Hello friends, so, welcome to the lecture
on Generalized Eigenvectors and Jordan Canonical
Form. So, as we know that, if a matrix of
order n by n having n linearly independent
Eigen vectors, then the matrix can be written
as P into D into P inverse or in other way
other words we can say, the matrix is diagonalizable
or the matrix is similar to a diagonal matrix,
where the diagonal entries are the Eigen value
of the matrix A.
So, if the matrix is not diagonalizable means
the matrix does not have n linearly independent
eigenvectors. Then, how to find out a similar
type of transformation so that, at least we
can write this matrix as P J into P inverse
where J is a block diagonal matrix. So, I
want to say that, if matrix is diagonalizable
then, I can write P into D into P inverse
where D is a diagonal matrix. And if matrix
is not diagonalizable, then write it P into
J into P inverse, where J is a block diagonal
matrix. So, it is a generalized similar transformation
where we are reducing a given matrix into
block diagonal matrix.
So, for doing this, we have to talk about
Jordan blocks. So, the definition of Jordan
block is a Jordan block corresponding to a
given eigenvalue lambda is a k by k matrix
with lambda on the main diagonal and one on
the super diagonal.
So, for example, if I want to write a Jordan
block of size 1 with respect to eigenvalue
lambda equals to lambda 0. So, j 1 lambda
0 is given as by this matrix. If I want to
write a Jordan block of size 2 corresponding
to eigenvalue lambda equals to lambda 0, then
it will be lambda 0 lambda 0 in the main diagonal.
So, 1 in the super diagonal and 0 will be
below.
.
Similarly, a Jordan block of size 3 corresponding
to eigenvalue lambda equals to lambda 0 can
be written in the same way. So, lambda 0 1
0 0 lambda 0 1and 0 0 lambda 0 here, you can
notice that lambda 0 are in the main diagonal
and 1 is in the super diagonal. So, in the
similar way, a Jordan block of size k corresponding
to eigenvalue lambda equals to lambda 0 can
be written as lambda 0 1 0 0 lambda 0 1 0
0 0 lambda 0 1 and finally, 0 0 0 lambda 0.
So, it is a k by k matrix. So, this is the
definition of Jordan block of different size.
So, if someone ask you write a Jordan block
of size 2 corresponding to eigenvalue lambda
equals to 3, so, it can be written as size
2 eigenvalue is 3 so, it is given as 3 1 0
3. Now, we will see some properties of Jordan
blocks; so, a Jordan block has only 1 eigenvalue
lambda equals to lambda 0. So, for example,
if you are having this particular matrix,
it is a Jordan block of size k. So, the eigenvalue
of this will be lambda equals to lambda 0
with algebraic multiplicity k.
So, algebraic multiplicity of this will be
k and the determinant of a Jordan block of
size k corresponding to lambda equals to lambda
0 will be given as lambda minus lambda 0 raised
to power k. The geometric multiplicity of
lambda equals to lambda 0 of a Jordan block
of size k will be 1 means, there will be only
1 linearly independent eigenvector corresponding
to a given Jordan block whatever be the size.
And third important property is, if e 1 e
2 e k denotes the standard basis in a k dimensional
vector space, then j k lambda 0 e 1 is given
as lambda 0 e 1 and j k lambda 0 e i is given
as lambda 0 e i plus e i minus 1 where i is
varying from 2 to k. So, suppose I want to
find out j k lambda 0 e 2. So, it will become
lambda 0 e 2 plus e 1.
Now, after learning about Jordan blocks, let
us divide the define the Jordan canonical
form. So, a Jordan canonical form is a block
diagonal n by n matrix given like this. So,
here, basically what we are having? We are
having these m Jordan blocks corresponding
to eigenvalues lambda 1 lambda 2 up to lambda
n and respective size are k 1 for the Jordan
block corresponding to eigenvalue lambda 1
k 2 is the size of the Jordan block corresponding
to eigenvalue lambda 2 and so on.
So, in this way, if a the matrix is n by n
matrix, then k 1 plus k 2 plus up to k m should
be equals to n. Please note it is a block
diagonal matrix and all these are 0 blocks.
So, block diagonal and we are having 0 blocks
above the block diagonal and below the block
diagonals. So, here we are having m Jordan
blocks as I told you and this complete matrix
is called Jordan canonical form. So, and if
I know the algebraic multiplicity and geometric
multiplicity of different eigenvalues for
a given matrix, then I can write the Jordan
canonical form of that matrix that we will
take some example of that.
Now, as you can see, the determinant of this
Jordan met canonical form is given as lambda
1 minus lambda raised to power k 1 lambda
2 minus lambda raised to power k 2 and so
on up to lambda m minus lambda raised to power
k m and this can be easily obtained by using
the concept of finding the determinant of
the block diagonal matrix where the determinant
of the complete matrix will be the product
of determinants of different block matrices.
Similarly, corresponding to each block as
I told you, when I was defining the Jordan
blocks, that corresponding to each Jordan
blocks there will be only one linearly independent
eigenvector. So, in that way, I will be having
this particular Jordan canonical form will
be having m eigenvectors given as X 1 X 2
X 3 up to X m each corresponding to lambda
on the Jordan blocks and this can be prove
by the method of induction.
Now, we are coming to a very important theorem
and this tells us about this is called Jordan
canonical form.
Similar to a transformation so, this particular
transformation tells us that every square
matrix of order n is similar to a Jordan canonical
form j of the similar size that is if A is
a matrix of order n by n then a is similar
to a Jordan canonical form J such that A can
be written as S J S inverse. So, where S is
the matrix containing the eigenvectors and
generalized eigenvectors of A.
Now, please note it here; if A is a diagonalizable
matrix, in that case, A will become P J P
inverse where P comes from the eigenvectors
of A because, then if A is diagonalizable,
A will contain n linearly independent eigenvectors
and if I write those eigenvectors as columns
of P, then I will get the model matrix P and
J will be D where diagonal entries are the
eigenvalues of A and P inverse. So, if A is
diagonalizable. So, this Jordan canonical
transformation will become diagonalization.
So, we can consider Jordan canonical transformation
as the generalization of classic diagonalization
transformation.
So, in this way, now question arise when A
is not diagonalizable, means A does not have
n linearly independent eigenvectors then,
how to write this matrix S? Because, if A
is a n by n matrix and I am able to find only
let us say some m linearly independent eigenvectors
corresponding to different eigenvalues of
A, then I will be able to write only m columns
of S. So, from where I will write raised n
minus m columns?
So, those columns I will write by finding
the generalized eigenvectors of the matrix
A. So, for writing this particular matrix
S, we need to learn how to find out the generalized
eigenvectors of a matrix. So, let me define
generalized eigenvector of a square matrix.
So, if A is a square matrix of order n, a
generalized eigenvector of A corresponding
to the eigenvalue lambda is a non 0 vector
X satisfying A minus lambda i raised to power
p into X equals to 0.
For some positive integer p such that A minus
lambda I raised to power p minus 1 X is not
equals to 0. So, A minus lambda I raised to
power p into X equals to 0 where X is a non
0 vector, but A minus lambda I raised to power
p minus 1 into X is not 0. So, we can say
that, a generalized eigenvector is a member
of null space of A minus lambda I raised to
power p. Let us take an example to find out
the generalized eigenvector.
So, find the generalized eigenvectors of a
matrix A where A is given as 1 1 0 0 1 2 and
0 0 3. Now, if I see here, the A is a upper
triangular matrix. So, eigenvalues of A will
become 3 1 1. If I calculate the eigenvector
corresponding to 
lambda equals to 3, then A minus 3 I into
X will become 0 and let us say it is X 1.
So, from here, I get an eigenvector X 1 equals
to 1 2 2 transpose. Now, similarly if I calculate
the eigenvector corresponding to 
lambda equals to 1, then A minus I into X
equals to 0 and from here I get only 1 linearly
independent eigenvector that comes out to
be 1 0 0.
So, here what we can say that the algebraic
multiplicity of A is 2 while geometric multiplicity
of lambda equals to 1 is only 1. So, hence
A is not a diagonalizable matrix. So, if A
is not a diagonalizable matrix and as I told
you, we can write A as S into J into S inverse
by the Jordan canonical transformation. So,
for writing the matrix S, I need to find out
1 generalized eigenvector corresponding to
lambda equals to 1. So, it means a generalized
eigenvector will be X 3 such that A minus
I and that is, I am talking about corresponding
to lambda equals to 1. So, A minus I X 3 and
A minus I square X 3 equals to 0.
And A minus I X 3. So, not be equals to 0.
So, since I want to take A minus I X 3. So,
not be equals to 0. So, if from here, I take
A minus I X 3 equals to X 2 because, as I
told you X 2 is an eigenvector. So, it is
a non 0 eigenvector.
So, this particular equation satisfy this
condition of generalized eigenvector if I
multiply both side by A minus I, then it will
become A minus I square into X 3 and it comes
out to be A minus I X 2 and from here, A minus
I X 2 is 0 because X 2 is an eigenvector.
So, hence I need to find out n eigenvector
or generalized eigenvector let me say X 3
which is satisfying this particular condition,
so that, I can calculate by using or by solving
this non homogeneous system of equations.
So, if I solve it, here it will become A minus
I X 3 equals to X 2. So, this gives me X 3
equals to 0 1 0 transpose. So, here X 3 is
a generalized eigenvector of the matrix A
corresponding to eigenvalue lambda equals
to 1. So, in this way, we can calculate the
generalized eigenvectors.
Once you find out the raised m minus n generalized
eigenvectors, then you are having m linearly
independent eigenvectors corresponding the
square matrix A of size n and then what you
have done you have calculated n minus m generalized
eigenvectors corresponding to different Eigenvalues.
So, what you can do? You will be having n
total eigenvectors and generalized eigenvectors
and those n vectors. You can write as the
columns of A matrix and that matrix will become
matrix S.
So, if X 1 X 2 X 3 X n is the set of all linearly
independent eigenvectors and generalized eigenvectors
of the matrix A, then S will be the matrix
having columns as these eigenvectors and generalized
eigenvectors. So, let us take an example to
write the Jordan canonical transformation
of a given matrix.
So, example is find the Jordan canonical form
of the matrix A equals to 2 2 1 0 2 minus
1 and 0 0 3 also find a matrix S such that
A equals to S J S inverse where J is the Jordan
canonical form of A. So, let me solve this
particular example. So, first of all, I need
to find out eigenvalues of A and again you
can see A is an upper triangular matrix. Eigenvalue
will be given by the diagonal elements.
So, here eigenvalues are lambda equals to
3 2 2. The algebraic multiplicity of lambda
equals to 2 is 2. Now, we will see what is
the geometric multiplicity of this.
If the geometric multiplicity of the eigenvalue
lambda equals to 2 is 2, then the matrix is
diagonalizable if it is 1, then it will be
we have to find out 1 generalized eigenvector
to write the matrix S. So, the eigenvector
corresponding to 
corresponding to lambda equals to 3 is given
as let us say X 1 and this comes out to be
minus 1 minus 1 and 1 transpose.
Now, eigenvector corresponding to lambda equals
to 2. So, it means a minus 2 I into X equals
to 0. So, from here what I got I got only
one linearly independent eigenvector which
is let me write as X 2. So, X 2 becomes 1
0 0 because, when I will write A minus 2 I,
the first equation will become 0 2 X 2 plus
X 3 equals to 0.
Second equation will give us that X 3 0. So,
from there, I will get X 1 is also 0 and third
equation will give me again X 3 equals to
0. So, here, I will get X 2 equals to 0 equals
to X 3 and X one is arbitrary. So, I have
chosen X 2 X 1 as 1 means X 1 X 2 X 3 are
different components of X 2.
Now, I need to find out one generalized eigenvector
corresponding to lambda equals to 2. So, if
I solve for that A minus 2 I square X equals
to 0 which is equivalent to solving A minus
2 I, let me write this X 3 equals to X 2.
So, if I do it, I will get X 3 as the generalized
eigenvector and this comes out to be 0 1 by
2 0.
So, after doing this, now I need to write
matrix S and the matrix J. So, here my matrix
J as I told you, I can write with only the
information about algebraic multiplicity and
geometric multiplicities of the different
eigenvalues of A. So, here 3 is having algebraic
multiplicity 1 geometric multiplicity 1. So,
there will be A 1 by 1 block of 3. Now, the
algebraic multiplicity of the eigenvalue is
2. So, algebraic multiplicity of a given eigenvalue
tells us that how many what will be the total
size of the sum of various blocks corresponding
to this eigenvalue.
So, here it is saying these that algebraic
multiplicity is 2. So, it will be a 2 by 2
blocks corresponding to eigenvalue lambda
equals to 2 the geometric multiplicity of
lambda equals to 2 is one. So, geometric multiplicity
tells us the total number of blocks corresponding
to that eigenvalue. So, algebraic multiplicity
tells size total size geometric multiplicity
total blocks.
So, algebraic multiplicity is 2 geometric
multiplicity is 1. So, only 1 block of size
2. So, I will be having a block of size 2
corresponding to eigenvalue lambda equals
to 2. So, in this way this particular matrix
becomes the Jordan canonical form of A.
Now, I can write the matrix S as the minus
1 minus 1 1 1 0 0. So, this vector as column
this vector as column and third column will
come from here. So, these are the matrices
J and S such that A equals to S J S inverse
and you can verify it later on.
So, this is the overall process for find using
the Jordan canonical transformation for finding
the matrix S and a Jordan canonical form J
of a given matrix.
Similarly, if I take this example so, find
a Jordan canonical form J of this matrix.
So, if I solve it here, the eigenvalues coming
out to be lambda equals to 3 3 3. So, algebraic
multiplicity of lambda equals to 3 is 3.
If I calculate the eigenvector corresponding
to eigenvalue lambda equals to 3 then A minus
3 I X 1 equals to 0 this gives me X 1 equals
to 1 2 0. So, hence the geometric multiplicity
of lambda equals to 3 is 1.
Now, calculate the generalized eigenvector.
So, A minus 3 I X 2 equals to X 1. So, from
that I got X 2 equals to 1 1 1 and the another
because I need to calculate 2 generalized
eigenvectors for writing the matrix S.
So, another generalized eigenvector can be
written as A minus 3 I cube X 3 equals to
0 which I can have A minus 3 I X 3 equals
to X 2 from this relation. So, from there
I got X 3 equals to 1 minus 1 1 transpose.
So, hence J will means total block is 1 total
block is 1 size is 3. So, a Jordan block of
size 3. So, 3 1 0 0 3 1 0 0 3 and S will become
1 2 0 is the first column 1 1 1 is the second
column and 1 minus 1 1 is the third column.
Hence we have A equals to S J S inverse. So,
I have taken a couple of examples for finding
the Jordan canonical transformation for a
given matrix. If matrix is diagonalizable,
then Jordan canonical form will be equal to
the diagonal matrix having eigenvalues as
the main diagonal entries.
So, let me explain the relation of Jordan
canonical form of a matrix with minimal polynomial
it given matrix A of order n, then the J C
of F of A the eigenvalues are the entries
on the main diagonal.
So, if the minimal polynomial of A is m lambda
and it is given as lambda minus lambda 1 raised
to power s 1 lambda minus lambda 2 raised
to power s 2 and upto lambda minus lambda
k raised to power s k where s i is the size
of the largest Jordan block corresponding
to lambda i in A.
So, powers in the minimal polynomial corresponding
to different terms, different factors will
give you the size of largest block corresponding
to that particular eigenvalue. And if lambda
minus lambda 1 raised to power r 1 into lambda
minus into lambda minus lambda 2 raised to
power r 2 up to lambda minus k raised to power
r k is the characteristic polynomial, then
r i is the number of occurrence of lambda
i on the main diagonal which is obvious. So,
the geometric multiplicity of lambda i is
the number of lambda i Jordan blocks in A
because each Jordan block will give you only
1 linearly independent eigenvector.
So, let us take an example corresponding to
this particular relation. So, consider a 6
by 6 matrix A having characteristic polynomial
lambda minus 3 raised to power 4 into lambda
minus 2 raised to is raised to power 2 and
minimal polynomial is lambda minus 3 raised
to power 3 lambda minus 2 raised to power
2.
.
So, here what I am having I am having.
An example here a is a 6 by 6 matrix having
characteristic polynomial as C of lambda equals
to lambda minus 3 raised to power 4 into lambda
minus 2 raised to power 2 and minimal polynomial
is m lambda equals to lambda minus 3 raised
to the power 3 and lambda minus 2 raised to
the power 2.
So, find J mean Jordan canonical form of A.
So, here as I told you, that these powers
will give you the size of maximum biggest
Jordan block corresponding to these eigenvalues
and these are the number of occurrence of
these eigenvalues on the main diagonal. So,
here I am having 4. So, the lambda equals
to 3 will occur 4 times on the main diagonal
out of which the biggest Jordan block will
be having size 3.
So, 4 equals to 3 which is the biggest Jordan
block, then what is rest one? So, from here,
I can get an information that, thus there
will be 2 Jordan blocks corresponding to lambda
equals to 3; one of size 3 another one of
size 1. So, hence I can have 3 1 0 0 3 1 0
0 3. This is the block of size 3 another one
of size 1. So, 3.
And then, here I am having 2-time occurrence
of lambda equals to 2 and the maximum Jordan
block will be having size 2. So, 2 equals
to 2. So, there will be only one Jordan block
of size 2. So, 2 1 0 2 and rest are 0 blocks.
So, in this way, this is the Jordan canonical
form of A if characteristic polynomial and
minimal polynomials are given in this way.
If minimal polynomial is lambda minus 3 raised
to the power 2 and another one lambda minus
2 raised to power 2. So, now, what is happening?
I am having total 4 size corresponding to
lambda equals to 3 and the size of the biggest
block is 2.
So, there will be 2 ways of writing 4 having
the biggest entry is 2; one is 2 plus 2 another
one is 2 plus 1 plus 1 and the other one is
2 which I have to have biggest one 2. So,
2 like this.
So, the possible Jordan block form will be
if I take this particular thing, so, there
will be 2 blocks corresponding to lambda equals
to 2 each of size 2. So, 3 1 0 3, then 3 1
0 3, then what I am having here? I am having
2 1 0 2. So, this is one of the possible Jordan
canonical form. Obviously, you can interchange
the Jordan blocks.
So, here I am not taking consideration of
reordering of the Jordan blocks. I am taking
them as the symmetrics. The other possibilities
if you use this combination , so, in this
combination, what I am saying one of the Jordan
blocks corresponding to lambda equals to 3
of size 2 and two are of size 1 1. So, it
I am having 3 1 0 3 and 2 are of size 1 1
and then I am having 2 1 0 2. So, this is
m and these are 0s. So, this is the 2 possibilities
for this the this these are the 2 possibilities
of the Jordan canonical form J of this matrix
A having this characteristic polynomial and
this minimal polynomial.
So, hence I can say that, the information
you can write the Jordan canonical form of
A matrix if you know either the algebraic
and geometric multiplicity of each eigenvalues
or you can get n information, if you know
the minimal polynomial as well as characteristic
polynomial of that particular matrix.
So, in this lecture we have learn about Jordan
canonical transformation and how to write
Jordan canonical form of a given matrix.
These are the references in the next lecture
we will learn evaluation of matrix functions.
Thank you very much.
