Okay, see we are discussing the notion of
minimal polynomials. We had looked at three
numerical examples first so let me recollect
this first example was that of a linear transformation
which did not have eigenvalues, second example
the matrix the linear transformation has eigenvalues
but it does not have enough eigenvectors,
third example it has eigenvalues in the field
and it has enough eigenvectors that is it
is diagonalizable, okay.
We are trying to calculate the minimal polynomial
when we were trying to calculate the minimal
polynomial we observed that for the first
case the minimal polynomial coincided with
the characteristic polynomial, second case
again not diagonalizable the minimal polynomial
was not the characteristic polynomial, third
case I am sorry second example minimal polynomial
was equal to the characteristic polynomial,
third diagonalizable case minimal polynomial
is just a product of distinct linear factors,
okay and it appears as though the bug stops
with the characteristic polynomial, okay it
appears as though we could do with characteristic
polynomials probably, okay.
The answer can we do is yes it is always we
will prove that today this is called the Cayley-Hamilton
Theorem, okay that is the characteristic polynomial
of any linear operator over a finite dimensional
vector space is an annihilating polynomial
that is Cayley-Hamilton Theorem, okay once
you prove we will prove the matrix case not
the linear transformation case once you prove
this what follows is that the minimal polynomial
divides the characteristic polynomial follows
that the minimal polynomial divides the characteristic
polynomial.
Let us also remember this one important connection
between characteristic polynomials and minimal
polynomials the zeros of both these polynomials
are the same the zeros of these polynomials
are the same, okay okay. Before I prove this
Cayley-Hamilton Theorem I also want to give
a little implication which I do not mention
when I was doing examples this could have
been useful the implications are following.
Suppose T is diagonalizable 
and let me say has before that lambda 1, lambda
2, etc lambda k be the distinct eigenvalues
be the distinct eigenvalues of T. See remember
I am not saying all the eigenvalues are distinct
I am only saying that the distinct eigenvalues
have been collected and I am calling them
lambda 1, etc lambda k each one goes with
a certain multiplicity lambda 1 goes with
D 1, lambda 2 goes with D 2, etc lambda k
goes with D k where D 1 plus D 2 plus D 3
etc upto D n is equal to the dimension of
the space, okay okay.
If T is diagonalizable then what I want to
prove it is a very simple proof the claim
is that remember I am using small m for the
minimal polynomial the claim is that the minimal
polynomial is a product of distinct linear
factors that is it is t minus lambda 1 into
t minus lambda 2, etc t minus lambda k. If
you go back to example 3 I simply ask you
to verify that the minimal polynomial is a
product of distinct linear factors the reason
why I told you was this result in my mind.
So let us proof this first, the third example
was that of a diagonalizable operator, okay.
I want to show that okay for one thing the
for one thing you cannot have the minimal
polynomial of degree okay I want to prove
the minimal polynomial is this just look at
this polynomial the minimal polynomial cannot
be of degree less than the degree of the right
hand side polynomial do you agree with that?
That is because if it were less than the degree
of the right hand side polynomial that is
k then it would mean what is the meaning of
that it would mean one of these factors have
been missing, okay but if one of the factors
is missing then I am in a situation where
I have an eigenvalue it is not the zero of
the minimal polynomial. So to begin with this
is the possible candidate for the minimal
polynomial, it cannot be less than this, the
degree of the minimal polynomial cannot be
less than the degree of this right hand side
polynomial which is k then the only thing
that I need to show is that this is annihilating
polynomial I will show that this is annihilating
polynomial then it follows that this is the
minimal polynomial, this is of course monic
the power of t the coefficient of t to the
k is 1. So this is the monic polynomial I
will only show that this is annihilating polynomial,
okay.
What is if that we must show we must show
that m of T is a 0 operator annihilating polynomial
so m of T is a 0 operator, I want to show
that an operator is 0 I will show that the
action of this operator on a basis is 0 then
it follows that the operator is 0 agreed,
okay T is diagonalizable that means there
exist a basis B 
with the property that each of the basis vector
is an eigenvector of T, let B be a basis of
V such that each x element of B is an eigenvector
of T corresponding to some eigenvalue I am
not too worried about that corresponding to
some eigenvalue but I know that there is at
least one such basis because T is diagonalizable.
I will show that m T of x is 0 it follows
that m T must be 0, do you agree m T is a
linear operator m of capital T that is the
linear operator I want to show that that is
the 0 operator I will show that m of T on
the basis vectors if 0 so for every x in B
I will show m T x is 0.
I will start with an x in B and consider consider
m T x I must show that this is 0, okay but
what is this m of T this is the expression
it is T minus lambda 1 I T minus lambda 2
I etc T minus lambda k I operating on x. I
want to show that this is 0, this is 0 vector
but observe that the property that these factors
have is that any two of them commute any two
of them commute.
T minus lambda j I into T minus lambda s I
this is equal to T square minus lambda s T
minus lambda j T plus lambda j lambda s I
this is (T minus) T square minus lambda j
T minus lambda s T plus lambda s lambda j
I and this is nothing but T minus lambda s
I into T minus lambda j I. Remember that in
general if s and T are operators s T is not
equal to T s, okay that is matrix multiplication
is not commuted but this is about special
matrices special linear operators, operators
of the form T minus lambda I these commute,
what is the advantage?
You go back here this x is x belongs to B
and each vector in B is an eigenvector so
this x is an eigenvector corresponding to
some eigenvalue lambda I push that to the
end operate on it you get 0. So let me say
that I know that x is in B and each vector
is an eigenvector so there is Tx equal to
lambda i x for some for some eigenvalue lambda
i. So I now that this equation holds for x
that is T minus lambda I of x equals 0 now
I use commutativity of this factors.
To write m T of x as T minus lambda 1 I lambda
2 I etc the Ith factor I can take T minus
lambda i minus 1 T minus lambda i plus 1 T
minus lambda k is the last one but I can push
T minus lambda i to be the last one because
of commutativity. Now I see that T minus lambda
x is 0 make use of that the rest of the factors
are linear so again 0, okay this is really
easy. If T is diagonalizable then the minimal
polynomial is a product of distinct linear
factors at the end of next lecture we will
be able to show that the converse is also
true, if the minimal polynomial is a product
of distinct linear factors then T is diagonalizable,
okay which will then give you another characterization
of diagonalizability, okay.
So the converse will be proved a little later
we need the notion of invariant subspaces,
annihilating polynomials I will do that later
but this is one result which could be used
in those examples if you know that it is diagonalizable
then calculating minimal polynomial is easy
if you know the eigenvalues, okay. As I told
you we will discuss Cayley-Hamilton Theorem
and let us proof for matrices, okay.
I just need one little observation before
proving the Cayley-Hamilton Theorem 
if I have a matrix with polynomial entries
then the matrix can be written as a polynomial
whose coefficients are matrices. I will just
give you an example if I have matrix with
polynomial entries let us say 1 plus t square
(t) 2t minus t cube 1 and let us say 1 minus
t cube I have a matrix with polynomial entries
then I am claiming that this matrix can be
written as a polynomial whose coefficients
are matrices as a polynomial whose coefficients
are matrices, what is the meaning?
Simply collect the coefficients of like powers
look at the constant for example constant
is 1 here, 0, 1, 1. So I can first write it
as 1 0 1 1 times 1 plus corresponding to t
there is only one term 0 2 0 0 corresponding
to t square, okay let me write this into t
this is the constant term of the polynomial
for A the polynomial is in terms of 1 t t
square t cube the coefficients are matrices
this is one coefficient, this is second coefficient
into t plus for t square I have only one term
1 0 0 0 t square plus finally for t cube I
have 0 minus 1 0 minus 1 t cube is it okay
please calculate verify this right hand side
is correct I have 1 plus t square is the first
term, second is 2 minus t 2t minus t cube,
the third goes with 1 in this entry, the fourth
one is 1 minus t cube, okay. So I have written
a matrix whose entries are polynomials this
matrix has been written as a polynomial in
in t the same variable t but with matrix coefficients
let us remember this.
Now next to get Cayley-Hamilton Cayley-Hamilton
Theorem I will give the statement for an operator
and also write down the statement for the
matrix and prove the matrix case T is a linear
operator on a finite dimensional vector space
V. Let okay I am using word p let p of (T)
let p be the characteristic polynomial for
T. Then Cayley-Hamilton Theorem states that
p of T is capital T that is the 0 operator
that is okay what this says is that the characteristic
polynomial is an annihilating polynomial but
we know that among all the annihilating polynomials
the minimal polynomial is a one with the least
degree so the characteristic polynomial must
be a product of the minimal polynomial and
another polynomial, okay. In other words the
minimal polynomial divides the characteristic
polynomial.
This is an important information and we must
compare this with the result that we proved
right in the beginning when we looked at the
definition of the minimal polynomial. So I
remember having made the statement that to
define the minimal polynomial first of all
you must know that there is an annihilating
polynomial show that there is an annihilating
polynomial remember we took the operators
identity t, t square etc and went upto t to
the n square and show that this polynomial
must be in a and showed that these vectors
these operators are linear independent that
gives rise to annihilating polynomial, okay.
Now look at that estimate for that annihilating
polynomial it is polynomial of degree n square,
okay what this theorem says is that you could
do with much less you do not have to go beyond
n that the degree of the characteristic polynomial
is n this theorem says you do not have to
go beyond n at most the characteristic polynomial,
is that clear? Okay proof as I told you I
will prove the matrix case let me just write
down this statement for matrices.
Let A be the matrix of T relative to some
basis 
then we would like to show that p of A is
the 0 matrix this is what we want to show.
See for proving this I will make use of of
course this observation and also another non
trivial result on determinants which I will
not prove this non trivial result on determinants
for the proof we need the following this non
trivial result is what I will write down next
we use the following for any square matrix
B for any square matrix B if you look at the
matrix B into the adjoint of B for any square
matrix B the matrix B into the adjoint of
B, what is this? Yes that is what I want this
is determinant of B times I this is not A
inverse you remember that see the determinant
of B could be 0, okay that is quite possible.
So there is no inverse coming here if you
know the determinant of B is non-zero then
you can remember this is an equation involving
matrices first B is a matrix adjoint of B
is a matrix, on the right hand side determinant
B is a number times identity matrix. So this
is an equation involving matrices and then
if determinant of B is not 0 we could divide
by that number and then write this as 1 by
determinant of B into adjoint of B, call that
as some T, okay T is a linear transformation
let us call it C then BC equals I, you can
also show CB equals I. So it follows that
this is a inverse, okay. But this is more
general than this determinant B equal to 0
non-zero case.
So I will make use of this result, this is
again a non-trivial result using the property
of determinants this is not easy but it is
not difficult again. So I will make use of
this result and then proof Cayley-Hamilton
Theorem I will apply this identity for the
matrix A minus lambda I in place of B, okay.
Before that let me call let me call determinant
A minus lambda I, I know that this is a polynomial
of degree n.
So let me call this as alpha not lambda to
the n plus alpha 1 lambda to n minus 1, etc
alpha n minus 1 lambda plus alpha n this is
the polynomial of degree n I am writing it
like this. Now this alpha not could be 1 or
minus 1 depending on the order but let us
keep it as it is what is that we need to show
we need to show that p of A is equal to 0.
So we must show that when I replace lambda
by A when I replace lambda by A I get a polynomial
I A I must show that this polynomial is the
0 polynomial I am sorry I get a matrix I must
show that this matrix is a 0 matrix, okay.
So we must show that alpha not A to the n
plus alpha 1 A to the n minus 1 etc alpha
n minus 1 A plus alpha n identity we must
show that this is 0 matrix, okay okay.
Let us now make use of A minus lambda I in
place of B and then see what we get so I am
using this formula and applying A minus lambda
I for B, okay in this I am applying A minus
lambda I. So I have this A minus lambda I
into adjoint of A minus lambda I is equal
to determinant of A minus lambda I into identity
matrix I will substitute determinant of A
minus lambda I the expression that I have
written down just now here and keep this right
hand side and then look at what I have from
the left hand side.
So let me write the right hand side the right
hand side is alpha not lambda n I is that
correct ya times identity plus alpha 1 lambda
to the n minus 1 identity etc alpha n minus
1 lambda identity plus alpha n identity, okay
just to remind myself the right hand side
is a matrix determinant of A minus lambda
I into I that is a matrix, okay. Now look
at the left hand side look at the second factor
adjoint of A minus lambda I how do you compute
the adjoint of a matrix?
The adjoint of a matrix B let us go back to
this is computed by using minors, cofactors
taking transpose first compute the minors,
what is the minor of an entry? The minor of
an entry is the determinant of okay if you
have an n cross n matrix the minor of an entry
Aij is a determinant of the n minus 1 cross
n minus 1 sub matrix obtained by deleting
the ith row jth column, okay okay.
We are calculating minor of A minus lambda
I, okay A minus lambda I along the diagonal
I have A 11 minus lambda, A 22 minus lambda,
etc. Let us take any arbitrary element calculate
its minor I will not calculate its minor I
will only say what this minor the form must
be. Can you see that okay I am deleting I
am deleting a row a row and a column and then
I am looking at an n minus 1 cross n minus
1 sub matrix I calculate the determinant of
that.
Now that is the determinant will be a polynomial
of degree n minus 1 that is what I want, that
is the polynomial of degree n minus 1 also
it is a determinant of a matrix whose entries
are polynomials so I can write this polynomial
like I will make use of this observation I
can write this polynomial in terms of in the
variable lambda but the coefficients are matrices.
So this is the point that is probably the
most crucial in this apart from this formula.
Adjoint of A minus lambda I, I can write this
as it is a polynomial of degree n minus 1
but the entries are the coefficients are matrices
this is the point. So I will use a similar
notation let us say B 1 lambda to the n minus
1 plus B 2 lambda to the n minus 2 plus etc
plus B n minus 1 lambda plus B n but I must
observe that where B 1, B 2, etc B n are matrix
coefficients. So they are matrices adjoint
of A minus lambda I adjoint of A minus lambda
I is a matrix whose entries are polynomials
in lambda adjoint of A minus lambda I is a
matrix whose entries are polynomials in lambda
but the maximum degree of each polynomial
in the adjoint of A minus lambda I is lambda
to the n minus 1 because you are calculating
the determinant of a n minus 1 cross n minus
1 sub matrix.
So these are matrices whose entries of course
depend of A but that is not important it is
mentioned only for completeness, these the
entries of these matrices depend on A. Now
substitute this into the left hand side equate
coefficients of like terms you get term you
do a one more operation and then you get p
of A equals 0, okay.
Let us substitute to calculate the left hand
side left hand side is A minus lambda I into
adjoint of A minus lambda I the expression
is available here, I have B 1 lambda to the
n minus 1 plus B 2 lambda to the n minus 2
etc B n minus 1 lambda plus B n this is the
left hand side. Let me collect the coefficients
and write down the terms accordingly there
is only one term corresponding to lambda to
the n that term is minus lambda B 1 sorry
minus B 1 there is only one term corresponding
to lambda to the n that is this.
There are two terms corresponding to lambda
to the n minus 1 one coming from this minus
B 1 the other one I am sorry one coming from
this and the other coming from this AB 1 minus
B 2 lambda to the n minus 1. Let me may be
write down one more term for lambda to the
n minus 2 you can verify AB 2 minus B 3, AB
2 there is a B 3 to the lambda n minus 3 into
that will give lambda n minus 2 so minus B
3. So it follows this order plus coefficient
of lambda coefficient of lambda will come
from contributions from second term in the
last term, last but one and the first term
AB n minus 1 minus B n lambda and constant
term is just AB n, okay these are the terms
corresponding to the degrees lambda to the
n lambda to the n minus 1 etc. So I must compare
these two.
So maybe I will call this equation 1 right
hand side formula 1, left hand side formula
2, okay look at right hand side I have formula
1, left hand side I have this expression these
two must coincide.
So I get the following equation coefficient
of lambda to the n on the right hand side
is alpha not I minus B 1 is alpha not I, AB
1 minus B 2 is alpha 1 I, AB 2 minus B 3 is
alpha 2 I, etc AB n minus 1 minus B n that
corresponds to what lambda that is alpha n
minus 1 times identity, the last equation
is AB n equals alpha not sorry alpha n identity.
So I have these equations n plus 1 equations
polynomial of degree n so there are n plus
1 terms these are the n plus 1 equations,
multiply the first one by A to the n minus
1, this by the n minus 2, this by this by
A, this by A square and then add you get term,
okay.
So let us say I multiply this by A to the
n, this by A to the n minus 1, n minus 2,
etc this will be multiplied by A square and
I will just leave it right so no this will
be multiplied by A, this is multiplied by
identity so that term ya I must go from A
to the 0 to A to the n multiply and add A
to the n maybe I will just do that quickly
minus A to the n B 1 pre multiply, matrix
multiplication is not committed pre multiply
minus A to the n B 1 plus A to the n minus
1 B 1 minus there is already n here minus
A to the n minus 1 B 2 plus A to the n minus
1 B 2 minus A to the n minus 2 B 3 plus etc
plus this one is multiplied by A A square
B n minus 1 minus AB n plus AB n on the one
side, on the other side alpha not A to the
n plus alpha 1 A to the n minus 1 etc plus
alpha n minus 1 A plus alpha n identity.
You can see that the left hand side is 0 these
two get cancel, these two get cancel etc.
You take any term the first term each term
has two terms you take any term the first
term gets cancel with the second term or the
previous term etc. So the left hand side is
0, the right hand side is what you have that
is Cayley-Hamilton Theorem this left hand
side is 0 matrix, right hand side is p of
A, okay.
Let us do one final problem in this section
which is purely for fun may be I want to look
at this 4 by 4 matrix. See remember calculating
the characteristic polynomial is not difficult
alright but calculating the eigenvalues is
difficult but in certain problems it will
be very clear. So I want you to look at this
problem let us look at the linear operator
T on R 4 whose matrix with respect to the
standard basis is this 0 1 0 1 1 0 1 0 0 1
0 1 1 0 1 0, I would like to calculate the
characteristic polynomial of this as well
as the minimal polynomial this matrix has
some nice properties which we will make use
of , okay.
Look at A square see I am not trying to advocate
this procedure for all matrices, for this
matrix it is nice to see how the minimal polynomial,
characteristic polynomial can be calculated
very quickly, okay. What is A square? A square
is 0 1 0 1 into 0 1 0 1 it will be 2 0 2 0
0 2 0 2 2 0 2 0 0 2 0 2 this is A square,
okay still there is no pattern. Look at A
cube A cube can be shown to be see it is a
product of this into this that is 4, so you
will have 0 4 0 4 4 0 4 0, mistake 
no problem this is correct, what is your question?
I am calculating A square here is that correct,
is something wrong here? 2 0 2 0 0 2 0 2,
okay this is fine this into this, okay and
A cube instead of this I will go back to this
pattern.
See before coming to the class I work it out
and come that is call preparation so this
is correct please check this this is equal
to 4 times A that is what happens in this
example, okay this is 4 times A which means
I have caught hold of a annihilating polynomial,
right A cube minus 4A is 0. So I will look
at I look at the polynomial I call it f this
time f of t is t cube minus t cube minus t
that is t into t square minus 1 sorry t cube
minus 4t that is the t square minus 4 that
is t into t plus 2 t minus 2 whichever be
if I call f as this polynomial then f of A
is 0 that much I know 
I know that f of A is a 0 matrix, okay.
I know that minimal polynomial is a polynomial
of least degree among the annihilating polynomials,
okay. So the minimal polynomial is see the
degree of this is 3, the degree of this is
3, can the minimal polynomial be a linear
polynomial? Can the minimal polynomial be
a linear polynomial? Linear polynomial is
a polynomial of the form t minus lambda linear
polynomial is t minus lambda, okay.
So can this be the minimal polynomial for
this matrix? Can it be linear? If the minimal
polynomial for this matrix is linear then
I must have something like A equals lambda
times I, A equal to alpha times I for some
scalar alpha. Obviously not so A is not alpha
times identity A is not which means can you
see that all these three individual factors
do not constitute minimal polynomial that
is minimal polynomial is not a linear factor
do you first agree. It is not linear it is
all that I want to say is that this is the
minimal polynomial, okay. I want to roll out
the possibility that it is not quadratic nor
is it linear, linearity is immediate.
The minimal polynomial is not linear what
is the reason? If it were linear then A would
be a multiple of identity and so that is rolled
out. If the minimal polynomial is linear then
the minimal polynomial is of the form t minus
alpha for some alpha but if that were the
case a minimal polynomial must be annihilating
polynomial. So I replace t by A I must get
A minus alpha I is 0 that is not the case.
So it is not linear, can it be quadratic?
If it is quadratic what are the choices? If
it is quadratic remember minimal polynomial
must divide this polynomial so this is like
a bench mark.
If it is a quadratic then it is either of
the form t into t (square) t minus 2 or t
into t plus 2 or t minus 2 into t plus 2,
okay. If it is of the form t into t minus
2 then t square minus 2t must be an annihilating
polynomial that is A square equals 2A that
is not the case. If it is of the form t into
t plus 2 then I must have A square plus 2A
equals 0 that is not of the form, the last
possibility is t square minus 4 t square minus
4 is a possibility for the minimal polynomial
in which case A square equal to 4I that is
again not the case t square minus 4 is the
third quadratic polynomial that is not a minimal
polynomial because that is not the minimal
polynomial because A square minus 4I is not
0, A square is not equal to 4I and so we have
already seen that this is a annihilating polynomial
so this must be the minimal polynomial, okay.
So please give the reasons for the quadratic
case.
It is not quadratic also and so the minimal
polynomial for this matrix is t into t plus
2 into t minus 2. Now the minimal polynomial
has the property that the zeros of the minimal
polynomial are the characteristic values.
So the characteristic values of the matrix
that we started with are 0 plus 2 minus 2.
Also 0, 2, minus 2 are the eigenvalues of
the matrix A there is a relationship between
the characteristic polynomial and the minimal
polynomial they have the same set of zeros.
The dimension is 4 I have three eigenvalues
so one of them comes twice which one? One
of the eigenvalues must come once more the
algebraic multiplicity the algebraic multiplicity
of one of this eigenvalues must be 2 which
one is that? If you give the answer you must
also give the reason, okay. Indians invented
0 so let us take 0 first I want to know the
dimension of the set of solutions of Ax equals
0x that is Ax equals to 0, okay.
What is the dimension of the set of solutions
of Ax equal to 0, A is given to you 
it should be 2 is okay but I want you to give
a reason, why is it 2? Wait wait wait rank
of A is 2, this row is same as this row, this
row is same as this row these two are obviously
independent rank of A is 2, nullity of A must
be 2 so dimension of set of all solutions
of Ax equal to 0 is 2, so 0 comes twice. Rank
of A is 2 and so the eigenvalue 0 appears
twice as an eigenvalue of this matrix and
so what is the characteristic polynomial?
So I will simply say appears twice so p is
the notation we have for the characteristic
polynomial. So p of t is t square into t minus
2 into t plus 2 the characteristic polynomial
should be of this form t square into t plus
2 into t minus 2, is the matrix diagonalizable?
See the answer is yes there are atleast two
reasons the answer is yes the matrix is diagonalizable
there are atleast two reasons one reason if
you assume my result that a matrix is diagonalizable
if and only if the minimal polynomial is a
product of distinct linear factors that has
not been proved but you could have taken the
statement.
The other thing is distinct eigenvalues have
linear independent eigenvectors, I am assured
of three independent eigenvectors 0, minus
2, plus 2 three independent eigenvectors come
alright, do I have one more eigenvector which
is independent with these three coming from
Ax equal to 0, okay the problem lies the question
is is 0 deficient that is do I have two independent
eigenvectors for the eigenvalue 0 that is
easy to see in this problem.
Ax equal to 0 there are two independent solutions,
rank is 2, nullity is 2 that is what we determine
just now the matrix is diagonalizable, okay
the matrix is diagonalizable and it is similar
even if you view this as a matrix over rational
field the field of rational numbers the matrix
A is similar I will use this the matrix A
is similar to this 4 by 4 matrix let say I
follow this order then it is this matrix,
okay this matrix I will come back this matrix
will that was here doubt I think 0 0 2 0 I
am taking it in this order 0 comes twice 2
and then minus 2.
So A is similar to this that means p inverse
A p equals this matrix, how do you get p,
p is a matrix whose first two columns are
the independent solutions of Ax equal to 0,
third column is the only solution of Ax equals
2x, fourth column is the only solution of
Ax equals minus 2x only non-zero solution
that is a matrix p that is invertible and
p inverse A p equals a diagonal matrix, okay
okay I think I will stop here.
I want to discuss the notion of invariant
subspaces, annihilating polynomials and how
it helps these two notions help in proving
the converse statement that if the minimal
polynomial is a product of distinct linear
factors then the matrix is then the operator
is diagonalizable. I will proof this in the
next class, okay let me stop here which argument
see you want to show that the matrix is diagonalizable
then you must show that it has 4 independent
eigenvectors, 3 independent eigenvectors are
already there because of 0 minus 2 plus 2
they are distinct so I look at a non-zero
solution of those eigenvalue equations, I
want one more that one should one more should
correspond to the eigenvalue 0 because the
characteristic polynomial is this that should
correspond to 0.
So do I have two independent solutions of
Ax equal to 0 is what the question is but
the rank of A is 2 we observed just now the
rank of A is 2 so nullity is 2, so there are
two independent solutions take those two independent
solutions put them in the matrix p you get
p inverse A p as this diagonal matrix, okay
so this is see this is just to illustrate
how certain problems can be do not apply this
for example in the exam you will waste time,
okay we will meet on Monday.
