So, let us start with the definition of the
Multiplicity of an eigenvalue, definition
of, let me put it in a bracket the word algebraic
multiplicity of an eigenvalue. So, let T be
a linear operator on a finite dimensional
vector space V, which has say dimension n
and let Lambda naught be a Eigen vector corresponding
or Lambda naught be an eigenvalue of our linear
operator T.
So, let Lambda naught be an Eigen value 
of T, then the multiplicity of Lambda naught
is the largest positive integer such that
Lambda minus Lambda naught to the power k
divides the characteristic polynomial of T.
So, let me define the multiplicity or maybe
let me just write a bracket algebra, this
many times also called algebraic multiplicity.
The algebraic multiplicity 
of Lambda naught is the largest positive integer
K such that Lambda minus Lambda naught to
the power k divides the characteristic polynomial
f of Lambda of T. So, let us go back and look
at our examples earlier.
So, the first example the characteristic polynomial
is Lambda minus 2 the whole square, the only
Eigen value here is 2 and the algebraic multiplicity
of the eigenvalue 2 here is equal to 2.
The same is the case with example 2, where
the characteristic polynomial is the same.
How about example 3? Example 3 has the characteristic
polynomial given by Lambda minus to the whole
square times Lambda minus 3, there are two
Eigen values in the Lambda naught equal to
2 and Lambda naught equal to 3, there are
two different Eigen values here, the multiplicity
of the eigenvalue 3 is equal to 1 that is
the largest number such that Lambda minus
3 to the power k divides our f of Lambda and
similarly, the algebraic multiplicity of Eigen
value 2 is equal to 2.
So, why are we concerned about this? The next
theorem tells us that the algebraic multiplicity
always dominates the dimension of the corresponding
Eigen space. So, let me prove that theorem
for you. So, let, T be 
a linear operator on a finite dimensional
vector space V and suppose, let Lambda naught
be an Eigen value of T.
So, the theorem tells us that, then dimension
version of E Lambda naught is less than or
equal to the multiplicity, so I will slowly
drop the adjective algebraic, so let us just
call it multiplicity of Lambda naught. So,
notice that if Lambda naught is an Eigen value
of T then there exists at least one vector
which is non-zero and such that it is the
Eigen vector corresponding to Lambda naught.
So, therefore the dimension is certainly greater
than or equal to 1. This theorem tells us
that the dimension has to be less than or
equal to the multiplicity of Lambda naught.
So, let us give a proof of this, so the fact
that, so let us give these things names so,
let k be equal to dimension of E Lambda naught
and therefore, what does that mean? Then there
exist v 1 to v k such that v 1 to v k or rather,
which is a basis of E Lambda. So, this is
a basis of the Eigen space E Lambda naught
corresponding to Lambda naught each of the
v 1, v 2 up to v k are Eigen vectors with
eigenvalue Lambda naught, but any linearly
independent set is contained in a basis so
we can extend it.
So, extending this, we get a basis which is
given by say v 1 to v k which are Eigen vectors
and then v k plus 1 to v n, where n is the
dimension of our vector space we have slowly
stopped putting n as being the dimension of
l always assume that n is the dimension of
v in the entire lecture.
So, let us try to look at how the matrix of
T so, let us call this Beta. So, interested
in what is T Beta, Beta what is the matrix
of T with respect to the basis Beta? What
will be the first column? The first column
will be T v 1, and what is that? T v 1 is
just Lambda naught times v 1, and similarly
T v 2 is Lambda naught times v 2, so other
bases do not contribute anything.
So, I would say that the first k columns will
just have a Lambda naught times I k and here
there will be a 0, the 0 matrix which is basically
n minus k cross k matrix. And then there will
be some matrix here which I would like to
split as something into k cross n minus k
matrix C and B which is n minus k cross n
minus k matrix.
So, what will be the characteristic polynomial
of T? So, let us see, then the characteristic
polynomial of T is given by, so recall that
the characteristic polynomial of a matrix
is invariant under similarity and therefore,
characteristic polynomial of a linear operator
can be defined with respect to any basis.
So, what is this? This is basically a determinant
of T Beta Beta minus Lambda times I n, which
will just turn out to be equal to the determinant
of the matrix Lambda naught minus Lambda times
I k, v 0 and the C are untouched, and there
will be a B minus Lambda times I n minus k.
This is precisely how the characteristic polynomial
will behave.
By one of the exercises, which was done in
week 6, this is just going to be equal to
the determinant of Lambda naught minus Lambda
to the power times I k times the determinant
of B minus Lambda I n minus k, determinant
of block matrices. Recall that this is a block
matrix of the correct sizes which was written
a bit above here.
So, what is the determinant of Lambda naught
minus Lambda times I k? That is just going
to be equal to Lambda naught minus Lambda
to the power k times some g of Lambda where
g of Lambda is the polynomial of degree n
minus k, which is basically the characteristic
polynomial of B here, so, it is going to be
a polynomial of degree at most n minus k.
And therefore, this is so what does this mean,
this means that k Lambda minus Lambda naught
to the power k so, hence, so, what is this?
This is f of Lambda and we have that f of
Lambda is Lambda naught minus Lambda to the
power k times g of Lambda therefore Lambda
minus Lambda naught to the power k divides
f of Lambda.
But what was the multiplicity of a given eigenvalue,
it is the largest positive integer l such
that Lambda minus Lambda naught to the power
l divides f of Lambda. Therefore, k has to
be less than or equal to the multiplicity
of Lambda naught, but what was k? K was nothing
but the dimension of the Eigen space and this
is less than or equal to the multiplicity
of Lambda naught and we have completed the
proof.
So, let me just show you the theorem we have
just proved. We have just shown that if you
start off with a linear operator and let Lambda
naught be an Eigen value of the given linear
operator T, then the multiplicity of the Eigen
value always bounds the dimension of the Eigen
space.
So, in the previous example, we saw that when
the dimension was equal to the multiplicity,
our linear operator was diagonalizable. So,
if the characteristic polynomial splits and
if this inequality which are circled in green,
if that turns out to be an equality, then
we had that, of course, we already know that
it was, we found out that it was diagonalizable
by explicitly computing a basis consisting
of eigenvectors
So, that yeah this I have already noted that
tempts us to conjecture that if in a general
n cross n matrix, if we can somehow show that
if we are given that the dimension of the
Eigen space is indeed equal to the multiplicity,
then for all the Eigen values Lambda i, then
maybe our linear operator is diagonalizable
and that is true and next goal would be to
exactly prove that. So, let us gather some
of the statements needed to prove what we
just mentioned.
So, the first one is a statement which we
have already proved in another case. So, let
T be a linear operator on a finite dimensional
vector space v. So, let Lambda 1 to Lambda
k be distinct eigenvalues of T. Suppose 
V i belong to E Lambda I, so that means V
i are elements in v or vectors in the Eigen
space corresponding to Lambda i.
If we have that v 1 plus v 2 plus up to say
v k so, this is for i equal to 1 to k. So,
for each these we have V i is in E Lambdas.
So, suppose v 1, v 2 up to v k all add up
to the 0 vector, then V i is equal to necessarily
equal to the 0 vector. So, I have slowly stopped
bothering about whether it is clear from the
context as to the 0 being written as the vector
zero or the scalar zero.
I think by now you should be able to quickly
catch from the context itself. So, you should
carefully look at what the context is and
conclude whether our vector is the zero vector
or the zero scalar. So, here it is the zero
vector as you can see, so I will leave this
as an exercise to you.
We have already noted that if Lambda 1, Lambda
2, up to Lambda k are Eigen values and distinct
Eigen values and v 1, v 2 up to v k are corresponding
Eigen vectors then they are linearly independent,
you should use that to conclude that this
forces V i x to be equal to be the 0 vector.
So, I will leave that as an exercise for you
and let me know make next maybe theorem.
So, let Lambda 1, Lambda 2, up to Lambda k
be Eigen values of distinct Eigen values of
a linear operator T. And suppose we get hold
of set say S 1 which are consisting of eigenvectors
with eigenvalue Lambda 1, and suppose S 1
is linearly independent and suppose S 2 is
a similar set corresponding to Lambda 2.
So, if you look at Lambda 1, union Lambda
2, the question is, is it again linearly independent?
The answer is yes in fact, for each Lambda
I, if we can have such an S I consisting of
linearly independent vectors of which are
Eigen vectors corresponding to Lambda I, then
the union will be linearly independent set,
let me write it down. So, let T be a linear
operator or maybe I should just consider matrices
and the same linear operators on V and suppose
and Lambda 1 to Lambda k be distinct Eigen
values of T.
What is that? I have always carefully written
Lambda 1 to Lambda k, k could be strictly
less than n that need not be n distinct eigenvalues,
if it is having n distinct eigenvalues, then
we already know that it is diagonalizable
and we do not need to do any of these things.
So, in fact, the interesting case comes when
case strictly less than n in whatever we are
doing right now. So, if suppose, we get hold
of S j be a linearly independent set consisting
of eigenvectors with eigenvalue Lambda j then
S equal to S 1 union S 2 union S k is linearly
independent.
So, let us prove this result so, let each
of these S j, just let us give some names
to the vectors in each of the S j. So, let
S j be given by v j 1, v j 2 up to v j n subscript
j, each of the S j consists of such vectors.
So, we would like to show that the union is
linearly independent. So, let there be a linear
combination of these vectors which is equal
to the zero vector.
So, let a ij be such that summation okay j
might not be a good idea so, let us be a little
careful with indices so use, yeah, that is
okay. j is going from 1 to k and say i is
going from 1 to n subscript j, and suppose
a ij v j. I should have just called it S I
and it would have been nice but that is okay.
Or let us call it a ji, does not matter, a
ji v ji, so i goes from 1 to n and so on.
This being equal to the zero vector, so this
is a linear combination. A typical linear
combination will be like this and we would
like to show that each of the coefficients
a ji are 0. To do that, what we will do is,
we will define w j to be equal to summation
i is equal to 1 to n j a ji vji.
So, notice that j is fixed and we are looking
at the thing in the bracket here, this is
what we are focusing on and we are looking
at sum here, what can we say about w j? Each
of the v ji belongs to S ji, S j rather this
means that w j belongs to span of S j which
is contained in E Lambda j.
Why is this? Since S j w j belongs to span
of S j, and S j is contained in E Lambda j,
because of this each of the W J's belong to
E Lambda j. And this expression above also
summation j goes from 1 to k, i goes from
1 to n j of a ji v ji is now equal to summation
j is equal to 1 to k w j, that is precisely
what we have written it as.
And we are given that this is equal to the
0 vector. But the previous proposition tells
us that if we have v 1, v 2 up to v k are
Eigen vectors corresponding to or rather elements
vectors in the Eigen space corresponding to
Lambda 1 up to Lambda k then each of them
have to be necessarily 0, by the previous
proposition we have w j is equal to 0 for
all j, but then what is w j?
Let us get back to what our w j was, w j was
the expression which I have put in a box and
this implies, so let me write it down this
implies 
summation a ji v ji that i goes from 1 to
n j is equal to the 0 vector, but what do
we know about S j, we know that S j is a linearly
independent set. So, there cannot be a linear
combination of vectors in S J which is equal
to 0, this implies that so for all j.
This implies that a j is equal to 0 the scalar
0 for all j and all i, and hence S is linearly
independent because we took an arbitrary linear
combination to be the 0 vector and we noticed
that this forces each of the coefficients
to be equal to 0. So, yes this is a linearly
independent set.
Now, let us state the main theorem. Main theorem
states that if we have a linear operator T
and suppose the Eigen values Lambda 1 up to
Lambda k, it satisfies the condition that
the dimension of the Eigen space. So, suppose
the characteristic polynomial of the linear
operator splits. So, we are given all this.
So, suppose we are in the situation where
the characteristic polynomial of our given
linear operators splits, then our linear operator
is diagonalizable if and only if the Eigen
space, the dimension of the Eigen space of
each of the Lambda i or each of the Eigen
values is equal to the multiplicity of the
Eigen values. So, let me state the theorem.
So, let T be a linear operator on a finite
dimensional vector space V such that the characteristic
polynomial 
of T splits. So, recall that that was the
entire context, we know that if T is diagnosable,
the characteristic polynomial splits. We are
now studying, given that the characteristic
polynomial splits, when can we say that our
linear operator T is diagonalizable. So, then
T is diagnosable if and only if the multiplicity
if the dimension of E Lambda i is equal to
the algebraic multiplicity of 
Lambda i for each Eigen value Lambda I of
T.
So, notice that it is if and only if statement,
it says that if T is diagonalizable, then
the dimension is the same, and if the dimensions
are equal, then T is diagonalizable. So, let
us prove both the directions, so let us first
assume that T is diagonalizable. So, we would
like to show that the dimension of E Lambda
i is equal to the multiplicity for each the
eigenvalues Lambda i. So, what does it mean
to say that T is diagonalizable? So, let Beta
be a basis of V consisting of eigenvectors
of T, so let us do one thing.
Let us start the proof one line ahead and
let n be equal to the dimension of V, let
us call the dimension of V to be equal to
m. So, and Lambda 1 to Lambda k be Eigen values
of T, so there are k distinct eigenvalues,
let me just add the word distinct, there are
k distinct Eigen values of T, k is less than
or equal to n. Okay that is an exercise for
you to show that there cannot be more than
n distinct eigenvalues of a linear operator
T.
So, we have a basis Beta and let Beta i be
equal to Beta intersected with E Lambda i.
So, Beta i captures those Eigen vectors in
Beta corresponding to Lambda i. So, let and
n i be equal to the number of elements in
Beta I. Now, n i is the number of linearly
independent vectors which are Eigen vectors
corresponding to Lambda i. And the first observation
is that n i has to be less than or equal to
the dimension of E Lambda i. So, let give
dimension of E Lambda I some name. So, let
d i be equal to the dimension of E Lambda
i and m i be equal to multiplicity of Lambda
i.
So, we know a few things, we know that d i
is by previous theorem or by a previous theorem
not the penultimate theorem we have d i is
less than or equal to m i. And the fact that
Beta i consists of linearly independent Eigen
vectors corresponding to Lambda i implies
that and the above observation and the fact
let me write down the reason, the fact that
linearly independent set in a vector space
of dimension d i has size less than or at
most d i
This implies that n i is less than or equal
to d i. But we know a few things about n i
namely, that summation n i, notice that summation
n i is equal to n, because summation n i is
the number of vectors in Beta, which is the
basis or which is a basis of V and therefore,
this is equal to n.
Since, Beta is a basis, also what is summation
m i that has to be equal to the degree of
f of Lambda, the characteristic polynomial
which is equal to n, this is from the explicit
form of the characteristic polynomial we have
seen in the last week. And what do we hence
have, we have that n is equal to summation
n i which is less than or equal to summation
d i, which is less than or equal to summation
m i is again equal to n, so there is a sandwiching
that has happened.
And therefore, summation m i or let me put
it this way m i minus d i is equal to 0, but
we know that m i minus d i is greater than
or equal to 0 by one of the previous theorems.
Even if one of them is greater than 0, the
sum cannot be equal to 0 because each are
non-negative quantities for all I, this implies
both star implies m i minus b i is equal to
0 and hence, we have proved one side of the
result for all i.
So, what have we proved? We have proved that
if we assume that T is diagonalizable then
we have shown that the dimension is equal
to the multiplicity. Let us now prove that
the dimension of E Lambda equal to the multiplicity
forces our linear operator to be diagonalizable
So, let us now assume that m i is or rather
di which is less than or equal to m i is in
this case equal to m i for all i. So, what
does this mean? This means that the dimension
of Lambda I, sorry dimension of E Lambda i
is equal to the multiplicity that is what
it means so, let us do one thing. Let Beta
i be a basis of E Lambda i, we know that Beta
i has size d i and by the previous theorem,
Beta is equal to Beta 1 union Beta 2 union
up to Beta k.
This is linearly independent because they
are linearly independent sets in different
Eigen spaces and therefore, Beta equal to
Beta 1 union Beta 2 up to Beta k are linearly
independent. What is the size of Beta? But
size of Beta notice that each of the Beta
i are mutually disjoint, it is equal to the
summation of the size of Beta i, which is
equal to the summation of the d i, but this
is now equal to the summation of a m i which
is equal to n.
Therefore we have a linearly independent set
which has size equal to the dimension of V
that forces it to be a spanning set and hence
a basis, hence Beta is the basis. What is
Beta? Beta consists only of Eigen vectors
of T therefore, basis consisting of eigenvectors
of T, hence T is diagonalizable. So, we have
obtained a necessary and sufficient condition
on, when T is diagonalizable given that the
characteristic polynomial splits.
