Ok, so ah we are talking about eigen eigen
value . Today we will start the eigenvector
. So, just to recap, so we have a square matrix
A is a square matrix n by n matrix. And we
are taking the characteristics polynomial
of this matrix which is defined as determinant
of A minus this is also n by n matrix . So,
this is a this is a characteristic polynomial.
And if we ah take the equation, if we get
to 0, then the this will give as characteristics
equations. And the roots of these equations
are the eigenvalues .
So, this is a n degree n degree polynomials.
So, it has n roots roots of these equation
are eigenvalues. So, if lambda is a eigen
value that means so this is the equation . If
lambda is a ah root, then x minus lambda is
a factor of this polynomial. Now, if the lambda
is a root of r times root that means A root
of multiplicity r, then it must write this
as x minus lambda to the power r and some
other polynomial, where lambda phi lambda
is not equal to 0.
Then it could be a another another multiple
then lambda is a root with multiplicity r,
and this is called algebraic multiplicity.
So, lambda is coming r times it has a n roots
among them r roots are lambda repetition roots
ok. So, this is called ah algebraic multiplicity
of eigenvalue .
So, then if this is happened, then r is called
algebraic multiplicity of lambda . Later on
will define the geometric multiplicity, which
is based on the eigenvector ok. So, ah this
is the thing, and we also have seen that this
lambda may not be from the same field.
So, this you have a matrix A which is a square
matrix where a i j is coming from the field,
which is a field on that this operation . Now
the a i j it could be real number or anything.
So, it is coming from the field . So, in the
last class we have seen, if you take the matrix
like this real 1 sorry 0 minus 1, 1 0 , then
the corresponding ah characteristics equation
is ah x square plus 1 characteristics equation
is this. So, x is plus minus i i is root to
the power of minus 1. So, this is a complex
number, but our field is a real field. So,
here field is real field . So, so eigenvalues
are not from this field here. So, this is
one example .
And in the last class we have also seen that
if the ah ah if we have a non-singular matrix,
then the there has to be a ah if you have
a singular matrix, if the determinant of this
matrix is 0, then there has to be eigenvalue
with 0 0 is the eigenvalue is an eigenvalue
of A. this is because we have seen from the
characteristics equation ah this product of
the eigenvalues. ah
So, the characteristics this we discussed
in the last class just to recap . Characteristics
polynomial can be written as C 0 x to the
power n plus C 1, x to the power n minus 1,
this is ending the polynomial then C n ok,
where C r is the minus 1 to the power n minus
r, and the sum of the all principle minor
of order r, sum of all principle minors of
order r, so that means, if you put C n, C
n is nothing but determinant of A, because
principle minor of order n is the determinant
And we know that this is a polynomial of degree
n. So, we know the product of the eigenvalues
. If there are n eigenvalues lambda 1, lambda
2, lambda n, then the product of the eigenvalue
will be written as this coefficients C n by
C 0.
Now, if the determinant is 0, then this is
0. So, product of the eigenvalues are 0 so
the that means one of the eigenvalue must
be 0 . And on the other hand if we have a
non-singular matrix, if this is no not 0,
then the all the there should not be a 0 eigenvalues
. So, this is the result we have seen ok.
Now, we will talk about something more on
the eigenvalues suppose we have a singular
matrix A . And if lambda is a eigenvalue,
then the 1 by lambda is eigenvalue of ah a
inverse .
So, let A be non-singular so that means it
has a inverse . This implies determinant of
A is not equal to 0 that means A has inverse
A inverse exists ok. So, non-singular means
there is no 0 eigenvalues. So, let lambda
which is not 0, because 0 cannot be eigenvalue
of this . Let lambda be an eigenvalue of A.
Then this theorem is telling this is a theorem
there we have to show 1 by lambda, which is
exists, because lambda is not 0, 1 by lambda
is a eigenvalue of A inverse, is an eigenvalue
of A inverse ok. It is a theorem .
So, how to prove this theorem? So, so to prove
this theorem we need to take a a A is a eigenvalue
so that means sorry lambda is a eigenvalue
of A that means determinant of A minus lambda
I n this is 0. So, lambda is a a root of the
characteristics equations. Now, we need to
check whether A in lambda inverse is also
an eigenvalue of this so, for that so will
use this .
So, prove we are doing. So, det of A minus
lambda I n is equal to 0 . Now, we take the
det of determinant of A inverse minus lambda
inverse I n . If we can show, this is 0, then
lambda is inverse is an eigenvalue of this.
So, this we have to show to be 0 .
So, what is this this is nothing but det of
A inverse (Refer Time:09:02) is and here will
write lambda inverse inverse is there, now
A inverse A which is basically the identity
. So, now, this will write as det of A inverse
minus A inverse lambda is a constant lambda
inverse is A constant we can write this as
this ok.
Now, we take the A inverse common from this
I minus lambda inverse A . So, this is nothing
but det of 
A inverse det of identity matrix I minus lambda
inverse A . So, this is nothing, but ah det
of A then the inverse . And we can take this
lambda common n times. So, this is lambda
inverse . So, this is lambda inverse n, and
determinant of lambda I n minus A .
Now, again we can take minus common over here.
So, this will det of A which is none of this
is 0. So, this is we can take minus, and this
det of A minus lambda I n. So, this is this
value is 0, because lambda is the is eigenvalue
of this . So, this whole thing is 0 so that
means det of A inverse ah.
So, this is the characteristics equation of
the A inverse. So, it has a root lambda inverse.
So, lambda if this implies, lambda inverse
. This implies lambda is inverse is an eigenvalue
of A inverse . So, this is the proof ok. Lambda
is the inverse is the eigenvalue of A inverse
ok. Now, we will see another result another
theorem . So, if we have a ah matrix A, and
if we have a square matrix sorry singular
all are square matrix . So, this is another
theorem .
So, I have a square matrix A, which is could
be singular and non-singular we have no we
we we do not know anything about it is on
singular, and non-singular, but and and this
is one matrix, and we have a matrix P, which
is non-singular matrix, which is non-singular
matrix . Then this theorem is telling that
P inverse A P, and A has same eigenvalue .
A and P inverse A P have the same eigenvalue
eigenvalues have the same eigenvalues so that
so how we can show that 2 matrix are having
same eigenvalue, the if their characteristic
polynomial are same so that we are going to
achieve. So, so these we have referred as
sum matrix, this is we have B matrix. So,
we have to show that characteristics polynomial
of A is same as characteristics polynomial
of B .
So, what is the characteristic polynomial
of B, characteristics polynomial of B is basically
determinant of B minus ah lambda x I n . So,
let us write B B is P inverse A P minus x
I n ok. So, this x I n we can write as ah
P P inverse form . So, P inverse x I n into
P, because P into P P inverse will give the
identity .
So, this we can write as determinant of we
can take P inverse common over here A minus
x I n into P . So, product is the ah ah determinant
of A into B same as determinant of A into
determinant of B. So, if we apply that, it
will give us determinant of P inverse into
det of this into det of P . So, this 
and this will give us one and this is nothing
but det of A minus x I n, which is the characteristics
polynomial of A .
So, characteristics polynomial of the characteristics
polynomial of polynomial of these, and these
are same so that means it has ah these two
has the these two has the same eigenvalues.
So, this is the proof these two has the same
eigenvalues . So, this is the proof of this
theorem. So, if we have a square matrix, then
we will be getting this the same eigenvalue
of this ok.
So, now we define eigen vector of a ah matrix
. So, we define eigen vector of a square matrix
ok. So, first let A is a square matrix n by
n matrix. We define the eigen vector eigen
vector is. So, ah this this square matrix
over A, ah so this matrix is coming from elements
are coming from the field . So, a i j is coming
from the field under the operations plus and
we have two operations in the field k . So,
this matrix is over field, this field could
be the real field complex field set of real
numbers set of complex number, but in general
this is a field ok.
Now, ah a non-zero vector X, which is a non-zero
vector, X is coming from F to the power n.
So, F to the power n nothing but F cross F
cross n time . So, this is a n tuple. X can
be written as x1 x 2 x n, I mean the transpose
of this is a n tuple . So, it is coming from
a so, ah X transpose is this we can say, or
yeah this is X, and X transpose is anyway.
So, this is coming from the this is a tuple,
either you write in row way or column way,
this is a n n tuple and elements are coming
from the field all all of these x L's are
from the field . ah
And X is non-zero non-zero vector . A non-zero
vector X is called a Eigen vector called and
eigenvector vector of A, ah if there exist
a lambda scalar, lambda eigen from the field
such that A X equal to lambda X, such that
AX equal to lambda. This is the definition
of this is the definition of eigenvector is
the definition .
So, what is your definition that if ah X is
a ah the a non-zero vector, X is called eigenvector
of a matrix A, if and only if there is a scalar
of the same field there is a scalar, such
that A X equal to lambda X, it is satisfying
this equation A X equal to lambda X. Then
X is called a ah X is called a eigenvector
of the matrix ok. A lambda is the corresponding
eigenvalue. So, we will come to that, so what
do you mean by A X equal to lambda X .
So, once we have A X equal to lambda X that
means, A of lambda I X equal to 0, you got
this equations . Now, we are looking for a
this this equation we are familiar with, so
this is nothing but the characteristics equations
. So, this is a homogeneous equation homogeneous
equations . And so, this is a ca[se] .
So, the how how we get the eigenvalues, eigenvalues
we are getting by the roots of these equations,
so so that means, this X is the non-zero solutions,
non-zero solutions of this homogeneous equations
are called eigenvector. And this lambda is
the ah corresponding eigen eigenvalue. So,
for each eigenvector, there is a ah there
for each eigenvector, there is a eigenvalues.
But, the question is is is the one eigenvector
can corresponding to eigenvalues, so that
we have to check . So, so this will give us
the eigenvalues. And the eigenvalues will
give us so, let us check that .
So, if you have a eigenvector, so the so an
eigenvector eigenvector sorry vector 
ah gives us a corresponding to a eigenvalue
. Now, the that means, if X is the eigenvector
that means, there is a lambda such that this.
Now, the question is is this unique, this
lambda? The question is if we ah can you get
two eigenvalue corresponding to a eigenvector?
Answer is no. Why, so this eigenvector will
correspond to the unique eigenvalue, so that
means, corresponding to a eigenvector, we
cannot have two eigenvalues lambda 1, lambda
2, where they are not same . This is not possible.
Why, how to justify that ok?
Suppose, it is possible, suppose for a given
eigenvector, we have two eigenvalues, now
one is lambda 1, and other one is lambda 2
that means, what that means, we have two equations
A x equal to lambda 1 X, and A X equal to
lambda 2 X. So, from here what we can do,
we can just and well lambda 1 and lambda 2
are coming from the field at the matrix elements
are there .
Now, what we do here, now we can just subtract
these two, and we get lambda 1 minus lambda
2 X equal to 0 0 vector. So, now this is not
possible, because X is non-zero, X is a non-zero
vector . And also this is a non-zero scalar,
because we . So, to happen, this we need to
a lambda 1 equal to lambda 2, so the that
is the uniqueness. So, for a given eigenvector,
we have a unique eigenvalue. So, one eigenvector
cannot corresponding to two different eigenvalues.
So, in a other word, the eigenvector ah is
giving the unique eigenvalues ok. So, ah this
is one observation. ah But, other way suppose,
we have given a eigenvalue, then how many
eigenvectors are there .
Let lambda be a be an eigenvalue of a matrix
A, which is n by n matrix, then then it will
leads at to a some eigenvectors, ah so at
least at least one eigenvector, so that means,
this this determine this characteristics.
So, lambda is the root of this characteristics
equation . So, this is this is a characteristics
equation .
So, lambda is a root of this characteristic
equation, so the that means, determinant of
A minus lambda n is equal to 0, so that means,
if we consider this homogeneous equation,
so A minus lambda I n equal to X . If we consider
this homogeneous system of equation, then
it has because this determinant is 0. If determination
non-zero, then it has only one root, which
is 0, or only one yeah a solution, which is
0, but determinant is equal to 0, so the that
means, it has many solutions non-zero solutions
infinite number of solutions.
And each of this is a eigenvector each of
this non-zero ah solution is corresponding
to the eigenvector is eigenvector corresponding
to this eigenvalue lambda . So, ah each X
not equal to 0 ah solution of this is called
eigenvectors eigenvectors corresponding to
to the eigenvalue to the eigenvalue lambda
ok, is in corresponding to the eigenvalue
lambda.
So, an an if if X is eigenvalue of if if X
is a eigenvector, suppose X is a eigenvector,
so the that means, ah sorry so, if X is a
eigenvector, so A minus lambda I n X is 0
. Then ah alpha X is also is a eigenvector,
because X minus lambda I n alpha X, alpha
is a scalar . This is also an eigenvector;
X is non-zero. So, we have infinitely many
eigenvectors corresponding to a eigenvalue
.
And if we consider this set, so we have infinitely
ah eigenvector corresponding to a to to ah
corresponding to an eigenvalue lambda. So,
given a eigenvalue, we have infinitely many
eigenvector, because that those are the non-zero
solution non-zero ah solution of this system
of homogeneous equations ok. So, we will continue
this in the next class.
Thank you.
