See 
we are discussing diagonalizability. In todays
lecture we will derive one necessary un sufficient
for diagonalizability of a linear operator.
So let us go back and look at this problem
diagonalizability of a linear operator T 
this means that I can write down the matrix
of T relative to some basis as a diagonal
matrix, okay. I am writing down this diagonal
matrix let me assume that by the way is it
clear that the eigenvalues of this diagonal
matrix must be the eigenvalues of T the eigenvalues
of the diagonal matrix.
So what is the meaning of saying that T is
diagonal? This means 
T u i I am just writing down the equation
that I gave yesterday instead of alpha let
me write lambda i u i, okay these are the
numbers that will figure here these lambda
i's are the numbers that will figure here,
okay. Now some of we have seen example the
second example where the eigenvalue repeats
some of these could repeat.
So what I will do is write down the distinct
eigenvalues and take care of multiplicity
when I write down the diagonal matrix. In
other words let me say lambda 1, lambda 2,
etc lambda k be the distinct eigenvalues be
the distinct eigenvalues of T with multiplicities
with multiplicities I will use this notation
different from what I used yesterday in the
last lecture with multiplicities let me say
d 1, d 2, etc d k that means lambda 1 comes
as an eigenvalue of the operator T d 1 times
in other words the characteristic equation
that characteristic polynomial is determinant
of (A min) determinant of T minus lambda i
is 0, characteristic polynomial is determinant
of T minus lambda i, characteristic equation
is determinant of T minus lambda i equal to
0 this is the polynomial of degree n.
When I write down this what I mean is that
lambda 1 appears d 1 times as a root of that
characteristic equation, lambda 2 appears
d 2 times, etc lambda k appears d k times,
okay. So what is clear is that d 1 plus d
2 etc plus d k is equal to n the degree of
the polynomial, okay. If this is the case
then this can be rewritten then I can write
this matrix diagonal matrix as lambda 1 now
appears d 1 times so the first block will
have lambda 1 appearing d 1 times is it clear
that I can write it as lambda 1 I 1, lambda
2 I 2, etc lambda k I k all other entries
are of course 0 with the convention that I
1 is the identity matrix of order d 1 cross
d 1, I 2 is identity matrix of order d 2 cross
d 2, etc I k is identity matrix of order d
k, okay.
So let me confirm where I i is the identity
matrix of order d i so I can first this is
the first observation the diagonal matrix
d can be written in this manner after listing
the distinct eigenvalues. So let us observe
that the statement that I made just now d
1 plus d 2 plus etc d k must be the dimension
of the space that I started with I will always
assume the dimension to be n in the rest of
the discussion, okay.
If this is the case what is the characteristic
polynomial of T? The explicit formula for
the matrix of T has been written down we knows
the characteristic polynomial does not change
when I change the basis, okay because it is
basically the determinant of some matrix which
does not change under change of basis, okay.
So the characteristic polynomial so I am using
p for the characteristic polynomial I will
use p here after for the characteristic polynomial.
So the characteristic polynomial of this operator
T can be written as it will be lambda will
be a variable lambda 1, lambda 2, etc are
the 0’s so I have lambda minus lambda 1
to the d 1 into lambda minus lambda 2 d 2,
etc lambda minus lambda k to the d k is it
clear that this is the characteristic polynomial
of T, okay. See in general this cannot be
done I am assuming if T, I am assuming the
T is diagonalizable in that case I can do
this.
For example look at the operator first operator
that we considered yesterday the rotation
matrix when the angle is 90, the characteristic
polynomial there is lambda square plus 1,
okay and lambda square plus 1 cannot be factorized
as lambda minus lambda 1 into lambda minus
lambda 2 for lambda 1, lambda 2 real because
this is an irreducible polynomial over the
polynomial ring F x, R x, okay irreducible
polynomial over R lambda square plus 1 is
one example we cannot factorize there that
is an example of a operator which is not diagonalizable.
So this cannot be done always if it is diagonalizable
then this can be done, if the operator is
diagonalizable then the characteristic polynomial
can be factorized into products of powers
of linear factors let me mention this is the
product of powers of linear factors, okay
okay. So this is one information that the
characteristic polynomial is a product of
powers of linear factors is one information.
I also have the other information look at
look at the eigens look at the dimension of
the subspace null space of T minus lambda
i I, I want to calculate the dimension of
the subspace null space of T minus lambda
i I, I want to calculate the nullity of T
minus lambda i, okay can you tell me what
it is?
What is what is this subspace first? See I
want T minus lambda i I, I have T here I can
treat like matrix for being specific let us
take lambda i to be lambda 1 i is 1 lambda
1 T minus lambda 1 I this first block is 0,
all the other entries will remain why because
they are distinct this will be lambda 2 minus
lambda 1 times I 2, this will be lambda k
minus lambda 1 times I k. I am doing T minus
lambda 1 I so these entries will not be 0
distinct eigenvalues so lambda 1 will not
be equal to lambda 2 etc lambda k.
But this block is 0, I want the set of all
solutions of the matrix equation let us say
some some A x equal to 0 where A is a matrix
whose first block is 0 all other diagonal
entries are not 0 so what is the dimension
of the solution space is it clear it is d
i the rank of T minus lambda 1 I will be the
rank coming out of these nonzero entries so
nullity corresponds to just this so nullity
of T minus lambda I is d i please check this,
so what is that clear first? The null space
has this dimension.
By the way what is the null space of T minus
lambda 1 I? Can you see that it is the it
is eigenspace corresponding to the eigenvalue
lambda I this I did not mentioned in the last
lecture. This is the set of actually the space
the set of all eigenvectors of T corresponding
to the eigenvalue lambda i this is the set
of all eigenvectors of T corresponding to
the eigenvalue lambda i first observation
this is a subspace I did not make this point
yesterday the set of all eigenvectors corresponding
to an eigenvalue forms a subspace because
it is a null space of a certain linear transformation
null space of T minus lambda i.
So this subspace has dimension d i remember
there are there are two objects coming here
one object is a polynomial the other one is
the dimension of some subspace, if T is diagonalizable
these two numbers are the same this is an
important observation because we will see
that you can go back to example 2, example
3, example 2 we discussed only two examples
we can go back to example 2.
Look at the dimension of the subspace corresponding
to the eigenvalue 1 the dimension is 1 there
whereas the multiplicity of 1 as an eigenvalue
was 2 in general these are different, okay.
For diagonalizability it is crucial that these
two numbers are the same, okay this is the
second important point first point is that
the characteristic polynomial is can be written
as a product of powers of linear factors,
second fact is that the dimension the number
of times lambda i appears as an eigenvalue
of A that is d i that number is the same as
the dimension of the eigenspace corresponding
to eigenvalue lambda i, okay okay.
Let us proceed and we will this is important
as you will see this is important for diagonalizability,
okay. As I told you I want to look at a characterization
let me give you one or two results before
I proof that result. First I want to demonstrate
the following 
suppose T is an operator on a finite dimensional
vector space V and f of lambda okay f of t
I use small t 
okay let me just say f be a polynomial 
be a polynomial over F, V is vector space
over F, f is a polynomial over F by which
I mean that this polynomial has its coefficients
coming from F, okay so if F is R this is a
real polynomial.
If Tx equals lambda x for lambda the underlying
field, then f T of x is f lambda x you will
need this little result when f T of x is f
lambda x what this means is that if lambda
is an eigenvalue for the operator T and if
x is eigenvector then look at f of T f of
T is also a linear operator see little f is
a polynomial f of T is another linear operator
for this linear operator we want to show that
f of lambda is an eigenvalue with a same eigenvector
x the same eigenvector x, okay.
Proof straight forward you just write down
f of t let say it is a not plus a a 1 t plus
a 2 t square, etc let us say a s t to the
s where the coefficients come from the underlying
field, then what is f of capital T wherever
little t comes I must replace it by capital
T the first term it is T power 0 that is identity
operator, okay so my f of T is this polynomial
a not I plus a 1 capital T plus a 2 capital
T square etc a s T to the s you remember T
is an operator on V it is a linear transformation
from V to itself.
So it is T square T cube etc make sense composition
this is f of T, what do I want to verify verify
this. So let me write f T x this will be a
not Ix plus a 1 Tx plus a 2 T square x plus
etc I need to calculate each term, okay but
it is easy to see that T is okay let us calculate
T square x for example T square x is T of
T of x T of x is lambda x, so this is T of
lambda x this is lambda T of x again T of
x equals lambda x. So this is lambda square
x, so T square x is lambda square x by induction
T power R x is equal to lambda power R x so
this can be rewritten so it is a simple result.
So if you look at f T x it is a not x plus
a 1 lambda x plus a 2 lambda square x plus
etc plus a s lambda to the s x, x is the vector
take that outside all the others are numbers
coefficients a not plus a 1 lambda plus etc
plus a s lambda to the s into x this is a
this is a number coming from the field but
this is precisely f of lambda instead of t
I have lambda. So this is f of lambda x, okay
so remember that f T is a polynomial in T
if lambdas and eigenvalue and x is the corresponding
eigenvector then we have found out an eigenvalue
for f of T that is f lambda for the same eigenvector
x, okay okay that is a simple computation.
I want one more result before I proof the
main theorem 
so 
the framework is the same as before lambda
1, lambda 2, etc lambda k are the distinct
eigenvalues of a linear operator T on a finite
dimensional vector space V. Let me say that
W i let W i be the eigenspace corresponding
to to the eigenvalue lambda i so I have these
k eigenspaces corresponding to eigenvalue
lambda 1, etc lambda k these eigenspaces are
subspaces I can talk about the sum of these
subspaces.
Let W equal to W 1 plus W 2 etc plus W k,
I take the sum of these k eigenspaces. Now
remember that dimension of W 1 plus W 2 is
general is not equal to dimension W 1 plus
dimension W 2 you need to remove the dimension
of W 1 intersection W, okay but in this case
the dimensions add up in this case the dimension
will add up that happens if the subspaces
are eigenspaces that is what we are trying
to do here.
So what is the conclusion then dimension W
is dimension W 1 plus dimension W 2 etc. In
particular what this means is that eigenvectors
correspond to distinct eigenvalues are linearly
independent, okay this is this statement is
encoded in this eigenvectors corresponding
to distinct eigenvalues are linearly independent,
okay okay.
So let us proof this then we will use this
in characterizing diagonalizability. All that
I will do is take a basis for W 1, take a
basis for W 2, etc basis for W k show that
the union is a basis for the sum W, is that
okay? We will take a basis for W 1, basis
for W 2, etc W k take the union I will show
that that is the basis for W then it follows
that dimension of W is the number of elements
in B 1 plus the number of elements in B 2
etc number of elements in B k that is precisely
this, okay.
So let us write down now basis explicitly
let B 1 equal for the subspace W 1 corresponding
to lambda 1. So B 1 I will call it u 11, u
12, etc u 1l 1 the first subscript denotes
that it corresponds to that eigenvalue first
subscript denotes it corresponds to eigenvalue
lambda 1, okay so this is corresponds to eigenvalue
lambda 1, B 2 is u 21, u 22, etc u 2l 2 this
corresponds to lambda 2 etc B k u k1, u k2,
etc u kl k this corresponds to the corresponds
to eigenvalue lambda k.
Let these be basis ordered basis for W 1,
W 2, etc W k. I must show that 
I must show that this is a basis for W, okay.
Spanning set we will (())(22:12) of quickly
linear independence is what is a little difficult
in this problem little more involved than
the other one spanning set. I want to show
that this union this is the spanning set for
W okay take anything in W little w that is
that is W 1 plus W 2 etc W k each W 1 in term
can be written in terms of these so it is
clear that this is a spanning set.
Clearly this is this spans W linear independence
is 
see if you show that this is the basis for
W, I am assure that it is a spanning set and
it is a linearly independent set. The fact
that this is a spanning set is straight forward
for the following reason take any little w
in W then by definition that W is sum W 1
plus W 2 etc W k but look at W 1 the first
term that is the linear combination of these
etc so this W is a linear combination of these
vectors that is those vectors in script B
and so this spans W linear independence.
Consider so we need to solve linear independence
so we need to consider a combination. So consider
alpha 11 u 11 plus alpha 12 u 12 I choose
the scalars also according to the superscripts
etc alpha 1l 1 u 1l 1 plus alpha 21 u 21 alpha
22 u 22 plus etc plus alpha 2l 2 u 2l 2 plus
etc last term alpha k1 u k1 alpha k2 u k2
etc alpha k what is that last l k u kl k suppose
this is 0.
I must show that each of the scalars is 0,
okay consider this combination I must show
that each scalar is 0 it would then follow
that these vectors u 11 u 12 etc u 1l 1 u
21 u 22 etc u 2l 2 u k1 u k2 etc u kl k they
are linearly independent, okay. Let us this
is this vector is 0 vector f of t is take
any polynomial for any polynomial f of t I
will consider f of capital T f of capital
T is also linear I will apply f of capital
T on this of this whole thing, okay.
Alpha 11 u 11 plus alpha 12 u 12 etc alpha
1l 1 u 1l 1 this time I will just write down
okay does not matter alpha 21 u 21 etc alpha
2l 2 u 2l 2 plus etc the last one 
last one is alpha k1 u k1 plus etc plus alpha
kl k u kl k, f of T is also a linear operator
so any linear operator has a property that
its action on the 0 vector this is 0 vector
so 0 is this, f of T is linear because T is
linear I will apply this to each term, okay.
So can you see that the first term will be
alpha 11 f T u 11 use the previous lemma it
is f of remember u 11 is an eigenvector corresponding
to eigenvalue lambda 1 all these u 11 u 12
etc u 1l 1 they comes from they come from
the eigenspace corresponding to the eigenvalue
lambda 1 so each of this is an eigenvector
corresponding to eigenvalue lambda 1 apply
the previous lemma f T of x equals f of lambda
x if lambda is the eigenvalue.
So the first set of terms is it clear? That
it is alpha 11 f lambda 1 u 11 plus f lambda
1 u 12 etc u 1l 1 first terms will go with
f lambda 1 plus the second ones will go with
the f lambda 2 the last one will go with f
lambda k do you agree with this? I forgot
the constants here, here also, okay is it
clear the first set of terms coming from this
bracket the first one here the first set of
terms here they will go with f lambda 1 because
each of those vectors u 11 u 12 etc u 1l 1
each of those vectors is an eigenvector corresponding
to the eigenvalue lambda 1 and I am appealing
to the previous lemma, okay.
This is true for any polynomial f this is
true for any polynomial f. I will make now
particular choices of f and then show that
this is 0, another choice I will show this
is 0, etc okay. Suppose I have one choice
for which the second set of terms, third set
of terms etc the last set of terms vanish
this only remains does it follow that those
scalars those scalars will be 0, okay I will
show such polynomial in such a way that this
is not 0.
So it will follow that alpha 11 alpha 12 etc
alpha 1l 1 is 0 apply the next polynomial
I will choose it in such a way that f lambda
2 is not 0 it will follow that these coefficients
are 0, etc I will go back and substitute into
this equation I am sorry I want to conclude
each scalar is 0 so it will follow that these
are independent, okay.
Now what is that polynomial? For the first
one I will do it for the first one the rest
is similar. For the first one I will choose
f of t to be see I want these to be 0 I want
a polynomial I want the polynomial to have
lambda 2 etc lambda k to be 0 give me one
choice t minus lambda 2, t minus lambda 3,
etc if you choose this then f of lambda 1
f of lambda 1 is lambda 1 minus lambda 2 lambda
1 minus lambda 3 etc product lambda minus
lambda k these are distinct so that is not
0, okay.
So f of lambda 1 is not 0 f of lambda 1 is
not 0 but f of lambda i equal to 0 for all
i greater than or equal to 2. So I will take
this polynomial apply it to this equation
then I get okay I take this polynomial apply
this polynomial to this equation then I get
the second set of terms are 0, etc I have
only these terms remaining, I can write it
as f lambda 1 outside into the rest of them
alpha 11 u 11 etc alpha 1(k) l 1 u 1l 1 only
the first set of terms will remain f lambda
1 is not 0 so this can be cancelled, look
at the rest that is 0 but these are linear
independent, right they form a basis for W
1 these are linear independent from this it
follows that the first set of coefficients
alpha 11 alpha 12 etc alpha 1l 1 first set
of coefficients must all be 0.
So you apply the second polynomial which is
T minus lambda 1 into T minus lambda 3, T
minus lambda 4, etc T minus lambda k then
you can show the second set of coefficient
0 etc. So it follows that each of these scalars
started with this equation follows that each
of the scalars is 0 and so these vectors are
linearly independent so this is the basis
for space W, okay.
As I mentioned earlier in particular this
means that eigenvectors corresponding to distinct
eigenvalues are linearly independent we have
proofed something more, okay let me proof
this theorem then theorem that characterizes
diagonalizability the framework is as before
T is the linear operator in a finite dimensional
vector space with dimension n over f V is
defined over f I have lambda 1, lambda 2,
etc lambda k as the distinct 
eigenvalues let W i be the eigenspace corresponding
to the eigenvalue lambda i then the following
statements are equivalent.
I will give two conditions that are necessary
and sufficient for T to be diagonalizable
first statement is T is diagonalizable. Second
is the condition involved in the representation
of the characteristic polynomial characteristic
polynomial I am denoting by p I will write
p of lambda it is lambda minus lambda 1 to
the d 1 lambda minus lambda 2 to the d 2 etc
lambda minus lambda k to the d k.
Where d 1 plus d 2 plus etc plus d k equals
n I am assuming that n is a dimension of the
space V. The last condition is in terms of
sums of the eigenspaces, if you look at the
sums of the eigenspaces call it W then this
W the subspace is the whole of space V this
is the last condition we have already observed
that a implies b, T is diagonalizable then
we have seen that the characteristic polynomial
has this form from b implies c, okay.
Proof a implies b, already observed b implies
c follows because okay you can set W to be
the sum of these subspaces then by the previous
lemma it follows that dimension of W is summation
i equals 1 to n dimension W i this comes from
the previous lemma because these are eigenspaces
corresponding to distinct eigenvalues but
what is the condition on look at condition
V the condition on these numbers d 1, d 2,
etc d k is that their sum is n.
So dimension W is summation i equals 1 to
n d i that is n. So I have a 1 to k there
are k subspaces ya so this sum is n so but
we know that the sum of two subspace is again
subspace, W is a subspace of V having the
same dimension so W must be the whole of V
that statement c the sum is equals to W, W
is equal to V we have shown so b implies c
also holds c implies a c implies a follows
from the definition of what we mean by diagonalizability
is that clear.
I want to show T is diagonalizable that is
I want to show that there is a basis script
b of V such that each vector from b is an
eigenvector of T that is diagonalizability
I saw this yesterday T is diagonalizable if
and only if there is a basis b for b for V
each of whose vector is an eigenvector for
T. I know that V is the direct sum of is a
sum of these subspaces look at the construction
that we did earlier take a basis b 1 for W
1, b 2 for W 2, etc b k for W k each of the
basis has the property that each of this basis
for the subspaces b 1, b 2, etc b k have the
property that their elements their vectors
in b 1 for instance is an eigenvector corresponding
to lambda 1, the vectors in b 2 are eigenvectors
corresponding to the eigenvalue lambda 2,
etc.
The combination is a basis for V the union
of these basis for the subspaces is a basis
for V and so each vector of this basis is
an eigenvector of V is an eigenvector of T
so T is diagonalizable. So can I just say
c implies a follows from really the definition
which we saw yesterday really the definition
that there is a basis b for V each of whose
vectors is an eigenvector for T, okay.
So this is one characterization that is necessary
sufficient condition for T to be diagonalizable
one is that the characteristic polynomial
can be written as a product of powers of linear
factors, second is that the whole space V
can be written as the sum of these subspaces
the subspaces being the eigenspaces, okay
this is one characterization, we will also
look at another characterization involving
the so called minimal polynomials, okay that
I will do in the next one or two lectures.
But before I conclude I want to atleast mention
what is the minimal polynomial we want to
know when precisely a linear transformation
is diagonalizable, okay one answer has been
given here two answers really look at the
eigenspaces, take the sum, verify if that
is the whole space V, the other thing is look
at the characteristic polynomial verify if
it is a product of powers of linear factors
then you know that T is diagonalizable.
There is another answer as I told you which
comes in terms of the minimal polynomial,
okay so let me atleast give the definition
of the minimal polynomial. Remember that we
are looking at an operator a single linear
operator T and we are analysing the operator
T. In this study what is important is to identify
classes of polynomials which have the property
identify classes of polynomials let us say
f of t such that f of capital T is 0.
What are all polynomials f of t such that
f of capital T is 0? Now why why is this statement
true that will be clear only a little later
you will get the connection between minimal
polynomial and the characteristic polynomial
then it will be clear as to why these polynomials
are important, okay but let me atleast give
the concept of the minimal polynomial coming
from this, okay remember f is a polynomial
its coefficients come from the underlying
field, okay.
Now if you look at if you look at F t call
that script A look at F t call that script
A then so what is this F t F t is the set
of all it is it has an algebra structure,
it is a set of all polynomials in a single
variable t in a single real variable t, I
should actually write F but for the sake of
convenience I am writing it R single real
variable t so I should actually write R of
t, okay may be let me go back and change this
to F.
So it is either for our discussion let us
say it is a set of all real polynomials over
a single variable t single real variable t,
the coefficients are real, the variable t
is also real I am using script A for that
it is what is called as an algebra you know
that it is a euclidean domain set of all polynomials
is a euclidean domain it is a commutative
of ring where you can do euclidean algorithm
it is a commutative ring where euclidean algorithm
can be applied.
It is it is an euclidean domain which has
a property that you know the concept of an
ideal concept of an ideal in subring, okay
does not matter if you do not know you will
learn it now this semester sometime. What
can be shown is that an ideal is a subring
of see this is what is called as an algebra
an algebra is an algebraic structure where
you can do multiplication of vectors, okay.
So an algebra is something more than a vector
space where there is also a possibility of
multiplying vectors. Now multiplication here
is multiplication of polynomials multiplication
of polynomials you know term by term multiplication
of polynomials is term by terms. So one could
do multiplication of polynomials it also has
one or two little (())(43:55) but let us not
worry about that this is an algebra this is
a euclidean domain where I can do where product
of vectors make sense, in such a euclidean
domain the notion of an ideal comes, an ideal
is a sub bring an ideal is a sub bring which
has a property that if I take an element from
the ideal and an element from outside the
ideal that is f of T then the product will
belong to the ideal, okay the product will
belong to the ideal.
Ideals in a euclidean domain have the property
that they are generated by a single unique
polynomial ideals in a euclidean domain are
characterized with the property that they
are generated by a single unique element which
means anything in the ideal is a multiple
of a specific polynomial anything in an ideal
given an ideal anything in that ideal is a
multiple of a unique element coming from that
ideal, okay.
I need this property one can also do without
this property for the minimal polynomial.
I wanted to find the minimal polynomial first
of all the question is given a linear transformation
T, does there exist a polynomial f such that
f of T is 0, okay. I will give two answers
one for a specific example 
linear transformation matrices they are equal
actually so I will take this matrix 1 1 1
1 I want you to consider this polynomial f
of t equals t square minus 2t, f of t is t
square minus 2t how do I get this?
At the end of the next two lectures you will
also be able to write down such polynomials,
t square minus 2t then you can verify that
f of A is a 0 matrix A square is equal to
2A, A square is equal to 2A for this matrix
that is the reason why I choose t square minus
2t then f of A equal to 0. So given a linear
transformation this make sense the question
does there exist a polynomial f such that
f t equal to 0 make sense I have given one
example.
I will actually prove it I will actually prove
that given a linear transformation on a finite
dimensional vector space there is a polynomial
which has the property that f of capital T
equal to 0 tomorrow and then may be define
the minimal polynomial and how it is related
to the notion of diagonalizability, okay let
me stop here.
