Okay, we are looking at invariant direct sum
decomposition characterisation of diagonalizability.
We are to proof the converse part right let
me write down that statement of the converse
there is a slight modification to the statement
that I gave in the last lecture.
So let me write down all the conditions conversely
suppose that suppose that the numbers suppose
that there exist I want to say suppose that
there exist distinct numbers lambda 1, lambda
2, etc lambda k and k non-zero linear operators
k non-zero linear operators I am calling them
E 1, E 2, etc E k these are operators on V
such that such that the following conditions
are satisfied.
First condition is T is lambda 1 E 1 plus
lambda 2 E 2 etc plus lambda k E k, T is a
specific linear combination of these operators.
Condition 2 identity is just a sum of these
k operators that is the second condition,
third condition I will write E i E j equal
to 0 for i not equal to j this is not the
third condition that I gave you yesterday,
third condition I gave yesterday is E i square
is E i.
Suppose I have these numbers lambda 1, etc
lambda k and operators k operator non-zero
operators that is important such that these
conditions satisfied then 
E i square equals E i this is really a consequence
not an assumption is a consequence of these
three conditions this happens and 5 also holds.
Now what is 5? 5 is range of E i equals the
eigenspace corresponding to the eigenvalue
lambda i.
So I will simply say eigenspace for lambda
i this is another consequence, condition 4
which was assumed yesterday is really a consequence
of 1, 2 and 3. Also what is more important
is that really this statement I want to prof
earlier than 4 and 5.
Lambda 1, etc lambda k are the eigenvalues
of T these are precisely the eigenvalues of
T that is what this means and T is diagonalizable
okay these are the 4 consequences. We will
first show that lambda 1, etc lambda k are
the eigenvalues and then show that E i square
is E i then show that T is diagonalizable
and finally show that range spaces are the
eigenspaces, okay.
So this is slight that is a slight change
from what I have given yesterday, okay first
thing is this that is easy I is E 1 plus E
2 etc plus E k so I multiply by E i E i is
E i into I that is E i summation j equals
1 to k E j i is fixed the subscript i is fixed,
the running index is j when j takes the value
I have condition 3 I have condition 3 that
the product is 0 when the subscripts are different.
So this goes inside and I have the product
summation j equal to 1 to k E i E j when j
equals i it is E i square all the other terms
are 0. So this is just E i square started
with E i I have E i square so condition 4
holds that is an immediate consequence of
1 and 3 sorry consequence of 2 and 3 that
is condition 4. I will show that range of
E i is eigenspace for lambda i not entirely
I will show that range of E i is contained
in the eigenspace corresponding to lambda
i later I will show that the eigenspace for
lambda i is contained in range of E i, okay
so this shows 4.
Next we show that okay let me write this we
show that 
range of E i is contained in the eigenspace
for lambda i which is null space of T minus
lambda i this is the eigenspace of the eigenvalue
lambda i, okay I will show that range of E
i is contained in this. Let x belongs to range
of E i then x equals E i x. Consider Tx minus
lambda i x I want to show that this is 0 I
want to show that this is 0, right. If I show
this is 0 then it follows that x belongs to
null space of T minus lambda i I started with
x in range of i so that will proof this, okay.
This can be written as T I will use representation
1 and then lambda i identity x I use representation
2. So I can write this as summation j equals
1 to k, for T I will have lambda j E j x for
T I have summation lambda j E j. For the second
term minus so I will use a bracket here, second
term goes with an i i is fixed i is fixed
so this goes with lambda i minus lambda i
into identity x identity is summation E j
identity is summation E j so this is E j x,
do you agree lambda j E j x lambda i E j x
lambda j E j minus lambda i so this is lambda
i. I can write this as okay okay this is 0
I want to show this is 0, okay.
Okay, let me write down this is equal to summation
j equals 1 to k lambda j minus lambda i E
j x is what I wrote x is E i x, okay. So when
again j is running index, i is fixed j is
running index, i is fixed whenever j is not
equal to i these terms are 0 whenever ya the
only term that is left is when j takes the
value i when j takes the value i this is 0
so this is 0, okay. So what we have shown
is that T minus lambda i x we have shown that
this is 0 yes and so this holds.
So range of E i is contained in null space
of T minus lambda i what also what this also
means is that see each of these operators
is non-zero each of these operators is non-zero
which means range must have at least one non-zero
element, one non-zero vector which means that
what is the consequence? Since E i is not
equal to 0 there exist x not equal to 0, x
element of range of E i that is from what
we have shown just now there exist x not equal
to 0 such that Tx equals lambda i x that is
these numbers are eigenvalues of this operator
T, right.
So lambda 1, lambda 2, etc lambda k these
are not the eigenvalues these are eigenvalues
presently these are eigenvalues of T the question
is does it exhaust all the eigenvalues is
it possible that we missed some eigenvalue
which is not counted because of the way we
have done the way we have got this inclusion.
Have we missed other eigenvalues, have we
missed other numbers which are possible eigenvalues
of T.
Then we will show that these are the only
eigenvalues so it will follow that these are
the eigenvalues is the problem clear, these
are eigenvalues alright are there other eigenvalues
which are not accounted here. Let me prove
that that will proof this statement then diagonalizability
will follow after that we will prove (condition
4) condition 5. So let me take a number which
is an eigenvalue and then show that this number
must be one of these one of these lambda i's
yes why am I using I suppose that is clear,
okay.
I want to show that if lambda is a number
if lambda is a number for which T minus lambda
x is 0 for some x not equal to 0 then lambda
is one of these okay. So let lambda be such
that Tx equals lambda x with x not equal to
0, I want to show that this lambda is one
of these lambda i's then it will follow that
this is the complete list of eigenvalues till
now we have not shown that this exhaust all
the eigenvalues of T, we have shown that these
numbers are eigenvalues we will show that
any other number that satisfy this equation
must be one of these, okay. So that this statement
would be proved, okay.
Lambda such that Tx equals lambda x, x not
equal to 0, so look at this equation this
means 0 equals T minus lambda I of x I will
now use the representation once again. Summation
j equals 1 to k lambda j E j minus lambda
E j of x lambda j E j is T, I is summation
lambda I is lambda summation E j this is summation
j equals 1 to k lambda j minus lambda into
E j x lambda j minus lambda E j x this is
0. I will use this part, okay now I will have
to make use of the fact that this is an equation
like u 1 plus u 2 plus etc u k equal to 0,
where u 1 is in range of E 1, u 2 is in range
of E 2 etc but we know that these are 
these are eigenspaces because range we have
shown range of E i is contained in null space
of T minus lambda I. So this is an equation
involving eigenvectors really is that clear?
What I have is 0 equals u 1 plus u 2 plus
etc u k where what is u i u j maybe u j is
the jth term that is lambda j minus lambda
E j x I am calling the first term u 1, second
term u 2 etc this is 0. Now this u j belongs
to range of E j and range of E j we have just
now shown any vector in range of E j if it
is not 0 then it is an eigenvector so these
spaces are independent range of E 1, range
of E 2, etc are independent because they corresponds
to distinct numbers lambda 1, lambda 2, etc
lambda k this is an equation involving a sum
where the each term comes from a subspace
the subspaces are independent.
So this means each must be 0 u j equal to
0 for all j that is lambda j minus lambda
E j x equals 0 for all j. Now is this clear
the fact that the fact that u j equal to 0
from this equation follows because I am now
really looking at eigenvectors corresponding
to distinct eigenvalues I am looking at subspaces
which are independent that is the reason why
each u j is 0. I have this equation for all
j suppose E j x is 0 for all j then what do
you know about x from this 
E j x equal to 0 for all j what do you know
about x? x must be 0 because this equation
tells me that x is equal to E 1 x plus E 2
x is the E k x so if each E j x is 0 then
x is 0.
But we started with x not equal to 0, we have
x not equal to 0. So if E j x equal to 0 then
x equal to 0 a contradiction and so E j x
cannot be the 0 vector for at least one j
this cannot be 0 for at least one j I go back
to this equation this number into this vector
is 0 for at least one j I am sorry this is
true for all j for at least one j this vector
cannot be 0 so this number must be 0 for that
j.
So lambda I want to really write lambda equals
lambda j for that j so what we have shown
is that these numbers exhaust all the eigenvalues
lambda 1, etc lambda k exhaust all the eigenvalues
of the operator T, is that clear? We started
with lambda being an eigenvalue so the equation
Tx equals lambda x with x not equals to 0
must be satisfied the rest is consistent with
throughout we have as assumptions, okay.
So what we have shown is that these are precisely
the eigenvalues now lambda 1, etc lambda k
are precisely the eigenvalues of T, why is
T diagonalizable that is almost a tautology
it does not need much explanation probably.
Any vector x in the space V can be written
as sum of vectors x 1, x 2, etc x k, x i coming
from E i but I know that range of E i okay
I know that anything in the range of E is
an eigenvector so long as it is non-zero.
All that I will do is take a basis for range
of E 1, take a basis for range of E 2, etc
combine them take the union that will be a
basis for V this basis for V has a property
that each vector is an eigenvector. So from
this equation and from the fact that anything
in range of E i belongs to the null space
of T minus lambda I it follows that T is diagonalizable,
is that clear? What do I want to show? I want
to show that I want to show T is diagonalizable
so I want to show that there exist a basis
for V each of whose vector is an equation,
okay all that I do is take a basis for range
of E 1, take a basis for range of E 2, etc
take a basis for range of E k the fact that
the sum is equal to identity means that I
have exhausted the the number of elements
in the basis for E 1 plus a number of elements
in a basis for E 2, etc the number of elements
in a basis for E k that number must be equal
to the dimension of the space because of this
condition so T is diagonalizable, okay. So
it is a immediate consequence of 2.
I will just write that by taking basis for
range of (E 1) range of E i okay let me just
give it for the sake of completeness namely
B 1, etc B k and by setting script B to be
the union of these basis it follows that the
matrix of T relative to B is diagonal that
is each element in this basis is an eigenvector
for T each element in the basis B is an eigenvector
for T because each element in the basis B
must belong to one of the B i’s anything
in B i is basically a subset of range of E
i anything in range of E i these are basis
vectors so they are non-zero vectors so each
of them is an eigenvector, is that clear so
T is diagonalizable that is the most important
part but it follows as an easy consequence
of what we have done earlier.
The final part is to show that this eigenspace
is contained in range of E i that is the last
part all other things have been proved, okay
we have shown that range of E i is contained
in the eigenspace we must show that the eigenspace
is contained in the range of E i, null space
of T minus lambda I will show that this is
contained in range of E i, okay let us take
an element I will call it u let u belongs
to null space of T minus lambda I then u is
T minus lambda I T minus lambda I u is 0 I
will again use that representation for T and
I. Summation j equals 1 to k this time I will
write down the simplified expression this
will be lambda j minus lambda i E j u 
lambda j E j u that is T u lambda i u is lambda
i E j u i is summation E j so I have this.
Now this is 0 again this is an equation involving
vectors that are in range of E i range of
E j so each term must be 0 because these are
independent subspaces. So I have lambda j
minus lambda i E j u this is 0 for all j 
not equal to i because when this is true for
all j is okay we will exploit this for all
j this is true 
all you have to do is ya conclude from this
that when j is not equal to i look at this
equation when j is not equal to i it will
tell you that E j u so just tell me if this
is clear so j not equal to i implies E j u
is 0 when j is not equal to i see I have this
product to be 0 about the ith equation I do
not know about the ith equation I cannot make
any conclusions but about all other equations
what I know is that this is not 0 so this
must be 0, okay I have equation from 1 to
k, i occur somewhere that ith equation is
lambda i minus lambda i into E i u equals
0 so there I cannot conclude E i u equal to
0.
While if I look at the other equations it
follows that E j u is 0 for all j not equal
to i, okay what is the meaning of this E j
u equals 0 for all j not equal to i it means
again any any vector can be written as any
vector x can be written as E 1 x plus E 2
x etc all terms are 0 except the ith term
so x belongs to range of E i ith term comes
from range of E i so sorry not x u u belongs
to range of E i.
So from this the last step is u belongs to
range of E i i is fixed. So I started with
u and null space of T minus lambda i I have
shown that u belongs to range of E i already
we have shown that range of E i is contained
in null space of T minus lambda i so these
subspaces are the same that is range of E
i is the eigenspace corresponding to the eigenvalue
lambda i, okay now that is the complete theorem
with its proof probably we look at an example
numerical example, okay I want to look at
a numerical example where we will calculate
all these an example where the matrix is diagonalizable
the operator is diagonalizable.
So let me quickly see if this this works by
the way before I work out this example there
is a fact that I will state without proving
we can discuss it if you want the proof later
outside the class. It can be shown that this
the operators, the projections, the E j’s
are in fact polynomials in T in particular
may be I will give this as an exercise. Let
me define p j of t to be the product of all
these polynomials p j of t so the product
is over i let us say i does not take the value
j look at t minus lambda i by lambda j minus
lambda i. I am defining a polynomial in this
manner this polynomial for one thing has the
property that p j of t i is delta i j this
polynomial has a property that p j of t i
is delta i j, p j of lambda i is delta i j.
I am defining for the numbers lambda 1, etc
lambda k that are distinct.
For every fixed j here the product the index
is i for the product index is i these are
these are really lagrange interpolating polynomials
we have encountered this before these are
lagrange interpolating polynomials what follows
what can be shown is that E j is p j of T,
what is the meaning of this? It says there
is a the formula for E j’s can be obtained
from the formula for the polynomials here,
okay why this is true is an exercise for your
real I will freely use this in the example,
okay. What is that example?
Let me take the matrix A as there are relationships
between these are called Newton’s formulas
there are relationships between the entries
of the matrix and the eigenvalues. For example
the sum of the eigenvalues is the trace of
the matrix the sum of the product of 2 eigenvalues
taken at a time is the sum of the 2 by 2 principle
minors the product of the eigenvalues is the
determinant of the matrix, okay I will use
these three properties I will not prove any
of these again these are exercises.
I want to write down the characteristic polynomial
for this matrix I use t right for that so
p of t I claim is 2 plus 3 5 plus 2 7 lambda
cube minus 7 lambda square I want to look
at the 2 by 2 determinants 6 minus 2 is 4
6 minus 2 is 4 what I want is really 4 minus
1 is 3 that is 11 it goes with a minus this
goes with a plus the final term is a determinant
what is the determinant? Determinant is 5
plus 7 plus 11 lambda minus 5 this is the
characteristic polynomial, okay this can be
shown.
What are the factors? 12 minus 12 1 is a factor
verify that this is lambda minus 1 the whole
square into lambda minus 5 5 is also a factor
p of lambda, is this diagonalizable, how?
Student is answering: lambda equal to first
get a eigenvector.
Can you repeat for this.
Student is answering: For lambda equals to
1 than null space of t minus i is 2.
Did you see this for lambda equals 1 the substitute
there is only one equation 1 2 1, 1 2 1, 1
2 1 only one row the rank of that is 1 so
nullity is 2 so this matrix is diagonalizable,
A is diagonalizable A is diagonalizable. I
want to compute see all that I want to do
is to illustrate that this previous theorem
by means of this example. So I will compute
E j’s and then verify sum of E j equals
identity and then t can the product is 0,
t can be written as lambda 1 E 1 plus lambda
2 E 2 that is enough really, okay.
E 1 is p 1 of T p 1 of T there is only one
factor left out p 1 corresponds to the first
eigenvalue T minus 5 identity by what is fixed
is 1 is fixed so 1 minus 5 minus 4 this is
E 1, 5 I minus T by 4 let me write this maybe
1 by 4 to be taken outside 5 I minus T 5 minus
2 3 5 I minus T minus 2 minus 1 minus 1 minus
1 minus 1 minus 2 5 minus 3 is 2 5 minus 2
is 3 there are many minuses let us take minus
1 by 4 outside minus 3 2 1 1 minus 2 1 1 2
minus 3 please check that my calculations
are correct.
What is E 2? E 2 will be T minus I by 4 I
am sorry I minus T minus I is correct yes.
So this is 1 by 4 T minus I 1 2 1 1 2 1 1
2 1 is that okay yes when you add do you get
what we want okay this one goes with a plus
sign 3 plus 1 4 4 by 4 1 ya this goes with
a minus sign the previous one would have been
better I will do it with the previous one
0 this is 0 this is 0, this is 4 by 4 is 1,
this sum is 0 this is 0, this is 0, this is
3 plus 1 by 4 so sum is identity.
E 1 plus E 2 equals identity. See I got a
once I got E 1 I could have got E 2 by using
the fact that E 2 is I minus E 1 but I wanted
to really verify that so I calculated E 2
independently by this formula, okay finally
you check that T that is A in this case lambda
1 E 1 plus lambda 2 E 2, okay that is 5 by
4 plus 3 by 4 8 by 4 2 that is the first entry
please check this is satisfied A is E 1 plus
5 E 2, please also write down the range range
spaces of these two operators they will be
precisely the eigenspaces.
For example it is immediately clear that the
range of E 2 range of E 2 is 1 1 1 (())(36:07)
by 1 1 1 range of E 2 (())(36:10) by 1 1 1
and E 2 range of E 2 is one dimensional, range
of E 1 two dimensional because lambda for
lambda equals to 1 the null space has two
independent vectors, okay so this is just
to illustrate. As I told you there are certain
properties I have just stated I have not used
the relationship between the roots of the
equation p of lambda equal to 0 that is eigenvalues
and the entries and why is this formula correct?
You please check this there is also another
result which at least I will state may be.
Let me state this result and then prove it
in the next lecture. Let us it is kind of
a summary of what we have done is the following.
We discussed the notion of diagonalizability,
we characterized diagonalizability in terms
of the minimal polynomial, okay an operator
is diagonalizable if and only if the minimal
polynomial is a product of distinct linear
factors. There is another characterisation
if the dimensions of the eigenspaces sum upto
the dimension of the space then the operator
is diagonalizable.
There is really another characterisation that
we have got just now this involves invariant
direct sums invariant direct sum decompositions.
If T is diagonalizable and if I can find out
certain certain operators then I know that
operators will satisfy these conditions and
conversely if there are certain operators
and certain numbers related in a specific
manner then T is diagonalizable.
If you look at the general problem then in
all these cases what is important is to realize
that the minimal polynomial can be written
as 
okay where if T is diagonalizable then r 1,
r 2, etc r k they are all 1 if T is not diagonalizable
then sum of these will be at least 2 at least
one of them will be 2 for instance but if
T is not diagonalizable we know that it is
always t triangulable if the minimal polynomial
is of this form any operator is triangulable
provided all its eigenvalues lie in the underlying
field which is a same as saying that the minimal
polynomial is of this form, okay.
So this kind of encompasses all these both
the case of diagonalizability and triangulability,
what happens if in in the case of r for instance
the irreducible polynomial the degree the
degree of an irreducible polynomial can be
2 in the case of r, in the case of c the degree
of an irreducible polynomial is precisely
1 for instance t square plus 1 is an irreducible
polynomial in the field in the space of polynomials
over r in the case of c these are linear polynomials,
okay.
What happens to a direct sum decomposition
in this case that is what we will refer to
as a primary decomposition theorem. In this
general case that is in the case when the
eigenvalues lie in the underlying field what
is the representation? What is the direct
sum decomposition? Okay I will write down
the statement and prove it tomorrow this is
called the primary decomposition theorem 
in other words this is kind of a most general
theorem that one has all the other results
are particular cases all the other results
that we have proofed can be shown to be particular
cases of this.
V is finite dimensional. Suppose that the
minimal polynomial is written as p 1 to the
r 1, p 2 to the r 2, etc p k to the r k I
am writing an expression which is more general
then what I wrote down here. Suppose if the
minimal polynomial can be written in this
manner where remember little m is minimal
polynomial for us that is a notation where
r i's are positive integers what are these
p i’s p i’s are monic irreducible polynomials
over the underlying field F.
Now this F is general r c rationals it could
be a finite field, could be any other field
then we have the following. Let me write it
like this let me use this notation this is
what I want to say V is the direct sum of
the null space of p i T power r i V is the
direct sum of the null space of p i T power
r i each of these subspaces is invariant under
T 
each of these subspaces I could have call
them W i each of these subspaces W i is invariant
under T T W i is contained in W i this is
called the primary decomposition theorem this
is the most general that one could do for
any operator.
Why is it call primary? Primary decomposition
is clear it is called primary decomposition
theorem because see it is monic polynomial
the degree of the coefficient of the highest
degree is 1 that is monic irreducible, what
is irreducibility? It is because of this irreducibility
that this is called the primary decomposition
theorem. If a polynomial is irreducible over
a field then it is called a prime polynomial
that is if a polynomial is irreducible when
do you say the polynomial is reducible if
it can be written as a product of two or more
polynomials each has degree at least 1 each
has degree at least 1, okay x square minus
1 is (x min) lambda square minus 1 is lambda
plus 1 into lambda minus 1 so lambda square
minus 1 is not irreducible over r lambda square
plus 1 cannot be written as lambda minus lambda
1 into lambda minus lambda 2 where lambda
1 and lambda 2 are real numbers.
So lambda square plus 1 is an irreducible
polynomial over r lambda square plus k where
k is a positive number is an irreducible polynomial
over r such polynomials are called prime polynomials.
If a field is such that the prime polynomials
are only linear such a field is called an
algebraically closed field this is the same
as saying that any polynomial equation has
all its zeros in the field any polynomial
equation has all its zeros in the field is
a same as saying that the polynomial can be
factored into a product of linear factors,
okay.
So in general if F is not r of c, an irreducible
polynomial is called a prime polynomial. So
these are prime polynomials they cannot be
factored into a product of polynomials of
lesser degree just like how you have composite
numbers and prime numbers. So this is called
primary decomposition theorem for any operator
this is the most general that one could have,
okay the decomposition corresponds to null
space of p i T r i, I will prove this theorem
in the next lecture let me stop here today.
