Let us now discuss an important concept of
matrices which all the Eigen value and Eigen
vectors of a matrix so let us understand what
is meant by an Eigen value an Eigen vectors
associated with a matrix.
Let us consider a matrix A we will start with
an example to find out given matrix A how
do you compute it is Eigen value and Eigen
vectors and look at their properties let a
given matrix we will take a very short small
size 1 or size 2 x 2 and let this be the matrix
for which we need to compute the Eigen value
and Eigen vectors and let a vector x which
as two components x1 and x1 the two components
here because the matrix is of size 2 x 2 so
we are talking of a square matrix here in
general okay.
However if the matrix A is of size m x n the
vector will be of size n okay so 11 components
n will be the dimensions of a inner vector
at this point we will not discuss Eigen value
associated with a non square matrix so let
us look at to compute this x okay so let x
be the of course Eigen vector the question
is how many of this access will be there that
depends on some properties of a basically
depends on the size of A.
We will now take that the number of possible
values of the Eigen vector and n number of
Eigen vectors x will be depended on the size
of A okay so let x be the Eigen vector associated
with 
an Eigen value for the time being we will
use the notation ? to denote an Eigen value
so if x is an Eigen vector and it is associated
Eigen value is ? you can write an expression
as A x which we can write by substituting
these two let us do that this is a simple
substation of A and then it is corresponding
Eigen vector x1, x2 okay and this is = to
?x okay in some we are Ax = ?x this also can
be rewritten in this particular form as A
– ?I times x by bring this to left hand
side this is = 0 okay.
So this is actually called the characteristic
equation for the matrix A and this will yield
some solutions for different ? okay how many
solutions I just talked about depending up
on the size of the matrix A so matrix A is
of size 2 x 2 for the size 2 because we are
talking about square matrices we can expect
there are 2 values of ?, ?1 and ?2 and also
there will be correspondingly two solutions
of this vector x well I will change the notion
here to be consistent.
Let us say this is the vector x okay with
these 2 components so now you will get in
some books the notation small x written here
with the vectors and which is indicated vector
so you can do that as well okay so it is all
right if you where actually written like this
then you had to put a x with a vector sign
but we will restrict ourselves the X which
indicates a vector okay so I am putting the
vector here yes x we will substitute x1, x2
are these 2 components of the vector x okay.
So the question now is given this characteristic
equation A – ?x Ix = 0 okay how many solutions
are possible okay and let us try to write
this in this particular form okay.
So A – I am writing this again it is called
the characteristic equation or characteristic
polynomial for the matrix A okay the degree
of the polynomial will be dictated by the
size of the A and hence the number of solutions
which are possible now depending up on the
value of A the question is how many solutions
are possible let us take a situation when
A – ?I this part remember I a magnitude
matrix introduce this earlier and ? is just
a scalar value okay.
So if we take this is matrix and look at it
is determinant value okay if you look at this
matrix and look this determinant value and
assume if this is singular that means if this
is = 0 that means this is singular what you
think will be the possible solutions of x
okay what you think will be the possible solutions
of x.
So we have an homogeneous system of equations
here and we look at the case when the determinate
of this matrix is singular and where we are
trying to find under this condition what are
the different values of ? which provide this
singularity condition within this matrix which
will actually then give this possible value
of ? and using those values of ? after we
substitute it here the value of the vector
x which satisfy this particular equation will
called the Eigen vectors.
So based on this fact that we have to find
out solutions for this characteristic equation
of the determinant of this term = 0 we will
substitute this value of A here on to that
expression and find out what are the value
? which satisfy this constrain to do that.
If we substitute this is what you will get
this is correct if you substitute it here
if you take the values of A the elements 2
x 2 matrix substitute it here this is what
you should get and this will in turn give
equation which is very simple to derive I
will leave this as a trivial exercise for
you to check that you can write it in terms
of this matrix okay what I mean is this particular
thing here is going to give this give me us
tow possible solutions for ? I will write
it simply 2 or 3 so hwy they are 2 values
because the size 2 x 2 if the size of A is
m x n you will have n different possible values
Eigen value, n different possible Eigen values
and correspondingly n different Eigen vectors.
We will use this 2 Eigen values to find out
the Eigen vectors in this case but if the
sides of A is very large say 10 x 10 matrix
are even higher in certain cases which we
care a lot in the field of pattern recognition
for clustering as well as classification there
are algorithm with the compute do what is
called an as an Eigen value become position
to find out what are the corresponding Eigen
values and then the corresponding Eigen vectors
we will talk of that later on as to what algorithms
one can use to actually to an Eigen value
decomposition to get the corresponding Eigen
value and Eigen spectrum.
And I will just name that one such algorithm
which is called the singular value decomposition
which you can use for large matrices but this
typical the case you can actually solve it
you know the matrix size is about 3 x 3 you
will get the polynomial of order 3 and you
can get 3 different diagrams but let us finish
this problem where you have 2 Eigen values
and using this so you chose the first one
which is ?= 2 did it here to get the corresponding
equations what you're looking at either A
– ? x = 0 or you can also write x = ?x is
given here I repeat again x = ?x / or you
can write something like this.,
A
And this will actually help you to form an
equation like this Ax – is let me take this
okay fine so if we substitute ? here let me
see what do I get ?= 2 will give me -1 , 1
, -2 what is this value tell me 2 multiplied
by say one such value of x which is given
by x1 and x2 = 0 so using the value of ?=2
as Eigen value you can substitute in this
characteristic equation and get this and you
have to find out a set of value for x1 , x2
which will stratify this constrain.
Actually what you will get is a the same equation
if you try substitute and write this in the
form of linear homogeneous equations we will
get 2 equations will are both identical and
the value of x1, x2 which actually stratifies
this can be given in this particular case
as this so this is an Eigen vector okay in
fact we very precise this can be multiplied
with an obituary value of K which is any scalar
quantities so K multiplied by this that means
you can replace 1, 1 by K and K here and all
such Eigen vectors with this quantity.
Fort all possible values of K will actually
give you possible Eigen vectors which will
satisfy this equation with this particularly
? now we will move to the other Eigen value
?= 3 and write the characteristic equation
there as well.
So if you do the equation which you will get
now is 
this is what you will get as A – ?I and
multiplied by x = 0 this is again giving you
the same is very clear that you will get the
same equation which will be stratified by
a value of a x = x1, x2 I am actually what
you are getting from this is a relationship
of this one okay what you will get is -2 x1
+ x2 =0 which in terms specifies that x1 = x2
/ 2 okay so any here are values x1 and x2
which satisfies this will be a possible solution
for the Eigen vector here one such example
could be well say 1 or 2.
Or you can also write this as ½ and 1 0.5
and 1 and multiplied by K because for any
obituary value f K here they will satisfy
this particular equation so this is a very
simple method where solve for the Eigen values
and look for the Eigen vectors which stratifies
this next we will take an example now here
the size of the matrix A is of size 3, 3 x
3 so now we will look at some examples of
obtaining Eigen values and Eigen vectors from
a matrix of size 3cross 3 okay and the dimension
is 3, so let us take an example.
3 cross3 matrixes so there are 3 rows and
3 columns very simple earlier we are taken
examples of 2 cross 2, so the matrix is – 21
– 9 19 0 6 0 -24 – 8 + 15, so we will
straight away look at the characteristic function
A – ?I which gives okay that should be,
so determinate of this matrix I leave it as
an exercise for you that when you write it
in this form you should be write a value in
fact then the dominant of diagonal and you
should be able to write and in this as an
exercise for you to write in terms of factors.
That is the third factor ?+9 we should be
able to write like this gives 3 correspond
Eigen value so the Eigen values are of course
you can ask me a question whether you will
get three such distinct factors or 3 distinct
Eigen values we will see an example next and
may not get 3 distinct diagonal values the
Eigen values are ? = 6 ?1 = 6 sometimes it
is called ?2 = 3 and ?3 = - 9 these are 3
corresponding Eigen values okay, so based
on base 3 Eigen values we will now try to
obtain the corresponding Eigen vectors.
So for each of this corresponding Eigen values
and it defined of the corresponding Eigen
vectors okay so what we need to do w need
to form equations where this is satisfied
okay we know the values ? so need to find
out corresponding these which will be satisfied
and their calling Eigen vectors but for this
solution of the equations they will be call
the non trivial solutions of this on system
or equation.
So let us start with the first Eigen vector
let us say which is – 9 to start which is
?3 = - 9 we can start with ?1 also does not
matter, so this will give A + 9I what will
you get that means the As is given here +
9 so you can see the diagonal elements will
only change correct, so that will give you
what will be the first element hones trial
– 21 + 9 which is – 12 the rest of the
non diagonal elements will remain the same
is not it, so this is 0 6 + 9 15 0 – 24
– 8 then 15 + 9 24, now 
you have to find out what value of V this
is satisfied so the method which is th allot
of for what is done is basically you try to.
Reduce this form of A – ?I we got a plus
because you have a – A this to – then
it is +9 okay so we call this is an row reduce
that means you try to operate row wise on
this matrix of course you can operate column
wise also and you try to get a row a current
form if your proper column wise will get a
column make in form also that is also possible
but here we are taking bout row radius to
get or to obtained the row Echelon form of
this matrix I will do in exercise to learn
to learn this.
If you do not know this what will do of the
end of the course will run, run one simple
example to show given matrix how to obtained
the row Echelon form the basic idea is that
they lead in entry of any particular row of
non zero rows should be 1 or it can be any
other non zero value all the zeros will be
at the bottom and correspondingly as we go
down the position of the starting element
of each such an non zero rows shift to the
right okay there are certain properties of
this row Echelon form.
We will see one simple example of trying of
the method to reduce m6 to row Echelon form
or row reduce form this is the example, so
I am giving this derivation here and if you
row Echelon form matrix and somebody work
out and tell me the values of this way Echelon
form t we have a 1 starting here, okay then
has to be 0 so the 1 could start this all
of the rest could have been 0 also so that
also valid in row Echelon form okay and then
you have, so this equivalent design that this
equivalent this form something like a scalar
multiple.
Because creating this row reduction form is
basically multiplying the matrix in which
the equivalents is established the equivalent
an exist and the corresponding Eigen vectors
have to be found out correspondingly say that
means what I will say some v1 v2 v3 in this
corresponding components of the Eigen vector
is should be equal to 0 okay, so actually
if you see here we can exploit the first two
rows of this matrix to get 1 constraint in
which you will get from here or may be should
I let I tell that.
Okay so from this I am proceeding here with
this let me write here, so you can see v on
– v3 = 0 so like that how do you get this
first row multiplied by this.
Similarly second row also you will get this
v2 = 0 so with the third elements any particular
constraint because it is all so based on this
you can set a constraint shear that let us
say v3 = some constant c1 arbitrary real value
one then what you will have is that this will
actually leads to v1 = - c1 or + because v1
= v3 here okay, so you will have something
like this 
some books will try o write Eigen vectors
as well as in that case they will poly Eigen
transpose okay so that will be this will be
one solution.
This will be one solution corresponding to
?1 = - 9 okay we started with this value,
so corresponding to this Eigen value so you
have any of these for any arbitrary values
of c1 is what you can put here multiplied
with this vector will give you the corresponding
the Eigen vectors and of you apply the same
process for the other two Eigen values 6 and
3 okay we will get two other different solutions
for Eigen vectors, so as I see here so this
solution comes from here ?3 = - 9 gives the
corresponding solution for first Eigen vector
okay so if you want I can write here as.
Say this sub script one indicating that this
is the first Eigen vector or a third on corresponding
to ? = 3 okay no similarly.
(Refer Slide Time: 26: 48)
You detect to Eigen values are ?2 = 6 and
?2 = 3 and ?1 = 6 will give me the solution
so ?1 = 6 will actually let me look at the
issue this is correct so that is the final
solution so these are the corresponding Eigen
values for this matrix and the corresponding
hanging vectors are 1 0 1 we actually take
c1 = 1 okay and the other two Eigen vectors
are given by this corresponding to okay and
these three Eigen vectors actually formed
a basic dimensions okay we will talk about
this basis.
Vector basis in span sub spaces and little
bit later on but for the time just remember
that these corresponding three Eigen vectors
are orthogonal dx span let us take another
example of a matrix of size three dimensions
3 and look at the Eigen values and Eigen vectors
whether we get first a fall whether we get
a unique set of Eigen values, so let us takes
example.
Let us say 
so straight away we will go to this which
will give us okay 5 – ? 0 – 1 0 8 – ? 0
– 3 0 then 7 – ? right so that if you
take the corresponding matrix are determinant
of this matrix and write in a polynomial form
and then factor it we will get I write this
as an exercise for you we should be able to
take okay may be you should get a summary
for pointing out right away that this is it
should be negative that is what you will get
okay so these gets as a sort of three Eigen
values where ?1 = 4 ?2 = ?3 = 8 this is one
way of writing this set of Eigen values there
are some is that ?1 = 4 is a multiplicity
1 and ?2 = 8 is a multiplicity to that means
it occurs twice.
Okay two Eigen values are same the question
comes is how many Eigen vectors well can expect
from this now that this will give Eigen vector
all right and this where of equal Eigen values
will give a model pair or Eigen vectors, so
you should get three Eigen vectors okay that
is what you should do so let us try the first
one A - 4 ? A – 4I that means what I have
done in this expression I substituted ?1 = 4
okay and if you do so if you look at this
express here what will you get tell me the
first row.
0 – 1 that is trivial than 0 8 – 4 this
is 4 0 – 3 0 this if you row reduce which
is left as an exercise and you solve it out
and tell me I will wait so what is the solution
we will get first row 1 0 – 1 last row 
so what we are looking is that 1, 0, -1, 0,
1, 0, 0, 0, 0 which is the same as the rows
here the correspondingly v1, v2, v3 the three
components of first correspond to ?1 this
should be equal to 0 okay and that will give
constraint which you have last time of did
you have v1 equals v3 or basically we get
v1-v3=0 and then v2=0 this is the solution
we adding in previous example as well.
So I am repeating that so it can look back
into a notes that means correspondingly for
this ?1=4 we will get the first remember what
I am suing this circuit indicates the dimension
or the component of the Eigen vectors or it
indicates the corresponding Eigen vectors
for this indexes or corresponding Eigen values
okay.
So what will you have last time a solution
v1 1, 0, 1 okay so we can write like this
if it is a row depth then you can write this
is a transpose so this is first row transpose
let start with the ?2 equals ?3 equals 8 what
is A-?i what is this matrix this will be -3
0 -1, 0, 0, 0, 0, -3,0, -1 you will get or
row column form that the leading in 3 will
be a non zero element in most cases this is
the consider as 1 but it can be more than
1 as well okay.
So this with the corresponding 
if I write in this form will actually give
you one condition that v1+v3 will be equal
to 0 with no constraint v2 or in other words
you can say that 3v1=-v3 that means you have
to know formula the play of Eigen vectors
from this constraint which will satisfy this
particular condition okay.
One of the sub conditions which I am going
to give you is the fall A which will be 
so this is one way of writing possible solutions
for this Eigen vectors under the constraint
you look at the conditions here 3v1=-v3 okay
which is the first and third dimension so
that constraint is satisfied here is a satisfied
here and v2 there is actually no constraint
so you set it to 0 and 1 so this two you can
choose arbitrary constants and they will give
you some dimensions of two Eigen vectors so
what is basically have now is that the first
Eigen values ?1=4 has given you the corresponding
Eigen vector here.
So this is v1 and v2 and v3 and the chosen
from here were chosen arbitrary values of
c1, c2 okay so choose arbitrary values of
c1 and c2 they can choose a 1 and 1 okay I
can choose 1 and 2 so such combinations will
give you two vectors v2 ad v3 correspond combining
with this v1 will actually give you three
set Eigen vectors which against three dimensions
okay.
So this is the simple example of a situation
where you have three Eigen values and the
Eigen vectors which you get are non specific
Eigen vectors but you can actually obtain
them from arbitrary constants let us take
is this always the case let us take another
example as a tree cause matrix which will
again have multiplicity but different results
of the Eigen vectors.
And that matrix an example which I am taking
now which as 0,1,3,0,6,0,-6,2,9 why I am taking
different examples of dimension three wee
have started with an example of dimension
two I amen vector 2 values to Eigen values
to Eigen vectors I am taking different examples
of dimension three because we are getting
three Eigen values some of them with more
them one multiplicity that means they are
duplicated.
And they are gain were is to three corresponding
Eigen vectors this Eigen vectors may or may
not span is talked about vectors specific
soon immediately after this in which the concept
of vector space sub space and the span will
be clarify so let us proceed with this and
the corresponding A-?i matrix will be –? 1,
3, 0, 6 -?, 0, -6, 2 then 9-? and need to
tell the determinant of this transpose.
And what you will get is a form this is an
exercise for you we should probably get this
again here multiplicity of 2 in 1 particular
case will start with this so this will give
you ?1=3 ?2=?3=6 is there a –sign here does
not matter4 actually of this is no – sign
here fine okay so let us start with the first
Eigen value which is ?=3.
So this was A-3?sorry A-3i what will the result
so you have -3,1,3,0,3,0,-6,2,6 correct tell
me what you left is basically 0 has operation
0 and 0’s at the bottom.
So the corresponding Eigen vector should deal
1, 0,-1, 0, 1, 0, 0, 0, 0 then of course v1,
v2, v3 like the process we have been doing
earlier find out the Eigen values substitute
into the expression with –?i times the equals
to 0 so this again will the constraint v1,-v3=0
v2=0 this we already have again which coming
back this constraint okay.
Then you know the solution by heart now I
think we have done there that this will give
you the solution first Eigen vectors tell
me 1, 0, 1 okay we want to prove it transpose
it to 1 row vector otherwise you can leave
it as a column here okay so this is 1 with
the corresponding ?1=3 so ?1=3 give as this
solution 3.
Let us take other one what is the multiplicity
here 6 ?2=?3=6 what is the matrix from A here
this is -6, 1, 3, 0, 0, 0, -2, 2 this is correct.
Yeah this 
is different from the previous cases correct
we did not have this probably so this gives
2v1-v3=0 on v2=0 into the first row will give
this on the second row will give this okay
so we can form the corresponding solutions
here is unlike a previous case which will
give us what? 1, -0,2 correct.
Because v3 equals 2v1 multiply by c1 if you
like but does not we do not have v3 in fact
v3 is same s v2 same Eigen vector okay so
this is the case where we can see that the
is one Eigen vector this is the second Eigen
vector and what e have form this Eigen vector
say this is substitute in three dimension
okay.
This two v1 and v2 put in together because
v2 will be actually same as v3 now because
if you substitute ?=6 you have gone for this
is different from the previous case when we
had multiplicity here as well ?3=8 if I kept
it on the board I have not rub it to show
you that this is equal to 8 then substituted
what will row reduce forms and then we got
this constraints.
We did not have constraints for v2 in the
previous example now we have it here this
constraint form the second row that was not
the case here for this free constraint from
v2 first step we write the solution in this
particular form for Eigen vectors which is
dependent on two arbitrary constants v1 and
v2 with satisfy the constraints as given here
and that gives as a corresponding to v2 and
v3 has functions of c1 and c2.
And then we have three Eigen vectors this
not case here in both case 6 for this particular
matrix example we have gone first solution
for three we have got this Eigen vectors for
the corresponding multiplicity 2 here for
Eigen values we have and this spans so here
v1 and v2 span we will say the subspace two
dimension I am going to talk about vector
spaces very, very soon.
