Now, we start with the Concept of Diagonalization.
In this we study introduction to diagonalizations,
mathematical analysis, the role of Vandermonde
matrix and example.
Now, about the introduction to diagonalizations;
in actual practice, we want to design a system
in such a manner that, we should get the desired
performance. So, if you’re having your original
model in that case sometimes it is difficult
to analyze that system.
Even sometimes, it is difficult to design
a controller or internal property of the system
that cannot be easily studied. Therefore,
we need to transfer a system into different
forms in such a manner that, the properties
should not change. So, here we are doing the
same thing, diagonalization means the same
thing that, we have one system and it has
been converted into another form and that
form is very much useful for control system
applications.
So, here the concept of diagonalization - This
approach is a method of transforming a general
state-space model into canonical form. It
is also called a diagonal form because the
system matrix A will be in a diagonal form
after transformation that is, original model
A is someone form. After transformation, it
has been converted into a diagonal form. That
diagonal form is we have only diagonal elements,
remaining elements are 0s. So, how they are
useful?
So, it is useful for the investigation of
system properties. It is useful for the evaluation
of time response. It is useful in checking
controllability and observability and finally,
control design and, and last it is useful
in model order reduction techniques. Now,
we will see the mathematical analysis, here
is state space we have always start with the
state-space model.
The state-space model is represented by X
dot equal to AX plus BU and Y equals to CX
plus DU. Now here is a problem, A matrix,
it is in general form, there are many elements
are involved in A matrix. So, our purpose
is that A must be converted into a diagonal
form. So, X is the state vector. Now what
we want here? We want to convert this state
vector into a new vector. So, this is possible.
So, we assume that A is non-diagonal and we
defined a new state vector K in such a manner
that so, we can write X equals to say M into
K.
Now, what happened we have X is the original
state vector, this K is an additional vector
new vector we have taken and M is some matrix,
model matrix. So, here M is operated on K
we got X, but now here we know X. So now,
in this case this M is nothing but n cross
n matrix. So, after this we can write down
this equation as K equal to M, M inverse X.
So, what we want we want state vector in terms
of K. So, what we can do? We can take K dot
K dot equal to M inverse X dot.
Now, after this, we can replace X dot equals
to AX plus BU in this equation. So, you will
get M inverse X dot equal to AX plus BU. And
after solving this we will get M inverse AX
plus M inverse BU. So, here we have an equation
in terms of K here we have got M inverse AX
M inverse BU. So, here K here is X. So, I
think we have to change it because we want
a complete equation in terms of K because
K is a new state vector. Therefore, what we
can do? We can write K dot equals to M inverse
A. Now what is X? X equals to MK M inverse
BU and we can write down as a M inverse M
K plus M inverse B into U. And now this can
be written as K dot equals to A bar K plus
B bar into U.
Now, we have got equation which is in terms
of K dot earlier it is in terms of X. Now
here A bar equals to M inverse AM and B bar
equals to M inverse B. And so, if you determine
this model matrix by determining different
eigenvectors and make inverse then multiply
AM. So, you will find that this M inverse
AM and this A have the same eigenvalues, same
eigenvalue means that; they have same properties.
So, though so once you got K dot now we can
see about Y output Y, Y equals to CX plus
DU and what is the X ,X equals to M K plus
D into U.
And here we can C bar K into DU. So, we got
B bar equals to M inverse B, A bar equals
to M inverse AM and C bar equals to CM. Now,
here a point has come, how to get the values
of M. M we called as the modal matrix. And
here the one important thing that the M inverse
AM and this A must-have same eigenvalues.
Now we can we want to determine the values
of M. So, what we can do we will assume m
1 m 2 m 3 m n be the corresponding is corresponding
to eigenvalues of lambda 1 lambda 2 lambda
3 up to lambda n.
Here we are saying that lambda 1 is the eigenvalue
and corresponding to these eigenvectors are
m 1, corresponding to lambda 2 eigenvectors
are m 2. Similarly, corresponding to lambda
n eigenvectors are m n. Now we can write down
the equation as A into M. So, A is a matrix
operated on M model matrix. So, we can write
down as A so, M involve eigenvectors m 1 m
2 up to m n. So, we can write down as m 1
m 2 this process repeated and m n.
So now we multiplied A m 1 A m 2 A M n. Now
we have seen that AX equals to lambda X, now
this equation we can also write down in terms
of lambda in terms of eigenvalues. So, we
can write down as lambda 1 into m 1, lambda
2 into m 2, equals to lambda n into m n. And
this can be written in terms of A bar M and
here this A bar equals to M inverse Am. And
M is the model matrix and this M can be written
as m 1, m 2 up to m n.
So, A and this A bar have same cast equations
their eigenvalues are same, therefore, there
cannot be changed in the properties, but here
one thing is has been observed that, A bar
which will come that we can get in a diagonal
form. Diagonal form means only few elements
are present remaining elements are 0. Now,
the problem will come that how to calculate
the eigenvectors, and we will find that the
calculation of an eigenvector sometimes is
difficult. But it is observed that if your
original matrix A it is some specific form.
So, we can easily determine the eigenvector
of this matrix.
That means if you have a system which is represented
by X dot equal to AX plus BU Y equal to CX
plus DU and this A it has some specific form
that is, 0 1 0 say 0 0 1 then here minus a
3 minus a 2 minus a 1. So, we have seen this
type of form earlier this is called controllable
canonical form or we can say as companion
form or phase variable form. So, if you have
A in this form and if you bought the model
matrix M so, how to get it because finally,
we want to achieve this M inverse A into M,
this we want. And this M to calculate the
M that is the eigenvectors are required. So,
to calculate the eigenvectors for A, we have
to calculate eigenvalues for respect to all
eigenvalues we have to calculate the eigenvector.
But if the given matrix A you know this particular
form then we can easily write down the eigenvectors.
So, how you will write down? So, in that case,
if the given matrix is like this. So, model
matrix M we can write down as. So, let us
say this third order. So, we can write down
1 1 1. First row we write down as 1 1 1. If
nth order, we can write down 1 1 up to n.
Now the second circuit row so we can write
down as eigenvalues are lambda 1 lambda 2
and lambda 3.
And now about third, it is lambda 1 square,
lambda 2 square, lambda 3 square this is the
whole matrix. No need to calculate the actual
eigenvectors. So, this M this matrix is called
basically a Vandermonde matrix. So, if you
know M we can calculate M inverse then multiplied
by A into M, we can go we will get A bar suppose,
if your matrix it is in generalized form.
So, A equal to say 0 1 0 0 0 0 0 1 0 0 0 0
0 0 0 1 and lastly let us say minus an minus
a of n minus 1 a n minus 2 up to say, let
us say the eigenvalues of A eigenvalues of
A; it is if it is written as lambda 1 lambda
2 up to lambda n eigenvalues has been written
as lambda 1 lambda 2 lambda n.
In that case, the model matrix M, it is written
as 1 1 1. Then let us write down the eigenvalues
are lambda 1 lambda 2 lambda n then lambda
1 square lambda 2 square lambda n square and
at the end we can write down as lambda 1 n
minus 1 lambda 2 n minus 1 up to lambda n
n minus 1. And so, corresponding to lambda
1 this is the eigenvector corresponding to
lambda 2 this is the eigenvector and correspond
to lambda n this is the eigenvector and as
told earlier this is called as Vandermonde
matrix.
Now, about another point invariance of eigenvalues,
eigenvalues under linear transformation cannot
change. Already we have seen that we have
given system matrix we have come into another
form now, we will see also in another way
that the eigenvalues cannot change if you
use the concept of transformations.
Now, our problem is we have a lambda I minus
A and lambda I minus M inverse AM. And our
problem is that both are 
the same. Now, how do you prove it? So, for
that purpose, we will write lambda I minus
M inverse AM equal to now this I identity
matrix we can write this I as M inverse M.
So, here we can write down as a lambda M inverse
M minus M inverse AM. Now, what we will do
here? We have lambda is there M inverse M
so, M inverse M in both way here it is at
this side and this side. So, we can take it
M inverse one side M on the other side. So,
we can write down this equation as M inverse
lambda I minus A into M.
Then this complete determinant; So, we can
write down as M inverse lambda I minus A into
M, and afterward we can write down this as
M inverse M to lambda I minus A. And therefore,
this is as M inverse M to lambda I minus A,
that is equal to lambda I minus A. So, here
lambda I M inverse M is same as lambda I minus
A; that means, you will find that eigenvalue
under transformation cannot change.
If you do it we saw like some of this example
you will find that A let us say this A matrix.
Then you take this M inverse AM that is, A
bar equals to M M inverse AM. You will get
same eigenvalues lambda 1 lambda minus 2 minus
3. Now the problem is there, if original system
has an eigenvalue minus 4 minus 1 minus 2
3 minus 2 minus 3 then we have come a different
form, what advantage we are getting here because
same eigenvalues what is the issue here. But
you will find that if you have a matrix A
we have minus 1 minus 2 minus 3; that means,
it is sort of type of decoupling effect you
can easily be observed.
So now the details of this we will see later
afterward, but now here we have shown that
by doing some transformations, the eigenvalues
cannot change. And we have original matrix
A and we can differ and another matrix A that
we have got the same eigenvalues. So, only
thing that we have to do some more calculations
ok; now we start solving one example.
Let us say it is equal to 0 1 minus 2 minus
3 X 0 1 U and Y equal to 1 0 X. Now here we
have to use the concept of the diagonalizations.
And we have to determine the model matrix
as well as the diagonal matrix. We will start
doing it. Now the first step we have to determine
lambda I minus A. So, that is given as lambda
0 0 lambda minus 0 1 minus 2 minus 3 and after
solving we will get lambda minus 1 2 minus
1 minus 1 2 and here lambda plus 3.
Now, the determinant of this lambda minus
A that is lambda square plus 3 lambda plus
2 equal to 0; So, we will get the lambda 1
equals to minus 1 lambda 2 equals to minus
2. Now, we have got 2 eigenvalues. Now our
purpose is to calculate the eigenvectors we
have seen earlier that if you have systemizing
in some different form, we can easily write
down the eigenvectors. So, you will find that
in this particular matrix A this matrix belongs
to a companion form or we call as controllable
canonical forms let us see here 0 1 minus
2 minus 3.
So now we can calculate eigenvectors, but
we no need to calculate the eigenvector because;
if this matrix in this form you can directly
write down a Vandermonde matrix. So, therefore,
the Vandermonde matrix is said MV model matrix.
Here we start it with 1 1 and what are the
eigenvalues minus 1 from minus 2 just here
earlier we have seen this concept 1 1 lambda
1 minus lambda 2 minus lambda 3 lambda 1 square
lambda 2 lambda 3 square, but see the order
is, but in our example, order is 2.
So, here 1 minus 1 1 minus 2; Afterward we
have to calculate the MV inverse that is inverse
of the model matrix. Now how to get the inverse
of this model matrix? So, the concept says
MV divided by the determinant of MV and what
is the adjoint of MV that equals to cofactor
of MV transpose divided by determinant of
MV. So now, what is the cofactor of MV? Here
is minus 2 1 then here we will get a 1 and
here we will get minus 1 this is transpose
and divided by determinant of this.
So, determinant of this we will get it minus
1. So, we can write down as these as minus
2 minus 1 1 1, but by minus 1 and a finally,
we can get this result as 2 1 minus 1 minus
1. So, we have got MV inverse, now the next
step what we want? We want MV inverse A into
MV this we want.
Now, this M V inverse A into MV; so, what
is MV? So, MV is 2 1 minus 1 minus 1 we can
check it again here ok. 2 1 minus 1 minus
1 then what is A matrix? A matrix we can check
it 0 1 minus 2 minus 3. So, we get write down
as 0 1 minus 2 minus 3 and what is value of
MV MV is 1 1 minus 1 minus 2. So, we can write
down as 1 1 minus 1 minus 2 we have to solve
it. So, first of all, we solve this and then
multiplied by this. So, you can get 2 1 minus
1 minus 1.
So, here 0 multiplied by 1 1 multiplied by
minus 1. So, we will get minus 1 then 0 multiplied
1. So, 1 multiplied by minus 2. So, we will
get minus 2 then you have a minus 2 multiplied
by 1 minus 3 multiplied by minus 1. So, we
will get it as. So, you will get minus 2 multiplied
by 1 minus 3 we will get as 1 and here minus
2 multiplied by 1 minus 2 plus 6 this we get
4.
So, 1 minus 2 to 1 4 now again we cross multiply
it. So, 2 multiplied by minus 1 and one multiplied
by 1. So, we will get it as minus 1 then 2
multiplied by minus 2 minus 4 1 multiplied
by 4 plus 4 we will get 0. Then here minus
1 multiplied by minus 1 is 1 and this minus
4 multiplied by 1 minus 1. So, plus 1 minus
1 we will get a 0. And similarly here minus
1 multiplied by 2 2 and here minus 4 2 minus
4 we will get as minus 2 we will get MV inverse
AM Mv.
Now, this is called a diagonal form. And we
will find that the, whatever the values here
they are the eigenvalues of the matrix or
eigenvalues of this A bar and these are the
also the eigenvalues of this A matrix.
Now, these are some references. And as I told
you that this diagonalization although we
will find them in a mathematical way, but
it is widely used in an application particularly
controllability, observability also in controller
design. So, we will see later on how this
is useful in control checking controllability,
observability and also in controller design.
Thank you.
