Welcome viewers. This lecture is an Eigen
values and Eigen vectors. The lecture
includes Eigen values, Eigen vectors, methods
for obtaining Eigen values and Eigen
vectors, characteristic equation and Cayley
Hamilton theorem.
.
We start with the definition of Eigen values
for this we consider a linear operator, which
is a linear transformation from R n to R n.
Let us say A which is n by n matrix associated
with this transformation. Then, the real number
lambda is called an Eigen value of the
matrix A. If there exist along the nonzero
vector X in this R n such that A X is equal
to
lambda X. So, every nonzero vector x satisfying
this equation A X is equal to lambda L
square an Eigen value of A associated with
the Eigen value lambda.
..
The Eigen values are also called proper values,
sometimes we call them characteristic or
some text call them latent values. The Eigen
values and an Eigen vectors occur in pairs
for a given lambda there are associated Eigen
vectors. So, for a given lambda there exist
a nonzero x satisfying this condition. Then,
lambda is an Eigen value and x is the
Eigenvector of the matrix or linear transformation
or linear operator A.
From this, you can say that x is equal to
0 always satisfies 1. But, it is not an Eigen
vector, because by the definition we have
included those values of x which are nonzero.
So, x is equal to 0 may this satisfying this
equation, that it is not an Eigen vector.
However, lambda is equal to 0 may be an Eigen
value corresponding to this equation.
And lambda is equal to 1 is an Eigen value
of identity matrix. Like, if you write down
lambda is equal to 1 this equation becomes
A x is equal to x. So, such a relationship
is
possible, then A is equal to I and that gives
us the result that lambda is equal to 1 is
an
Eigen value of identity matrix. Further every
nonzero vector in R n is an Eigen value of
I
associated with lambda is equal to 1. So,
whenever I X is equal to X, then whatever
value I assign to x this equation will be
satisfied. So, I say every nonzero x will
be an
Eigen value of I.
..
If lambda is an Eigen value and corresponding
Eigen vector is x, then one can establish
that r x is also an Eigen vector corresponding
to the given Eigen value lambda. One can
easily prove this, let us say lambda is an
Eigen value that simply means, there exist
x
such that A x is equal to lambda x. And, this
imply if I multiply both the sides by a scalar
r, then r times A x is nothing but A times
r X is equal to lambda times r X and that
simply establishes this result.
Now, we start with the equation A x is equal
to lambda x rewriting equation 1 as A x is
equal to I am writing x as I x, so it becomes
A x equal to lambda I x. Then, I take this
on
the other side, then it is A x minus lambda
I x is equal to 0 or I can say it is A minus
lambda I times x is equal to 0. So, this is
nothing but a homogeneous system of n
equations, because A is n dimensional, so
this is the homogeneous system.
..
And this homogeneous system will have a nontrivial
solution. Why I am talking about
nontrivial solution? Because, I am interested
in x is equal to 0, x is not equal to 0. So,
lambda be an Eigen value, we should have a
nontrivial solution, which is possible only
when determinant of lambda I minus A is equal
to 0. This is the result which we have
discussed earlier. So, this homogeneous system
will have a nontrivial solution, under this
condition. So, this gives me the scalars lambda,
which may satisfy this equation. So, the
Eigen value of A will be the scalars for which
the matrix A minus lambda I is singular.
.
.I can write down this determinant lambda
I minus A is equal to 0, in this form one
can
notice that, in this determinant each and
every term. The term in the diagonal they
are of
the form lambda minus a 1 1 lambda minus a
2 2 and lambda minus a n n. This is
because of fact that I am writing lambda I
minus A.
So, we need a diagonals are affected and because
it is minus A, so all these elements will
become negative. So, when I expand this determinant,
then I can use any row or column,
let us say I use this column. Then, it is
lambda minus a 1 1 multiplied by this
determinant, so 1 factor is lambda minus a
1 1.
Then, I can of course have terms corresponding
to this. But, if I consider this particular
term, then this is again a determinant of
order n minus 1. So, one more term like lambda
minus a 2 2 will come. And that way we will
be having terms like lambda minus a 1 1
multiplied by lambda minus a 2 2 multiplied
by lambda minus a n n. So, we will be
having n such factors. And that means, this
equation when expanded will be a
polynomial of degrees n. So, we say this determinant
which is a polynomial in lambda
and it is of degree n. So, this characteristic
equation is a polynomial in n, and since the
polynomial of degree n.
.
So, f lambda can be written this form lambda
raise to power n plus c 1 lambda n minus 1
plus c n. This polynomial is called the characteristic
polynomial. And see this
polynomial of degree n, so it will have exactly
n roots. The roots may be distinct, means
.that all of the roots may be different or
some of the roots may be repeated, if a root
is
repeated k times. Then, the Eigen value is
said to be of multiplicity k. So, we call
it
repeated with if k is less than equal to 2.
So, if all the roots having multiplicity k
is equal
to 1, then the roots will be distinct. So,
if k is equal to 1, then Eigen value is a
simple
Eigen value in that case.
.
Now, one can observe that, if I put lambda
is equal to 0, then determinant minus A is
equal to coefficient c n. And if determinant
A is 0 or the matrix is singular, then c n
will
be 0. And that means, my characteristic equation
will come in this form, lambda raise to
power n plus c 1 lambda m minus 1 plus n minus
1 lambda. The last one will be receive
from here. And, this means I can take lambda
outside and we can have lambda multiplied
by polynomial of degree n minus 1.
And that simply needs, that lambda is equal
to 0 is an Eigen value of the matrix A. So,
determinant A is 0. Then lambda is equal to
0 is an Eigen value of the matrix A or
lambda if the matrix is singular, then lambda
is equal to 0, then be a root of the
characteristic equation. Therefore, we have
following result and we can write down in
the form of a theorem, that an n by n matrix
A is singular, if and only if 0 is an Eigen
value of A. I have to proved one part, but
the detail proof can be taken out later on.
..
The set of all solutions to this equation
is called the Eigen space of A corresponding
to
the characteristic value lambda. As I told
you, that this will have number of solutions.
So, they all the solutions will form an Eigen
space corresponding to a value lambda. The
Eigen space is a set of all Eigen vectors
and zero vectors, mind here 0 vector is not
an
Eigen vector. And without zero vector it will
not be a Eigen space or a vector space. So,
it is important that Eigen space is a set
of all Eigen vectors and a zero vector. The
Eigen
space of A is a kernel of this matrix lambda
I minus A.
.
.Now, we have discussed the meaning of Eigen
values and Eigen vectors. But, now we
will discuss of the procedure of finding Eigen
values and Eigen vectors for a given
matrix A. So, the first step towards this
is like we form the matrix lambda I minus
A for
the given matrix A and then calculate its
determinant equate it to 0.
And then we solve this equation as I told
you, that this is polynomial of degree n.
So,
solution of this equation means finding the
roots of this equation. And depending upon
the value of n we will be having number of
Eigen values. So, let us say the Eigen values
are lambda 1 lambda 2 lambda n for A to B
and n by n matrix it has to have n roots some
of them may be repeated, but now the total
number of roots will be n.
Now, for each lambda I we solve the system
lambda I minus A X is equal to 0. So, we
put this value lambda 1 here and we solve
this equation. This as system of equation
this
X simple equation actually represents system
of equation, so if A is n by n. So, there
are
n such equations, there will be n simultaneous
equations. And this is the matrix equation
and this equation is to be solved. The solution
of the system gives the Eigen vector
associated with the Eigen value lambda.
.
So, let us illustrate this procedure with
the help of examples. So, in the first example
I
have I am considering a 3 by 3 matrix 4 minus
1 5 0 6 0 1 minus 2 0, while the second
example I am considering a 2 by 2 matrix.
So, first matrix A as this the first example,
so
.according to the method which I have illustrated,
I have to first obtain the solution of this
characteristic equation.
So, the characteristic equation is determinant
lambda I minus A is equal to 0. So, I
substitute this value of A in this equation.
So, the characteristic equation becomes
lambda minus 4, because it is I means only
diagonal terms will be effected. So, lambda
will appear in the diagonal only and because
of minus A, all these terms are negated
here.
So, 4 plus, so it is minus 4 minus 1, so I
have 1 here 5 the corresponding term is minus
5
0, this is minus 6 and 0 1 becomes minus 1
and we have 2 and lambda. So, this
determinant equal to 0 is the characteristic
equation for the given matrix A.
.
So, this is the latest equation this is to
be solved. So, what I am doing here is let
us
consider that, we will expand this along this
column. So, it is lambda minus 4 multiplied
by lambda minus 6 into lambda multiplied by
minus 5 into lambda minus 6. So, this is
the characteristic equation.
So, when you expand this determinant we will
have this characteristic equation, which
can be simplified. So, these two terms are
multiplied will have this expression. And
then
minus 5 lambda plus 30 is equal to 0 further
simplification will give me a cubic
polynomial, it is a 3 by 3 matrix. So, we
expect characteristic polynomial to be cubic
.polynomial. So, we have lambda cube minus
10 lambda square plus 19 lambda by
combining these two terms plus 30 is equal
to 0.
So, this characteristic equation is to be
solved, so we factorize this equation. So,
lambda
is equal to minus 1 is a factor of this equation.
So, we take it out this factor and what
remains is this. And simplifying this will
have lambda plus 1 lambda minus 5 into
lambda minus 6 is equal to 0. And this gives
me three distinct Eigen values as minus 1
from this factor this 5 and from this factor
6. So, this 3 by 3 given matrix has 3 Eigen
values minus 1 5 and 6. So, once we obtain
Eigen values, the next step is to obtain Eigen
vectors.
.
So, we have to obtain Eigen vectors corresponding
to each of the Eigen values 1 by 1, so
I start with lambda is equal to minus 1. So,
for this I have to form this matrix equation
A
minus lambda I X is equal to 0 and lambda
is equal to minus 1 in this and A is this
given
matrix.
So, if I substitute this A and this lambda
is equal to minus 1 in the given equation
I have
minus 5 1 minus 5 this will be negative and
0 minus 7 0 minus 7 2 minus 1 multiplied by
x 1 x 2 x 3 is equal to 0 0 0. This 3 by 3
system is to be solved and to solve this
homogeneous system, I form the augmented matrix
for this system and it is minus 5 1
minus 5 0 minus 7 0 minus 7 2 minus 1 the
coefficient matrix are and this column is
appended here.
.And then I like to apply linear transformation,
so that this can be reduced in a diagonal
form or in an echelon form. So, this can be
made 0 by suitable liner transformation and
this matrix reduces to this matrix.
.
And from this I can further apply a linear
transformation and what I have is 5 minus
1 5
0 0 7 0 0 and in fact I can have, but I can
multiply this by 9 and this by 7. So, after
subtracting this will become 0. So, one row
becomes 0 in this augmented matrix and by
this we can say that rank of the matrix A
is 2 and from rank nullity theorem nullity
A is
equal to 1. And that means, the solution vector
of A X equal to lambda X is the Eigen
vector which can be obtained as minus k 0
and k, so nullity is equal to 1.
So, what can I say, what can I do is, I can
write down the third it in the form of in
the
third equation I can write it write down the
third variable as k. Then, this equation will
automatically be satisfied for y is equal
to 0 and by writing is y equal to 0 and z
is equal
to 5 will give me X is equal to minus 1.
So, this is a solution for the given equation
and we will have in fact infinite many
solutions of this equation and each solution
will be represented in this form k can take
infinite many values. So, we will be having
all the solutions will be of this particular
form, so we can say this is the basis for
the Eigen vectors associated with lambda is
equal to minus 1.
..
Once we have solve this problem, then we go
to the next Eigen value lambda is equal to
5 the same X as steps can be repeated. That
means, you again start with the matrix
equation A minus lambda I times X is equal
to 0. For the given matrix A and then we
form the augmented matrix we apply linear
transformation is series. And once, this
matrix is reduced to be echelon form which
is this particular example for lambda is equal
to 5. And one can notice that the again this
row becomes 0. And that means, rank of this
coefficient matrix is 2 and nullity is equal
to 1. And in that case, if you solve this
equation the system of equation k can be taken
as arbitrary value y can be 0. So, when
we substitute it here, this is k is arbitrary,
k and this is 0, so X becomes 5 k. So, the
Eigen
vector corresponding to the Eigen value lambda
is equal to 5 will be of the form 5 k 0 k.
..
So, Eigen vector the third Eigen value was
lambda is equal to 6. Again the whole thing
is
repeated A minus lambda I into X is equal
to 0 for this given value of A. The augmented
matrix is obtained first we apply linear when
we apply a element transformations and
this time these element transformations lead
to this matrix here this row becomes 0.
So, again the rank of the coefficient matrix
is 2, but the nullity is 1 when lambda is
equal
to 6. And that means, we can write down the
solution of this matrix as if you say X 3
arbitrary. Then, y becomes minus 7 by 5 x
3 and we substitute x 3 as x 3 here and minus
7 by 5 X 3 here. Then, this can be X can be
obtained as 16 by 5 x 3. That means, X y and
z they are written in terms of the component
x 3 or let us say x 3 is equal to k then this
can be written in the form sixteen k minus
7 k and 5 k. So, this vector is an Eigen vector
corresponding to Eigen value lambda is equal
to 6, so we have three different Eigen
values in this example and we have obtained
three different Eigen vectors in this case.
..
In the next part we have been given a 2 by
2 matrix. So, we apply the Eigen we have to
first find the characteristic equation. So,
we write down lambda minus 5 1 minus 1
lambda minus 3 determinant of this is equal
to 0, we solve the determinant lambda minus
5 into lambda minus 3 plus 1 equal to 0.
And this gives me lambda square minus 8 lambda
plus 15 and 1 16 equal to 0. This
equation gives me lambda minus 5 lambda minus
4 whole square equal to 0. So, this case
gives me lambda is equal 4 is an Eigen value
and it is multiplicity is 2. So, it is a case
when that Eigen values are repeated. So, when
the Eigen values are repeated, then you
have to solve this augmented matrix to obtain
the Eigen vectors corresponding to lambda
is equal to 4. And one can notice that this
row is a same as this row, so rank of this
matrix is 1 and nullity is also 1 from the
rank nullity theorem.
..
So, nullity is also 1 and that means, it has
any one linearly independent solution, which
can be obtained from the homogeneous system.
So, although the multiplicity root is 2,
but we have only one Eigen vector corresponding
to the Eigen value lambda is equal to
4. So, in the earlier example was have three
distinct values and we got three Eigen
vectors. But, in this case we got two we got
repeated Eigen values and we have only one
Eigen value corresponding to that. So, what
about Eigen value vector corresponding to
the lambda is equal to 4 it is k and minus
k.
.
.Now, we estimates some result related with
Eigen value and Eigen vectors. So, let us
say
A is a square matrix of order n with Eigen
values lambda 1 lambda 2 lambda n, then
determinant of A is the product of Eigen values.
Now to prove this, let us say square
matrix of order n. Then, the roots of this
characteristic equation the determinant lambda
I
minus A is equal to 0 are lambda 1 lambda
2 lambda n is been given to us. Then, since
we have roots n this is the characteristic
polynomial at degree n. So, we can write down
that characteristic polynomial which has roots
lambda 1 lambda 2 lambda n as the
product of this.
So, we can say determinant lambda I minus
A is equal to lambda minus lambda 1 lambda
minus lambda 2 and the factor lambda minus
lambda n. Now in this case, if we put
lambda is equal to 0 in this, the determinant
minus A is equal to this minus this minus
this minus these lambdas are put to 0. So,
they are n minus signs, so we will have minus
1 raise power n lambda 1 lambda 2 lambda n
and this is equal to determinant of minus
A,
this is lambda is put to 0.
Now, we know that determinant of minus A is
equal to minus 1 raise to power n
determinant of A. So, this minus 1 and this
minus 1 will get cancel and we have
determinant A is equal to lambda 1 lambda
2 lambda n. So, we have proved the result.
.
So, the example which we have done earlier,
we have obtained the Eigen value as minus
1 5 and 6. So, this is this was the matrix
in that example. So, if you substitute if
you use
.this result then determinant of A is the
product of this Eigen value. So, determinant
of A
is equal to minus 30.
.
The next result is if A is a triangular matrix.
Then, the Eigen values will be the diagonal
entries a 1 1 a 2 2 a n n. Like, if A is a
triangular matrix let us assume that it is
an upper
triangular matrix the similar result, similar
way the result can be true follow a triangular
matrices also. So, let us assume at the moment
that A is a upper triangular matrix.
Then, the characteristic equation will be
this. So, it say is the upper triangular matrix
is
upper triangular matrix. So, all the elements
in the lower part is 0 and only these
elements are nonzero. And then characteristic
equation is lambda I minus A. So, will
have minus a 1 1 minus a 2 2 and so on in
the diagonal terms and will have lambda
minus a 1 1 and all these terms will be simply
negated.
So, if you expand this determinant it comes
out to be lambda minus a 1 1. Now, we are
expanding the determinant along this column.
So, lambda minus a 1 1 into this
determinant and lesser terms will not contribute
anything, because they are 0. So, lambda
minus a 1 1 into determinant this is lambda
minus a 2 2 into this determinant and so on.
And that means, it is lambda minus a 1 1 lambda
minus a 2 2 lambda minus a n n.
So, this nth degree polynomial and all these
are factors. So, if it is 0 means all these
factors have to be 0. And that means, lambda
is equal to a 1 1 is one Eigen value lambda
.is equal to a 2 2 is other Eigen value lambda
is equal to a n n is the nth Eigen value of
the
given matrix A, what are a 1 1 a 2 2 a n n,
they are nothing but the diagonal elements
in
the given matrix A and that was to be proved
in this theorem.
So, this result is particularly important,
because to solve an Eigen value problem 1
has 2
first obtain the characteristic equation.
And that means, this determinant is to be
obtained. Finding the nth degree polynomial
corresponding to the given matrix, if you
have to expand this determinant is any difficult,
and then finding Eigen value and finding
the solution of that characteristic equation
is further complicated. But, if we can have
a
lower triangular matrix or upper triangular
matrix, then we do not have to worry about
the rest of the elements you can simply write
down the Eigen values as a 1 1 a 2 2 a n n
that Eigen elements. So, this result helps
us in finding out the Eigen values of matrices.
.
There is one more result it states that v
1 v 2 v n be the Eigen vectors corresponding
to n
distinct Eigen values of lambda 1 lambda 2
lambda n respectively. Then, v 1 v 2 v n will
be linearly independent. So, if v 1 is an
Eigenvector corresponding to lambda 1, v 2
is an
Eigen vector corresponding to lambda 2 and
v n is the Eigen vector corresponding to
lambda n. Then, these vectors are distinct
if these vectors are linearly independent
if
these Eigen values are distinct. So, this
is the theorem.
So, let us prove this result. So, let us say
that lambda 1 lambda 2 lambda n are distinct
Eigen values corresponding to a given square
matrix A and lambda 1 v and v 1 v 2 v n be
.the Eigen vectors. Then, we can prove this
result by induction. So, we start with v 1
is an
Eigen value corresponding to lambda 1, so
only one vector, so it is always linearly
independent. Next, we assume that v 1 v 2
v n minus 1 corresponding to different lambda
1 lambda 2 lambda n minus 1 are linearly independent.
.
Then, we have to prove that when v n is added
in this linearly independent set v 1 v 2 v
n
minus 1, then the complete set is linearly
independent. So, to prove the linear
independence of this set v 1 v 2 v n minus
1 v n. Let us consider the linear combination
c
1 v 1 plus c 2 v 2 plus c n minus 1 v n minus
1 plus c n v n is equal to 0. And if we can
prove that the c 1 c 2 c n minus 1 and c n
are all 0, then we have a we can say that
these
vectors are linearly independent.
So we apply A, we cancelation A linear transformation
A to this A is a matrix. Let us
multiply this equation by matrix A. And since
it is a linear combination and A is a
matrix. So, we can say c 1 into A v 1 plus
c 2 A v 2 plus n minus 1 A v n minus 1 plus
c
n A v n and A multiplied by 0 is 0.
Now, since v 1 is an Eigen value corresponding
to Eigen we since v 1 is an Eigen vector
corresponding to Eigen value lambda 1. So,
A v 1 is equal to lambda 1 v 1, so that is
we
are replacing A v 1 by lambda 1 v 1. Similarly,
A v 2 is to be replaced by lambda 2 v 2
as lambda 2 is the Eigen value and v 2 is
its corresponding Eigen vector.
.Similarly, this term will become lambda n
minus 1 v n minus 1 and the last term will
be
c n lambda n v n equal to 0, let us call this
equation as 2.
Then, multiplying the equation 1 by lambda
n and subtracting from 2, what we get is
this, is multiplied by lambda 1 and we multiplied
with by lambda n subtract from this.
So, what will have is this equation, c 1 v
1 is common and this is lambda 1. But, this
is
being multiplied by lambda n and subtraction
is taking place. So, it is lambda 1 minus
lambda n v 1 plus c 2 times lambda 2 is coming
from here and lambda n from here. So, it
is c 2 times lambda 2 minus lambda n v 2 and
will have all factors up to these terms. But,
when we see this last term it is c n lambda
n v n and this is also c n lambda n v n. So,
these 2 terms will be cancelled out. So, we
will have terms c 1 up to c n minus 1.
.
Now, given that all the roots are distinct.
That means, lambda i is not the same as lambda
n for any i 1 2 to n minus 1 lambda n is different,
then all these lambda i is... So, they are
not 0 and further v 1 v 2 v n minus 1 are
linearly independent. And that means, this
is
equal to 0 this is not 0 this is not 0 this
is not 0 this is not 0. So, sum of this means
a
combination is 0 simply means that c 1 c 2
and c n minus 1 has to be 0, because v 1 v
2 v
n minus 1 are linearly independent.
So, c 1 c 2 and c n minus 1 they are all going
to be 0, so when we substitute these values
in 1, then we have the expression c n v n
is equal to 0 v n is not 0 its given to as
it is a
vector which is nonzero. So, what we have
is that, c n is equal to 0 is only alternative,
so
.we have obtained a linear combination which
is 0 only when c 1, c 2, cn's all are 0. And
that proves that v 1, v 2, v n will be linearly
independent. So, if we have n distinct value
Eigen values then the corresponding Eigen
vectors will be linearly independent. So,
this
was the theorem.
.
Now, if the characteristic root lambda of
a matrix A is repeated k times. Then, A may
have k independent Eigen vectors it may happen
it may not. If all that values are distinct
definitely we will have independent Eigen
vectors. But, if they are not distinct they
are
repeated. Then, the corresponding Eigen values
will be dependent or independent both
the things are possible.
We have discussed an example in which A is
a 2 by 2 matrix 5 minus 1 1 3 we have
obtained its Eigen value as 4 and 4. So, the
Eigen values are repeated and we could find
only one independent Eigen vector in this
example as k minus k, so for the roots repeated
we got only one Eigen value.
..
However, if we consider A as this matrix,
then the characteristic equation is determinant
lambda I minus A X is equal to 0 gives me
this determinant equal to 0. And when you
solve this determinant the characteristic
equation comes out to be lambda lambda minus
1 lambda minus 1 equal to 0. That means, this
is a singular matrix and it has as Eigen
value 0, this result we have already established.
And the matrix A has a repeated Eigen
value 1, so it has 3 Eigen value 0 1 and 1.
.
.So, if we consider the repeated value lambda
is equal to 1. Then, lambda I minus A X is
equal to 0 is to be solved for lambda is equal
to 1 and to solve this we consider
augmented matrix. We apply elementary transmissions
let us say this and this row has to
be added. So, this is become 0 row, so this
matrix has 2 0 rows.
So, the rank of the coefficient matrix comes
out to be 1 and nullity is 2 and this means
the solution of this system is 0 r and s see.
Any value 0 r and s will satisfy this equation
will satisfy this equation and we write down
X is equal to 0, y is equal r and z is equal
to
s will also satisfy this equation. So, any
solution will be of this form.
Any solution of this system will be of this
form by this involves two arbitrary constant
r
are in s. That means, there are two independent
vectors as a solution of this system they
are 0 r 0 and 0 0 r. So, in this example we
have I repeated Eigen value 1 with multiplicity
2. And when we solve for Eigen value Eigen
vectors for corresponding to this Eigen
value, we find that there are still two independent
Eigen vectors associated with the
given matrix.
So, I have compute different examples where
the in one example there is one Eigen
value which is repeated. But, we could get
only one independent vector and in the
another example there are two independent
Eigen vectors associated with an Eigen value
of multiplicity 2. So, if the roots are distinct
we are sure that the vectors are going to
be
independent. But, if the vectors if the roots
are not distinct. Then, we cannot say the
vectors will be independent or not they may
be independent in some examples and some
other examples they may be dependent.
..
Now, another theorem is says that the Eigen
values of a matrix is the same as its
transpose. So; that means, if I have a matrix
A having some Eigen values and if I take the
Eigen values of the corresponding transpose
matrix the Eigen values between matrices
will be the same. So, let us consider the
characteristic equation for A transpose. So,
it is
lambda I minus A transpose is equal to 0.
Then, lambda I minus lambda I minus A transpose
can also be written as lambda I minus
A transpose. Because, I is a symmetric matrix
and determinant will not be affected. So,
we can take determinant of lambda I minus
A transpose is equal to 0 as simplification
to
this.
And this means lambda I minus A is equal to
0. Because, determinant of this and
determinant of this they are same. And that
means, you have same the characteristic
equation for A as well as for A transpose.
And that means, the solution will be the same
and that implies that the Eigen values of
A and it is transpose will be the same.
..
There is another property of Eigen values
and Eigen vectors and according to this let
lambda be an Eigen values of A. Then, lambda
square is an Eigen values of A square. To
prove this, let us say lambda is an Eigen
values and X is this Eigen vector. So, A X
is
equal to lambda X is given to us. I pre multiply
this equation by A, so it is A times A X
is equal to A times lambda X. Since, lambda
is a scalar and matrix multiplication is
associated.
So, left hand side becomes A into A X is equal
to right hand side becomes lambda is
lambda can be taken outside from the right
hand side and right hand side becomes
lambda times A X. And that means, A square
X is equal to lambda time A X and X into
the lambda X is been given to us we can have
A square X is equal to lambda X and from
here one can say that lambda square is an
Eigen values of A square. So, lambda square
is
an Eigen values of the matrix A square, this
is most this is to be proved.
..
Then, the next theorem says that if lambda
is not equal to 0 is an Eigen value of a
nonsingular matrix A, then 1 upon lambda will
be the Eigen value of A inverse. The
same lines we can prove this result. So, we
say that lambda is an Eigen value of A; that
means, A X is equal to lambda X and X is not
equal to 0. Because, this is the equation
for determining lambda Eigen value lambda
and Eigen vector X.
So, X is not to be 0. For lambda not 0, we
can write down this equation as 1 upon lambda
A X times X. For lambda not 0, A is nonsingular
this is been given to us A is
nonsingular. So, lambda is not zero. Therefore,
A inverse exist for nonsingular matrix A
inverse exist and we can multiply this equation
by A inverse. So, will have 1 upon
lambda multiplied by A inverse A X is equal
to A inverse X. And A inverse A is
identity, so 1 upon lambda into I X is X is
equal to A inverse X and from here one can
conclude that A inverse has an Eigen value
1 upon lambda.
..
Now, we say that if lambda is equal to 0 is
an Eigen value of a matrix A, then A is
singular. To prove this, let lambda is equal
to 0 is an Eigen value of A and X is not equal
to 0 is an Eigen vector associated with lambda.
That means, A X is equal to 0 into X is
equal to 0. So, this system A X is equal to
0 will have a nontrivial solution if and only
if
A is singular. This is the result which we
already developed and this proves that lambda
is equal to 0 is an Eigen value of a matrix
A, then A is singular.
.
.Now, let us consider this expression determinant
lambda I minus A times identity matrix
is equal to ad joint of lambda I minus A multiplied
by lambda I minus A. This result we
have already developed. Now, this lambda I
minus A is determinant is a polynomial of
degree n. Then, this multiplied by I is a
matrix equation is a matrix expression.
And we can write down this determinant lambda
I minus A as an lambda n plus an minus
1 lambda n minus 1 plus an minus 2 lambda
n minus 2 plus a 1 lambda plus a naught you
swap polynomial of degree n. Each term in
ad joint lambda I minus A this 1 the right
hand side is the polynomial of degree n minus
1 in lambda. Why, because in the ad joint
matrix each term is a determinant and determinant
of 1 or the lower. So, it is a
polynomial of degree n minus 1 in lambda.
So, we can write down ad joint of lambda I
minus A is equal to B n minus 1 lambda n
minus 1 plus B n minus 2 lambda n minus 2
plus B 1 lambda plus B naught. What we
have done is that? We have written B n minus
1 as a n minus 1 order square matrix. B n
minus 2 is a n minus 1 order square matrix.
So, each term is written separately, so we
write down this ad joint lambda I minus A
as this expression here B naught is identity.
.
Then, determinant lambda I minus A I is equal
to B naught lambda n minus 1 plus B n
minus 2 lambda n minus 2, last term has to
be B naught multiplied by lambda I minus A.
What we can do is, on this side we have and
polynomial expression lambda I minus A.
.This is multiplied by matrix I and here also
we have polynomial terms lambda n minus 1
lambda n minus 2 and they are multiplied by
matrices.
So, let us equate terms related to lambda
raise to power n and it is various powers.
So, if
I compare coefficients of lambda n, then on
this side it is a n I and on the right hand
side
it is B n minus 1, B n minus 1 and this is
multiplied by lambda I. So, this becomes
lambda raise to power n. So, on the left hand
side I have a naught I, there is the
coefficient of lambda n, in this characteristic
equation. And in the right hand side we
have B n minus 1. Secondly, lambda n minus
1, so on the left hand side I have a n minus
I is equal to from here, I will be having
two terms one corresponding to this, another
corresponding to this.
So, we will have this term, similarly other
powers. So, we have this expression. Now,
if I
multiply the first equation by A n and this
equation by A n minus 1 and then I add. So,
what will happen, this is A naught n this
is A raise to power n and here I have B n
minus
1. B n minus 1 this is A n minus 1, this B
multiplied by this, so this is A n minus 1
and
here also have A, so this becomes A n.
So, if this way if I multiply each and every
term and add, then this term will get cancel
with this, this term will get cancel with
the other terms and this term will get cancel
with
this. So, if I add these expressions, then
we will have a naught A plus a n minus 1 A
n
minus 1 plus a n minus 2 a n minus 2 and so
on, a 1 A plus a naught multiplied by I in
the left hand side. But on the right hand
side will not any term. In fact, all the term
will
get cancel, so this equation is obtained and
this is equal to 0. This is the matrix equation
equal to 0.
..
So, this is the characteristic equation which
we have a n lambda n plus a n minus 1
lambda n minus 1 plus a 1 lambda plus a naught
equal to 0. And what we have a obtain
is, a n A n plus n minus 1 multiplied by the
matrix A raise to power n minus 1 and so on.
So, if you compare these two expressions what
we can say that this lambda is replaced
by A.
So, if this is the characteristic equation,
then the matrix A will satisfy it is own
characteristic polynomial. So, this is the
result which we have that a square matrix
satisfies it is characteristic polynomial.
Now, this is an important result, we call
it Cayley
Hamilton theorem. Let us illustrate this result
with the help of this example. So, find the
characteristic equation for the given matrix
A and verify the Cayley Hamilton theorem
for the matrix.
..
So, let us say A happens to be 1 2 1 0 simple
2 by 2 matrix for this we can find the
characteristic equation determinant lambda
I minus A equal to 0. As this determinant
equal to 0, simplify this determinant it is
lambda times lambda minus 1 minus 2 equal
to
0. And lambda square minus lambda minus 2
equal to 0 is the characteristic equation
for
this given matrix.
So, according to Cayley Hamilton theorem,
this matrix A satisfies it is characteristic
polynomial. That means, A square minus A minus
2 is equal to 0. So, to prove this let us
compute A square, A square is this matrix
A multiplied by A. So, if you simplify it
is 1
plus 2 that is 3 1 into 2 is 2 this is 0.
And similarly, when we multiply this row by
this we
will have 1 this row multiplied by column
is this 2.
Substituting the values of A square and A
in the matrix equation A square minus A
minus 2 I gives 3 2 1 2 minus the matrix A
minus 2 times identity is equal to 0. And
check, 3 minus 1 minus 2 is 0. And similarly
other terms and will have A square minus
A minus 2 I equal to 0. So, the we have proved
in this example, that matrix A satisfies it
is characteristic polynomial.
..
We can use this result for finding the inverse
of a given matrix. So, if you start with the
characteristic equation lambda n plus a 1
lambda n minus 1 plus a 2 lambda n minus 2
and so on. And we write down the corresponding
matrix equation according to Cayley
Hamilton theorem will have A n plus a 1 A
n minus 1 plus a 2 n minus 2 and so on equal
to 0.
Then, multiply this equation by A inverse
if it exist. Then, it is A inverse is pre
multiplied equal to 0. And then this a naught
multiplied by A inverse I is this term on
the
left hand side rest of the terms are taken
on the other side. So, we have an expression
involving A inverse in terms of powers of
A. This will can be used for finding A inverse.
Let us illustrate this with an example. So,
find inverse of given matrix A using Cayley
Hamilton theorem A is 1 2 1 0 which is given
to us it is a 2 by 2 matrix.
..
So, we first find it is characteristic equation,
which we have obtained as in the case of
earlier example, lambda square minus lambda
minus 2 equal to 0. Then, the
corresponding matrix equation according to
Cayley Hamilton theorem is A square minus
A minus 2 I equal to 0. We substitute the
value of as a pre multiply this equation by
A
inverse. So, will have 2 A inverse is equal
to this two terms can be taken on the other
side is A minus I.
So, we can substitute the value of A and I
in this equation. And this gives me A inverse
as half, this minus this 0 2 minus 0 is 2,
1 minus 0 is 1 and 0 minus 1 is minus 1. So,
this
is A inverse. Let us, verify that we have
got the right result. So, we say A into A
inverse
is equal to half into this 1 2 1 0 multiplied
by the inverse 0 2 1 minus 1 is equal to 1
0 0
1. So, A A inverse comes out to be identity
I have tried this example with pre
multiplying this equation by A inverse. But,
the same result can be obtained. If you
multiply post if you perform post multiplication,
then again we will get the same result.
..
To summarize, we have discuss Eigen values,
Eigen vectors, characteristic equations.
We have discussed the method for finding Eigen
values and Eigen vectors of a given
matrix. This method depends upon in the evaluation
of determinants finding
determinants is not a simple problem. In fact,
we have to develop more methods of
finding Eigen values and Eigen vectors for
a given problem. We have already discussed
Cayley Hamilton theorem. We have discussed
how to use Cayley Hamilton theorem for
finding inverse of a given matrix, with this
we come to the end of the lecture.
Thank you.
.
