Now what is the advantage of doing all these
kind of things.
Why I am bothering and doing such kind of
a thing you can ask a question.
It is nice it looks nice mathematically, but
why one should bother about it.
See look supposing A is a matrix for that
matrix that we had taken -5 right -5, -7 and
2 and 4.
Suppose I want to compute the 100 power of
that matrix right that was the matrix.
What was the matrix we have.
A was the matrix which we had -5, -7, 2 and
4 right.
These are the and for this matrix A have got
P inverse AP= 2, 0, 0, 3 there exist P such
that this one is right.
I want to compute what is A raise power 100
right.
So from here what is A= I can write A= from
this equation so that is A= 2, 0, 0, 3 on
the right I multiply it by inverse so P inverse
and here I multiply by P is it okay.
So A=this so what is A 100 that is P this
matrix 2, 0,0, 3 P inverse whole thing raise
to the power 100 what is this= looks surprising
what it is= let us compute it for square first
of all and then we will see what is happening.
So let us look at P 2, 0, 0, 3 P inverse square
so what is that= P this matrix P inverse multiply
by itself right.
So what is it P inverse what is that.
So what I will get P this square P inverse
this is 2, 0, 0, 3 what is a square of this
it is 2 square, 3 square 0, 0 right.
So what will be P to the power 100 what will
be A to the power 100 that will be P 2 to
the power 100, 0, 0, 3 to the 100 P inverse.
See the computation of the powers of a matrix
become very easy because that is a diagonal
matrix now on the right hand side.
And for which powers are very easy to compute
and multiplying by I have to only find out
what is P and what is P inverse only 2 matrices.
I see only one matrix and it is inverse right.
So such a result is what is called diagonalizability
of a matrix A. So what we have done is given
a matrix A which was our special one we have
said we can find a matrix P which is invertible
such that P inverse AP is a diagonal matrix.
A diagonal matrix is known a matrix P is also
known right.
So the question comes for what kind of matrices
one can do that.
For this particular one we are able to do
it, but we may not be that lucky so we have
to find a condition which ensures if this
matrix satisfies this property then it can
be diagonalized not only it can be diagonalized
that means I can find a matrix P such that
P inverse AP will be diagonal and I know what
is a diagonal matrix also right.
So those are the scalars which are going to
appear on the diagonal are nothing, but determinant
of A- lambda I=0 solution of that.
So it becomes important to solve determinant
of A- lambda I right so let us write that
as a definition.
So these are the question so let us write
lambda is called an eigenvalue.
So such a scalar is called an eigenvalue if
there exist a non-zero vector oh gosh it is
a typo.
There is a vector X such that AX= lambda X.
So if there is a non-zero vector which is
scaled by that vector by the matrix A then
that is called an eigenvalue for and this
vector is called an eigenvector for that eigenvalue
it is called an eigenvector.
And just now I said for a given Eigenvalue
in this example the previous example Eigenvalue
is there too, but there are many eigenvectors
possible for you if one is a eigenvector scalar
multiple of that also is a eigenvector right
the same eigenvalue.
So there could be more than one actually we
can put them all together in one box you can
call it as Eigen subspace of with respect
that Eigenvalue so we will come to that slowly.
So that is called an Eigenvector.
So the problem is this is what you called
the Eigenvalue problem we call it and given
a matrix A find its Eigenvalues and find corresponding
Eigenvectors right at least the process is
quite clear.
There is nothing we have already done one
example of that.
So what is a process take the matrix A it
is a square matrix look at for a scalar lambda
look at A- lambda I because you want AX= lambda
X.
So look at A-lambda I that matrix you want
a non-zero solution for that right with that
matrix as a coefficient matrix A- lambda I
apply to X=0 for a non-zero X you want a non-zero
X that means you are looking for a non-zero
solution for the homogenous system that is
true if and only if the matrix is not invertible
right.
Nullity should be at least > or=1 right the
rank should not be full that means the matrix
should not be invertible.
And in terms of determinant that says determinant
of A- lambda I should be=0 right.
So let us give that determinant of A- lambda
I also a name okay.
So that is called the characteristics polynomial
of the matrix.
Given a matrix A look at determinant of A-
so what will be A-lambda along the diagonal
you subtract lambda right everything else
remains as it is.
So determinant of this what it will look like
try to expand.
See in the example 2/2 it was a quadratic
in 2/2 it came out as a quadratic lambda square
-right whatever that was.
So here let us says suppose you expand by
this right you expand by first column I am
saying so this will be gone and this will
be gone.
So we will have a11- lambda into determinant
of the remaining right.
So when you write determinant the remaining
again we will have to expand by that.
Let us again expand by this column then it
will be a22- lambda coming out as a factor
into determinant of the remaining one.
So what will be the highest power of lambda
that will come lambda to the power n right+
some lower powers+ some lower.
So essentially we are not exactly proving
it, but giving you a hint that intuitive we
are saying that this determinant is a polynomial
in lambda of degree n.
So the characteristics polynomial is a polynomial
that is why it is called a characteristics
polynomial because it actually turns out to
be polynomial and it is characteristics of
that matrix is a property of the matrix is
a determinant A is a polynomial of degree
n.
And what will be the coefficient of the highest
term lambda to the power n it will be- lambda
or - lambda - lambda it will be -1 to the
power n that factor will come out anyway right.
So it will be lambda 1 to the power n* lambda
n+ something, something it will be=0 right.
If you want the problem is how to solve that
polynomial find the roots of that polynomial.
These are the values which are going to be
called as the Eigenvalues of the matrix right.
So let us look at this so the roots of this
are the eigenvalues right.
Roots of the characteristics polynomial of
the Eigenvalue.
Now here is a theorem in algebra called fundamental
theorem of algebra which says here the coefficient
are all going to be real right real matrices,
but the roots of a real polynomial may not
be always real they can be complex right.
So the characteristics polynomial will be
a polynomial with real coefficient.
It may not have any real root right all the
roots maybe complex.
So Eigenvalue need not exist that is a possibility.
So for a given matrix the Eigenvalue no Eigenvalue
may exist.
So gone case for that you can do nothing you
cannot diagonalize that matrix at all, you
are gone at the first step itself.
It may have Eigenvalues right some of the
roots may be real, some of the roots maybe
complex how many all will be there total is
a polynomial degree n.
So a fundamental theorem of algebra says there
will be n roots for this.
Some of them maybe real some maybe complex,
but the complex roots always occur in pairs
right.
So if it is a polynomial if the matrix n is
of odd order then at least one real root will
exist right.
So then a hope is there you start with that
at least.
So that is coming from algebra basically okay.
So let us we will slowly come to so a root
there is something called the multiplicity.
If a root is repeated right then how many
times it is repeated that is called the multiplicity
of that root.
So Eigenvalue may or may not exist it may
exist with some multiplicity that are possibilities.
Now let us look at this matrix looks slightly.
If you calculate it I think this example just
to show you computation determinant of this
you get lambda=2 okay.
How do you get for a cubic how do you solve
a cubic even for a 3/3 matrix finding roots
is not a easy job.
There is no definite algorithm like for a
quadratic.
For a quadratic you know, that gives you all
the roots real or complex whatever it is for
a quadratic for a cubic one does not know.
So the hope is you find one by hook or crook
divide by that and get a quadratic and try
to solve it kind of thing.
For 4th degree there is no hope at all right.
It is very difficult to find roots of a polynomial
okay, but here anyway for this by just looking
at it you can see that there is a possibility
of saying that there is a root lambda=2.
Sometimes if I can look at 1-1 2-2 and something
this is for just the sake of examples nothing
more right.
In life in mathematics life is not easy to
find roots of a cubic even it can be tricky
right.
So here you find 2 is root and all other root
are negative so probably one should.
Let us look at only for 2 what is a Eigenvalue
eigenvalue 2 what is a.
So I am just giving a process for lambda=2
to find an Eigenvector.
This you will do for all Eigenvalue A-2I apply
to a vector X=0 so you apply that you get
the homogenous vector right homogenous system.
So how do you solve reduce it to the row echelon
form right.
So you get this right.
You see all along in all our course till now
the only one idea which is important what
is the single most important idea.
Given a matrix find its row echelon form find
it reduced row echelon form everything comes
back to that right.
Everything gets related in that because that
is a way you will compute things on machines
okay.
So find this so that means what if this is=C
so what you get x, y, z you solve it right.
This decimal you can remove so you will get
25 z, z is arbitrary there is a one row which
is right rank is less one row is 0.
So what is our variable you will get arbitrary
value.
The third one z right you give z arbitrary
value you find xyz so find y in terms of z
from this equation put back the value you
get the value of x so comes out= to this.
So you can take z out okay so that is a scalar
okay times 25, 5 and 1 so that is a eigenvector
for eigenvalue.
So the Eigenvalue for lambda=2 is this is
an eigenvector so is clear.
How to find Eigenvalues and Eigenvectors corresponding
to that solve the characteristics polynomial
and for each root solve the corresponding
homogenous system okay.
Here is a roots of see what is a determinant
of A it is same as determinant of A transpose
right.
So finding the characteristics polynomial
for A is same as finding characteristics polynomial
for A transpose Eigenvalues will be same,
but Eigenvectors may not be the same for both
because you will be taking A- lambda I or
A transpose -lambda that will change.
So for computation of Eigenvalues you can
replace A by A transpose if you want.
So here is one example of that so look at
this matrix okay some funny looking matrix
to look.
We want to find a vector u so that u times
B=u what does that mean?
It does not seem to be related with Eigenvalue
right, but supposing you take transpose of
this equation what you will get B transpose
of light to u transpose= u transpose that
means for B transpose you are saying that
for an eigenvalue=1 find an Eigenvector that
will be u transpose is it okay.
Once you have found u transpose you can find
u is it okay so that is the idea of application
of that.
So B apply to here is some observation which
is not really important, but for this matrix
if we look at the transpose of this so what
is the sum of the rows of this matrix 0.8,
0.1, 0.1 that is 1 some of the rows=1.
So if we take transpose some of the columns=1
such matrices are called stochastic matrices
whose column each column add up to 1.
They basically arise in probability theory,
statistical analysis because the probabilities
are distributed so total=1 essentially is
something like their probability of something
happening kind of thing and you want to this
is called the stochastic matrix because it
describes the system in probability and there
you will like to know if you apply that process
again and again what happens.
So you will be required to compute the powers
of that matrix and that is why for such a
matrix diagonalizability becomes important
when can you diagonalize and one shows a theorem
that for a stochastic matrix number one is
always an Eigenvalues that can be proved as
a theorem.
We are not bothered about that.
So here one is an Eigenvalue so we look at
B transpose so B transpose-1 I v=0 solve that
system right.
So that will give you the vector v and what
is that vector v that is the transpose of
the vector u that you are looking for right.
So this was basically just brought into say
that for determinant of A is same as determinant
of A transpose that can be used in finding
Eigenvalues of a vector.
So that is what you find solve that computation
and you find you can choose if you want a
unit vector.
These are eigenvector right 0.2, 0.4, 0.1
okay.
If you do not want decimal you can multiply
everything by 5 that also is an Eigenvector
so that is 1 to 5 you can normalize that If
you want.
If you have a normalized Eigenvector which
is non-zero anyway divided by norm you will
get a Eigenvector of length 1 right.
And this is what is important you go and applying
there is a something called equilibrium of
a stochastic system and after some stage it
does not change the system at all you have
reached equilibrium in probably theory so
let us do not bother this line okay.
So important thing is for the Eigenvalue lambda=2
and Eigenvalue lambda=1 for a stochastic matrix
you will find the Eigenvector so this has
importance in other subjects.
So basically let me summarize finding characteristic
polynomial and its roots Eigenvalue problem
is basically.
Find a characteristics polynomial, find its
roots, find its eigenvector by solving the
resulting singular homogenous system right
that is so as I said you can put them altogether
in one box the null space of A-lambda I that
depends on lambda Eigenvalue that is null
space of this matrix.
You can also call it as Eigen subspace right
vector space subspace corresponding to the
Eigenvalue lambda.
Now there is something which probably will
bother next time.
So given even if a lambda an Eigenvalue exist
there is a Eigenvalue that means you have
to find a Eigenvector for that, but sometimes
one allows even complex Eigenvalues the complex
roots which are coming for the characteristics
polynomial.
Even one takes into account that there are
then the corresponding matrix that you will
get the solution A- lambda I then you do not
restrict yourself to real entries your entries
will change to complex also right.
So you may get a vector X which is a complex
vector.
Complex vector means what entries are complex
number right.
So instead of looking at matrices over real
one start looking at matrices over complex
numbers also because it is forced by that
problem of Eigenvalue problem right.
So we will start looking at this next time
more.
Let us stop here I think.
