenvector for lambda
2. Now, if you equate it, what will get lambda
1 X equal to lambda 2 X, then if you solve
it, will get lambda 1 minus lambda 2 into
x equal to 0 and what about these X? X is
Eigenvector, it can never be 0, and Eigenvector
cannot be 0.
So, a possibility is that then lambda 1 minus
lambda 2 equal to 0; that means lambda 1 equal
to lambda 2. But we have taken different Eigenvalues;
that means, this is not possible therefore,
we have written the same as Eigenvector cannot
be associated with different Eigenvalues.
Now, we will solve the example and from this
example, we will see how to determine the
Eigenvalue and Eigenvector.
Now, we have taken an example, a matrix A
equal to 0, 1, minus 2, minus 3. This is matrix
and our problem is, we had to calculate the
Eigenvalues of this matrix.
So, how we determine this? So, first of all
we write this A matrix as lambda I minus A
, lambda a minus A, here, I is the identity
matrix. So, hereafter replacing identity matrix
and lambda we get lambda minus 1 2 lambda
plus 3. Now, what is the next step? We have
to take the determinant of this matrix. So,
determinant of lambda minus A, that is equal
to you solve it lambda into lambda plus 3
that is this multiplication minus this is
minus 2 equal to 0. So, we will get lambda
square plus 3 lambda plus 2 equal to 0. So,
if you solve it, we get lambda plus 1 lambda
plus 2 equal to 0.
So, lambda equals to minus 1 lambda equals
to minus 2; that means, we have two Eigenvalues;
one is a lambda 1 equals to minus 1 and lambda
2 equals to minus 2. These are two Eigenvalues.
Now, our main purpose is to determine the
Eigenvectors. Now, how we determine the Eigenvector?
Now, first of all, we determine the Eigenvector
for lambda 1 equal to minus 1. So, with equation
available to us, we have to take the same
equations that are lambda 1 I minus A into
X equal to 0 lambda 1 because earlier we have
taken this lambda, but now, we have got two
Eigenvalues lambda 1 and lambda 2.
Therefore, I have change lambda 2, lambda
1. So, lambda 1 I minus a multiply by X equals
to 0. So now, we write equations as lambda
1, 0, 0, lambda 1 minus A, what is A matrix?
A matrix is 0, 1, minus 2, minus 3 complete.
Now, how many states; X 1, X 2 equal to 0,
0. Now, we have to solve these equations.
So, after solving, what will get will get
equation as lambda 1, minus 1 2 lambda 1 plus
3, X 1, X 2 equal to 0, 0. Now, in this equation
we replace lambda 1 equals to minus 1.
So, this lambda 1 equal to minus 1 we have
to replace. So, what we will get? We will
get it as minus 1 minus 1 2 2 and X 1, X 2
equal to 0, 0. Now, if you solve it, what
you get minus X 1, minus X 2 equal to 0, second
2 X 1 plus 2 X 2 equal to 0; both are the
same equations. That means, if you take X
1 equal to 1 and what is the X 2. In that
case, X 2 equals to minus 1.
Because here X 1 plus X 2 equal to 0, therefore,
lambda 1 equals to minus 1. Our Eigenvectors
are X 1, X 2 that is equal to 1, minus 1,
that is X 1 is 1, X 2 equals to minus 1. So,
Eigenvectors are 1 minus 1. Now, next what
you want? We want Eigenvectors when lambda
1 equal to minus 2.
Now, for lambda 2 equal to minus 2, so, we
write the equation as lambda 2, minus 1 2
lambda 2 plus 3, X 1 X 2, 0 0 and now, replace
lambda 2 equal to minus 2 in this equation.
So, what will get? We will get minus 2, minus
1 2 1, X 1 X 2 equal to 0 0. Now, we solve
this equation. So, after solving we will get
as a 2 X 1 minus X 2 equal to 0 and second
equation 2 X 1 plus X 2 equals to 0; that
means, these equations for this.
And this is for this and we will find that
again. They are the similar equations. That
means 2 X 1 plus X 2 equal to 0. Suppose,
if you take X 1 equal to 1, so, what is the
X 2? X 2 equal to minus 2. Therefore, when
lambda 2 equals to minus 2 your Eigenvector
is 1 minus 2. Therefore, for lambda 1 equal
to minus 1, we have got Eigenvector as 1,
minus 1 and when Eigenvalue lambda equal to
minus 2, we have got Eigenvector as 1 minus
2.
So, this is a second-order system. We have
solved easily, but if you, if you increase
the order of the system, this process it quite
cumbersome. Therefore, we want method we can
give easily the Eigenvectors. So, the literature
this one method which is called cofactor method
which you have already seen in the linear
algorithm that that methodology can be useful
to get the, we can say the Eigenvector of
system or Eigenvectors can easily be determined
by the concept of the cofactors.
So, how to determine it? So, further purpose,
I am taking in the same example. See here,
for lambda 1 equals to minus 1, we have got
this equation and after that we solve. So,
what we are doing; we are not doing these
steps. So, what we are doing here when lambda
1 equals to minus 1, we have got a matrix
as minus 1 minus 1, 2 2, just you can check
it; so minus 1 minus 1, 2 2. Just you can
see here, minus 1 minus 1, 2 2.
Now, we want the Eigenvector for this one.
So, what we can do, the Eigenvector. Now,
the cofactor means we have to take the minors.
So, for minor of this 1 is 2 whereas, the
minor of this is 2, but because here it is
in the second-row first column, so, it is
odd. So, will get is as minus 2; that means,
we divide by 2 both, this 2 and 2 will get
1, minus 1. So, you find that you get the
same 1, minus 1 say, without doing any calculations,
we have directly get the Eigenvector. Similarly,
when lambda 1, say lambda 2 equals to minus
2. So, what matrix we have got we have got
matrix is that is we have got minus 2, minus
1, 2 1. Now, we want Eigenvector for this.
So, what we do? We have do the principle minors
of all this, minus 2, it is 1 and for this
2, but because here 2 plus 1, 3 this odd number
that is 2 rows first column, therefore, it
becomes minus 2; 1 minus 2 1 minus 2 1 minus
1 here, 1 minus 1; that means, we have easily
determined the Eigenvectors of a given system
matrix.
Now, here we observed that here Eigenvalues
are distinct, but it may possible that sometimes
the Eigenvalues are repeated. They in that
case, how to calculate the Eigenvector? Now,
we start solving the example of generalized
Eigenvector.
So, I have written a matrix A as an equal
to 0, 1, minus 1 minus 2. So, if you solve
it. So, what will get lambda I minus A, if
you solve it, will get it as lambda square
plus 2 lambda plus 1 equal to 0. So, we will
get it as lambda 1 equals to minus 1 and lambda
2 equals to minus 1.
So, these are two Eigenvalues. We have got
for this matrix 0, 1, minus 1, 2 1, minus
1 minus 2. Now, but Eigenvalues are repeated.
Now, how to proceed further to get the Eigenvector
for such type of problems? So, for that purpose,
what I am doing here will take. So, here lambda
2 we have taken, we can say, lambda 1 equals
to lambda 2. Now, how will we proceed further,
so, what I will do here? I will write equation
as lambda 1 I A equal to lambda 1 minus 1
1 lambda 1 plus 2. Next, we write this as
for lambda 1 equals to 2 say minus 1. So,
when what will do; when lambda 1 equals to
minus 1, we can write down this as minus 1
minus 1 1 1. Now, in order to get the Eigenvector,
what will take? We will take this part as
a cofactor part.
So, we can write down as 1 and for this, we
can write down minus 1 that is; that means,
we have got Eigenvector for lambda 1 equals
to minus 1. Now, the problem is that we want
to determine the Eigenvectors for another
one. So, what we can do here, we can take
what we do we can, we can for lambda 1 equals
to minus 1, this cofactor for this is lambda
1 plus 2.
And cofactor for this minus 1 is minus 1;
that means, for lambda 1 equals to -1, by
conventional method, we have got it. But to
get the Eigenvector for another lambda 1 which
is equal to minus 1, what we have done? From
the physical variable itself, I have to lambda
1 plus 2 and for this minus 1 I have to minus
1. Now, what we do? I will differentiate this
equation that is what we will do differentiation
with respect to lambda 1 that is lambda 1
plus 2 and differentiation of lambda 1 with
respect to this minus 1.
So, if we differentiate it, so, what we will
get. Here you get 1 and this is 0, 1 0. This
is Eigenvectors for lambda 1 equals to minus
1 that is second Eigenvalue and original Eigenvalues
lambda 1 equals to minus 1. We have got Eigenvector
as 1 minus 1; that means, whenever we need
to determine the Eigenvectors, what you do?
We have to write the equation in terms of
generalized form, then express this, write
the minors and that minors.
We have to differentiate and then replace
the values you can get the Eigenvectors for
a repeated Eigenvalues. Now, there are some
references, I.J. Nagrath and Gopal and N.S.
Nise.
Thank you.
