Hello friends, welcome to lecture series on
Matrix Analysis with Applications. Now, today’s
lectures based on Eigenvalues and Eigenvectors.
That if we have a given matrix A, then how
can you find this eigenvalues and the corresponding
eigenvectors.
So, first of all what eigenvalues and eigenvectors
are. So, let V be a vector space over the
field F, and T be a linear operator on V that
is T is a linear transformation from V to
V. A non-zero vector v belongs to V is called
an eigenvector of T, if there exists a scalar
lambda belongs to F such that T v equal to
lambda v. So, if we have a linear operator
T, and we have a non-zero vector v in V, and
there exist a lambda belongs to field such
that T v equals to lambda v, then we say that
lambda is eigenvalue, and the corresponding
eigenvector is v.
Now, a square matrix A may be viewed as a
linear operator T, and is defined as T x equals
to A x, where x belongs to V ok. So, if you
have a linear operator the linear operator,
then A matrix can be viewed as a linear operator
also. Now, now let us define a matrix A, which
is a square matrix of order n. Then if we
consider a system of homogenous equation A
x equals to lambda x.
Then the value of lambda for which the system
of linear equations has non-trivial solution
are called eigenvalues or the characteristic
values of A, and the corresponding non-zero
vectors x are called eigenvectors or the characteristic
vectors of A. Since, since every matrix can
be viewed as a linear operator T, so in the
previous definition we have replace this T
by a matrix A here. And this x is in V, this
is a scalar lambda in field ok.
So, what I want to say basically, suppose
you have a matrix A, which is a order of n
cross n. Then if A x is equals to lambda X,
where X is not equal to 0. Then this lambda
is called this lambda is called eigenvalue
of a and X is called corresponding eigenvector
corresponding eigenvector of A, or eigenvector
corresponding to lambda.
Now, we are having a A X equal to lambda X,
so this can be written as A X minus lambda
X equal to 0 ok. Now, this can be written
as A minus lambda into I times X equal to
0, where I is an identity matrix. Now, what
we are having basically, we are having we
are having A minus lambda I, X equal to 0
means, we are having system of homogeneous
linear equations you know, because other side
is 0.
Now, now since X is not equal to 0 as we already
defined a definition X is not equal to 0.
So, this means if we are taking this as a
linear operator T, if we take A minus lambda
I as a linear operator T, then T X equal to
0, that means, that means and X is not equal
to 0, that means, the nullity the nulls space.
The nullity of this matrix or this operator
is greater than or equal to 1 ok, because
X is not equal to 0 means this means nullity
is not 0, and nullity is not 0 means, it is
greater than or equal to 1.
Now, what is what is the dimension of dimension
of A minus lambda I of course, dimension of
A minus lambda I is n. So, rank by rank nullity
theorem, the rank of this A minus lambda I
plus nullity of A minus lambda I must be equal
to dimension of matrix or operator, which
is n So, this implies rank of A minus lambda
I must be equals to n minus nullity of A minus
lambda I.
And since, nullity is greater than equal to
1, this may this is less than equal to n minus
1. And since, rank is less than equal to n
minus 1 less than equal to n minus 1, this
means in the echelon form of this matrix at
least 1 0, 1 row is there, which is having
all 0 elements, and that means, this implies
determinant of A minus lambda I must be 0.
Now, this equation, determinant of a minus
lambda equal to 0 is called a characteristic
polynomial, this will give a characteristic
polynomial in lambda characteristic polynomial
of lambda . Now, now this is a matrix of order
n cross n. So, determinant, so this determinant
will give a polynomial of degree n in lambda.
So, what we are having now? So, now if we
if we visualize the determinant of A minus
lambda I equal to 0, that means, that means
this determinant is equal to 0 you see.
If we are taking determinant of A minus lambda
I equal to 0, so it implies that determinant
of A is this matrix you see, a 1 1, a 1 2
if you are taking this matrix as a 
and a and 1 up to a n n minus lambda times
I is identity matrix. So, when you take determinant
put it equal to 0, so this is nothing but
determinant of you see, only this lambda will
be subtracted from the diagonal elements,
because all other entries are 0. So, this
will be determinant of a 1 1 minus lambda
a 1 2 and so on up to a 1 n, then a 2 1, a
2 2 minus lambda, then a 2 n and so on a n
1, a n 2, and so on up to a n n minus lambda,
this determinant is equal to 0.
So, this will give us a polynomial in lambda,
and that polynomial is called characteristic
polynomial ok. So, this is basically gives
minus 1 raised to power n lambda raised to
power n minus say c 1 lambda raised to power
n minus 1 plus c 2 lambda raised to power
n minus 2 minus and so on plus minus 1 raised
to power n into c n is equal to 0 ok. Because
when you when you expect this in terms of
lambda, so you will get a polynomial in of
degree n in lambda.
Now, now if you carefully observe this determinant
you see, when you open from the first element
you see, if you take a 1 1 minus lambda, then
you take then you delete this row and this
row and this column. So, this will be the
remaining I mean determinant 
minus a 1 2.
Now, when you take, when you expand from this
element, you delete this column and this row,
that means, the two values of two linear factors
of lambda are not coming in this in this expansion,
so that means, this expansion will containmaximum
lambda raised to power n minus 2, because
when you take, when you expand from this,
you are deleting this row and this column.
So, two factors of lambda are not coming in
are not coming in that expansion, that means,
the maximum power of lambda, which is coming
from this term is maximum power of lambda,
which is coming from this term is lambda raised
to power n minus 2.
Similarly, if you take a 1 3, so here will
be something a 3 3 minus lambda, and when
you delete this column and this row, then
again two factors of lambda are not coming,
so that means, maximum power of lambda, which
are coming from this term is lambda raised
to power n minus 2. Similarly from the other
terms so, what we can conclude that the remaining
terms will contain say alpha 1 lambda raised
to power n minus 1, and so lambda raised to
power n minus 2 and so on, so that means,
the power of lambda raised to power n, and
power of lambda raised to power n minus 1
are coming only from the first determinant.
Now, if you see here, again if you see here,
now if you open from this is ok, if you open
from this, here is something a 3 3 minus lambda
is again here, if you open from this element,
again you will delete this column and this
row. So, two powers of lambda are not are
still not here I mean not coming, so that
means, you see if you take a 1 1 minus lambda,
if you expend, this determinant then it is
a 2 2 minus lambda, and again some determinant
plus and minus a 2 3, and determinant of some
other thing, which in which 2 power 2 powers
of lambda are not there ok.
So, that means, that means, it will contain
maximum lambda raised to power n minus 2,
and minus 3, because this determinant is having
lambda raise to power n minus 1 1 1 factor
is already outside. So, this determinant is
having lambda raised to power n minus 1. And
two factors are not coming, that means, the
maximum power, which this is this element
is having is lambda raised to power n minus
3.
And when you multiply this with lambda, so
it will be having lambda raised to power n
minus 2, so that means, what I want to say
basically that when you expand similarly,
the entire determinant. So, lambda raised
to power n and lambda raised to power n minus
1 are only coming in the product of the diagonal
elements. When you similarly try to expend
this entire determinant, so lambda raised
to power n, and lambda raised to power minus
1 is coming only from product of the determinant
diagonal elements ok.
So, and all other and the lambda raised to
power n minus 2 lambda raised to power n minus
3 are from the all from all the components,
may be from all the components, all the elements,
so that means, that means coefficient of lambda
raised to power n is simply coefficient of
lambda raise to power n in simply a 1 1 minus
lambda a 2 2 minus lambda and so on a n n
minus lambda. And it is simply minus 1 raised
to power n, so carefully see it is simply
minus 1 raised to power n.
And similarly, if you want to see coefficient
of lambda raised to power n minus 1, it is
coefficient of lambda raise to power n minus
1 in again this product a 1 1 minus lambda
a 2 2 minus lambda and so on up to a n n minus
lambda. So, so what is the coefficient of
lambda minus lambda raised to power n minus
1 in this, it is simply a 1 1 plus a 2 2 plus
and so on a n n. You can you can simply see,
the sum of these elements a 1 1 plus a 2 2
and so on up to a n n will be the coefficient
of lambda raised to power n minus 1.
So, what basically now we are having, we are
having determinant of A minus lambda I is
something like minus 1 raised to power n,
and it is lambda raised to power n minus c
1 lambda raised to power n minus 2 and minus
1 plus c 2 lambda raised to power n minus
2 minus and so on minus 1 raised to power
n c n, so this is equal to this.
Now, let so how many roots this equation will
be having, this equation is will be having
n roots, because the it is of n degree polynomial.
So, let lambda 1 lambda 2 and so on up to
lambda n are the roots be the roots of this
equation.
So, if these are the roots, then what is the
sum of the roots, you can simply see, the
sum of the roots is simply this is the equation.
So, some of the roots of this will be simply
negative of minus c 1. It is minus negative
of negative of minus c 1, and this is equal
to basically this is equal to you see the
coefficient of lambda raised to power n minus
1 is simply this, so this is a 1 1 plus a
2 2 and so on up to a n n. And this is called
a trace of a matrix, trace of a matrix A.
So, what we have concluded, we have concluded
that sum of eigenvalues is nothing but the
trace of the metrics, that means, some of
the diagonal elements, the first property.
The second property is now this result that
this is equal to this whole for every lambda
for any lambda ok, so this whole for lambda
equal to 0 also. And when you substitute lambda
equal to 0 we will get determinant of A as
equal to c n as c n.
Now, if you find product of roots for this
equation, product of roots. So, product of
roots are lambda 1, lambda 2, up to lambda
n, and that is simply equal to you see it
is minus 1 raised to power degree of equation
last term upon first term, which is nothing
but c n ok, c n upon minus 1 raised to power
n, and that is simply equal to determinant
of A. So, this implies product of eigenvalues
is nothing but determinant of A. So, we have
noted here, two important properties of eigenvalues,
the one property is that the sum of the eigenvalues
is equal to trace of the matrix that is the
sum of the diagonal elements, and the second
property is the product of the eigenvalues
is equal to the determinant of the matrix
A.
So, the sum of the eigenvalues simply equal
the trace of the matrix. And the product of
the eigenvalues simply the determinant of
A.
Now, let us let us discuss this problem. The
first problem is let us consider this matrix
A, which is 1 minus 1 1 1, let us find this
eigenvalues and the corresponding eigenvectors.
So, the matrix A is simply 1 minus 1 1 1.
So, you can simply write A minus lambda I
determinant this must be 0 for the characteristic
polynomial, or finding the roots of the equation,
roots of the latent roots or characteristic
roots of this matrix A, so this is nothing
but 1 minus lambda minus 1 1 1 minus lambda
determinant equal to 0. And this implies on
minus lambda whole square plus 1 should be
0, or 1 minus lambda square is equal to minus
1 that implies on minus lambda is equal to
plus minus iota, or lambda equal to 1 plus
minus iota so that two roots of two eigenvalues
of this equations R 1 plus iota and 1 minus
iota, so these are the eigenvalues.
So, let us find corresponding eigenvectors.
So, first you find eigenvector eigenvector
corresponding to lambda equals to say 1 minus
iota. So, how you will find that you can simply
see that A minus lambda I times X must be
equal to 0, this we have already seen that
this X is a is the eigenvector correspond
to this lambda.
So, this implies, now you substitute lambda
as lambda as A 1 minus iota, so this is A
minus 1 minus iota times iota times identity
matrix into x equal to 0. So, this implies,
now when you when you take 1 minus lambda
here, so it is you can simply take lambda
here as 1 minus iota, so it is iota 1, and
it is 1 and it is again iota, it is minus
1 it is iota, and x is x 1, x 2 equal to 0
0.
So, this implies, now you can you can apply
some elementary row operation this in this
matrix. You simply first interchange these
two rows, it is 1 iota. Iota minus 1 x 1 x
2 remain as it is, interchange these two also
0 0. Now, this implies you can make 0 here,
with the help of this by applying the elementary
row operation R 2, this means R 2 minus iota
times R 1.
When you apply this elementary row operations
here, it is 1 iota it is 0 it is again 0,
and it is x 1 x 2 0 0. So, this implies x
1 plus iota x 2 equal to 0. So, this give
infinitely many solutions of x 1 and x 2 ok.
You can substitute any value of x 1, you can
find corresponding value of x 2 such that
x 1, x 2 is not equal to 0 of course,.
So, 1 such value is 1 such x 1 is you can
simply take you see x is what x is x 1, x
2. You can take x 1 as minus iota x 2 from
here, and it is x 2, so it is x 2 times minus
iota and 1, where x 2 is anyreal number.
So, so if you are talking about number of
linearly independent eigenvector corresponding
to lambda equal to 1 minus iota, so it is
1. So, you can pick any 1 linearly independent
eigenvector from this, because because it
gives infinitely many eigenvectors. You can
pick out any one eigenvector, so we can said
that vector is linearly independent eigenvector
corresponding to lambda equals to 1 minus
iota. So, we can say that corresponding to
corresponding to lambda equal to 1 minus iota,
the linearly independent eigenvector is say
minus iota 1 transpose.
Now, similarly you will takelambda is equal
to 1 plus iota, and we can you find out the
corresponding linearly independent eigenvector
on the same lines. Now, let us consider second
problem. Here the matrix B is 0 0 2 0 2 0
and this matrix. It is given towards at the
one of the eigenvalue of this matrix B is
4, then we have to find the remaining two
eigenvalues, and also the corresponding eigenvectors
of B.
So, let us discuss this problem now. Here
the matrix A is simply 0 0 2 0 2 0 and 2 0
3. Now, one lambda is 4, it is given to us
say lambda 1 is 4. Now, we know that the trace
is equal to the sum of the eigenvalues. So,
let us suppose the other two eigenvalues are
lambda 2, lambda 3, it is of order 3 cross
3, so it will be having three eigenvalues.
So, some of the eigenvalues is equal to trace
of the matrix that is 5 0 plus 2 plus 3. Now,
lambda 1 is 4, so this implies lambda 2 plus
lambda 3 is 1.
Now, the product of the eigenvalues, we know
that product of eigenvalue is simply determinant
of A. Now, the determinant of A, simply when
you when you simply open this matrix, I mean
determinant, then you simply get it minus
8. Now it is 4, so this implies lambda 2,
lambda 3 will be minus 2. So, solving these
two equations, we can easily find lambda 2
and lambda 3. So, it is clearly as we as we
are seeing it is lambda 2 is 2, and lambda
3 is minus 1. So, sum is 1 and the product
is minus 2. So, the other two eigenvalues
are 2 and minus 1.
So, now we can say that this matrix are the
eigenvalues 4, 2, and minus 1. Now, we have
to find out the corresponding eigenvectors.
Let us suppose, we have to find out the eigenvector
corresponding to lambda equal to 4. Similarly,
we can find eigenvector across point lambda
equal to 2, and lambda equal to minus 1.
So, let us find eigenvector corresponding
to lambda equal to 4. How you will find that
this is A minus lambda I X equal to 0 again,
so this implies A minus 4 I into X equal to
0. So, this implies minus 4 0 2, this is 0
minus 2 0, this is 2 0 minus 1, and X is X,
which is equal to 0, this implies.
Now, we will try to convert this into its
echelon form. So, we will make 0 here with
the help of this, this is already 0, we will
make 0 here with the help of this. So, which
operation we will apply R 3 2 R 3 plus half
of R 1. So, this is minus 4 0 2, this is 0
minus 2 0, this is 0 0, this plus half of
this is against 0. Now, this implies, this
implies minus 4 x 1 plus 0 into x 2 plus 2
into x 3 equal to 0. If you are taking X as
x 1, x 2, x 3, it substitute it here, and
we multiplied these two matrices, then we
simply get these equations, and minus 2 x
2 equal to 0. So, this implies x 2 equal to
0, and x 3 as 2 x 1.
So, what is the corresponding eigenvector,
corresponding eigenvector will be you can
say, you can see, it is x 1 0 and 2 x 1, so
it is x 1 times 1 0 2. So, again there are
infinitely many eigenvectors corresponding
to lambda equal to 4, but if you are talking
about number of linearly independent eigenvector
correspond to lambda equal to 4, so that is
1. You can pick any one eigenvector from this
space, so we can say that is a linearly independent
eigenvector correspond to lambda equal to
4. Similarly, we can find out eigenvector
corresponding to lambda equal to 2, and lambda
equal to minus 1.
Now, if lambda is an eigenvalue of matrix
A, and x is a corresponding eigenvector, then
we have the following properties. Now, it
is very easy to see these properties, you
see here.
See if A has a eigenvector eigenvalue lambda
and corresponding eigenvector is X, that means,
A X equal to lambda X. Now, if you want to
calculate the eigen, eigenvalues of alpha
A, where alpha is a non-zero scalar. Then
how we can find this, you simply multiplied
these two by alpha, both sides by alpha. So,
we can simply see that we can simply see that
this alpha A has an eigenvalue alpha lambda,
and the corresponding eigenvector is A corresponding
eigenvector is X.
Now, if you want for A square, you see A X
equal to lambda X, X not equal to 0. If you
multiply both sides by A, it is A into X equals
to A into lambda X, which is lambda times
A X. And A X is lambda X, so it is lambda
X. So, we can say that A square X is equals
to lambda square X. So, what we can say that
the if A has an eigenvalue lambda and the
corresponding eigenvector is X, then A square
has an eigenvalue lambda square and the corresponding
eigenvector is X.
Similarly, similarly we can say that A raised
to power K is equals to lambda raised to power
K into X, you can simply find similarly find
A cube, A raised to power 4 and so on. So,
what we have concluded, we have concluded
that if A has an eigenvalue lambda and the
corresponding eigenvector is X, then A raised
to power K has eigenvalue lambda raised to
power K, and the corresponding eigenvector
is X.
Now, next property is A and A transpose, both
have same eigenvalues that is very easy to
show, you can see that eigenvalues of or given
by this expression. Now, A minus A this matrix,
the transpose of this matrix is equals to
A minus the determinant of this is same as
determinant of this, because by changing interchanging
rows and columns will not change the value
of the determinant. This is equal to 0, it
is given to us, and this implies A transpose
minus lambda I whole determinant is equal
to 0 so that lambda is not changing here,
whatever lambda we are we are having here,
same lambda we are having here, that means,
the eigenvalues of A and A transpose are same.
The next is the next is if you say that A
X equal to lambda X, and suppose determinant
of A is not equal to 0, that means, A is invertible.
Then you can take A inverse both the sides,
then a inverse of A X will be equals to lambda
times A inverse of X. And this implies, identity
into x equals to lambda into A inverse of
X. And this implies, A inverse of X will be
1 by lambda time X, this is this is this is
one or identity here. So, we can say that
if A has an eigenvalue lambda and the corresponding
eigenvector is X, then A inverse has an eigenvalue
1 by lambda and the corresponding eigenvector
is X.
So, these are few of the properties of eigenvalues,
which is stated here. And if g lambda is an
eigenvalue of g A, where g lambda is a polynomial
of lambda, where x and x is an eigenvector
of g A corresponding to eigenvalue g lambda.
That that is very easy to show again, because
the third property itself is states that if
g lambda is a polynomial of lambda, then g
lambda is an eigenvalue of g A.
Now, let us let us discuss these few problems
quickly. You see, if A is a singular matrix
of order 2 and having trace 3. If it is singular
matrix means, determinant is equal to 0, determinant
is nothing but product of eigenvalues, product
of eigenvalues equal to 0 means at least one
of the eigenvalue is 0. Now, it is order to,
that means, it has it is having only two eigenvalues,
say lambda 1, and lambda 2.
So, so we can say A is a order 2 cross 2 having
two eigenvalues, lambda 1, and lambda 2. Then
product of eigenvalues is equal to 0, because
matrices singular, and trace is 3, trace is
3. So, here from here, we can say that one
of the eigenvalue is 0 say lambda 1 is 0.
If lambda 1 is 0, so lambda 2 will be 3. So,
the eigenvalues are 0 and 3 of A; eigenvalue
of A transpose will be again 0 and 3, and
eigenvalue of A transpose square will be 0
and 9 ok.
The next one is if A is a 4 cross 4 cross
4 matrix of eigenvalues 1 minus 1 2 minus
2. Then the eigenvalues of this how we can
find, you see eigenvalues of A are 1 minus
1 2 minus 2, and B is given as it is 2 A it
is 2 A plus A inverse minus I. So, it is some
polynomial in I mean, it is some polynomial
in I mean B, I mean A. So, the eigenvalue
of B will be nothing but if A as an eigenvalue
lambda, then the eigenvalue of B will be nothing
but 2 lambda plus 1 by lambda minus 1 2 has
2 lambda eigenvalue A inverse as 1 by lambda
eigenvalue and this eigenvalue is 1.
So, simply, simply you can substitute here,
corresponding 1, where having 2 plus 1 plus
1 minus 1 is 2 for B. Corresponding minus
1 if it substitute minus 1, it is minus 1
minus 1 that is minus 4. Corresponding 2,
it is 1 plus 1 by 2 minus 1 that is that is
4 minus 1 by 2 that is 7 by 2. Corresponding
minus 2, if I talking about minus 2, because
minus 2, it is minus 4 minus 1 by 2 minus
1 that is minus 4 minus 3 by 2 that is minus
11 by 2. So, these are the eigenvalues corresponding
to B.
So, the third problem is simple, if A is A
matrix of order n, and such that A raised
to power k is 0, where k belongs to a natural
number. Then all this eigenvalues are 0, because
if one of because if any of the eigenvalue
is not equal to 0 of A, then A raised to power
k will not be 0 ok. So, therefore, all the
eigenvalues of A must be 0.
So, this is a result that if A is A square
matrix of order n having k distinct eigenvalues
lambda 1 up to lambda k. Let v i be the eigenvector
corresponding to the eigenvalue lambda i,
i from 1 to k. Then the set this is linearly
independent.
So, let us try to prove this result. You see,
lambda 1, lambda 2 up to lambda k are distinct
eigenvalues, that means, that means, lambda
i is not equal to lambda j for i not equal
to j number 1. Now, the set we are considering
this v 1, v 2 up to v k, we have to show that
this set is linearly independent. v 1 is a
eigenvector linearly independent eigenvector
correspond to lambda 1, v 2 is a linearly
independent eigenvector correspond to lambda
2 and so on. So, we will prove this by the
method of induction ok.
Let us for n equal to 1 for k equal to 1,
we are having v 1, and we v 1 a single set
is I mean, it is it is linearly independent.
I mean singleton set is always linearly independent,
singleton non-zero vector, and v 1 is a non-zero
vector is always linearly independent. Now,
we will assume that it is true for k equal
to r I mean this result hold for k equal to
r, that means, the set of k r vectors are
linearly independent. And we will try to showed
that this also holds for k equal to r plus
1.
So, take the linear combination of these vectors
now, to in order to showed that these are
linearly independent. Now, you multiply both
sides by matrix A, so we can simply take A
into alpha 1 v 1 plus A into alpha 2 v 2 and
so on A into alpha r plus 1 v r plus 1 equal
to 0.
Now, A into v i is lambda i into v i for all
i, because lambda i is the eigenvalue and
the corresponding eigenvector is v i. So,
we can simply write lambda 1 A v 1 is a lambda
1 v 1, similarly alpha 2 lambda 2 v 2 and
so on up to alpha r plus 1 lambda r plus 1
v r plus 1. Now, say this equation 1, and
this is equation 2. You can multiply 1 by
lambda r plus 1 into 1 and subtract with 2,
so what we will obtain, we will obtain lambda
alpha 1 into lambda r plus 1 minus lambda
1 v 1 and so on up to alpha r lambda r plus
1 minus lambda r times v r equal to 0.
Now, we have already assumed that a set of
r vectors are linearly independent. So, this
is some linear combination of this linearly
independent vector, which is equal to 0. So,
this implies, alpha i is lambda r plus 1 minus
lambda i equal to 0 for all i. But, eigenvalues
are distinct, so this is not equal to 0. So,
this implies, alpha equal to 0 for all i for
all i means for i from 1 to r.
And when you substitute alpha from 1 to r
here with all 0, then you will get you will
get alpha r plus 1 v r plus 1 equal to 0.
And since, v r plus 1 is not equal to 0 is
eigenvector, so this implies, alpha r plus
1 equal to 0. So, we have shown that all alpha
i’s from i from 1 to r plus 1 equal to 0
that mean, this set of vectors are linearly
independent. So, in this lecture, we have
seen that that what eigenvalues and eigenvectors
are, and some of the important properties
of eigenvalues and eigenvectors.
So, thank you.
