So let us now focus our attention on developing
some techniques to evaluate the eigenvalues
and the eigenvectors of a given n cross n
matrix. So let us begin by key proposition
in this direction. So let me write down the
proposition.
So a scalar lambda is an eigenvalue of an
n cross n matrix A, if and only if the determinant
of A minus lambda I n is zero. Let us give
a proof of this statement. So proof, so let
us assume that scalar lambda, let us call
it lambda here itself. Let us assume that
lambda is an eigenvalue of the matrix A. So
let lambda be an eigenvalue of A.
That means that there exists a non-zero vector
v, i.e. there exists a non-zero vector v such
that A v is equal to lambda v or given lambda.
We write it, this implies that A minus lambda
I n, where I n is the n cross n identity matrix
times v is equal to 0. And v is a non-zero
vector. This just implies that A minus lambda
n is not an injective linear transformation.
Hence A minus lambda I n is or the linear
transformation corresponding to this is not
injective. Because (that) this means that
v is in the null space of A minus lambda I
n. Because this means that v is in the null
space of A minus lambda I n.
But what can we say about linear transformation
from a finite dimensional space to itself,
if it is not injective? We can say that a
linear transformation from finite dimensional
space v to itself is injective if and only
if it is surjective and hence invertible.
So if it is not injective, A minus lambda
I n is not invertible, i.e. A minus lambda
I n is not invertible because if it were,
so the reason is that this is R n has dimension
n, finite dimensional and any invertible linear
transformation any injective linear transformation
is necessarily invertible. And any invertible
linear transformation is obviously necessarily
injective. So let me write dot dot dot for
you to complete down the reason. So A minus
lambda I n is not invertible.
But what had we proved in the last week? We
have proved that a vector a matrix is invertible
if and only if the determinant is non-zero.
Here we have that A minus lambda I n is not
invertible. And therefore, that forces by
a theorem from the last week, from the previous
week, we have determinant of A minus lambda
I n is equal to 0. If it were not 0, then
A minus lambda I n would have been invertible.
And that is precisely what we have set out
to prove. If you look at the preposition,
it is said that determinant of A minus lambda
I n is equal to 0.
Now let us try to prove the converse. So suppose,
it is just the backtracking of the argument.
Suppose A minus lambda I n has determinant
0, that implies that A minus lambda I n is
not invertible. And if it is not invertible,
it is not injective. Because here invertibility
and injectivity are the same. It is finite
dimensional. This implies that, there exists
v not equal to 0 in V such that A minus lambda
I n v is equal to 0 v in the null space of
A minus lambda I n, which implies that A v
is equal to lambda I n v which is equal to
lambda v.
Therefore, lambda v is an eigenvector and
lambda is an eigenvalue corresponding to v
of the matrix A. So we have effectively found
out a (character) characterization of when
we can say that scalar is a eigenvalue. So
this prompts us to give a definition for the
number or sorry, the expression determinant
of A minus lambda I n. Notice that this is
going to be a polynomial expression in lambda,
so let me give a definition here.
So we call the expression, call the polynomial
in lambda or rather let me call it expression
as of now, determinant of A minus lambda I
n as the characteristic polynomial of lambda
of A. Characteristic polynomial of the matrix
A. 
And we will denote it by f of lambda. And
we shall denote in the rest of this lecture
this by this polynomial by f of lambda. So
what or the above preposition just said is
that, lambda is an eigenvalue if and only
if it is the root of f. So hence the eigenvalues
of A are precisely the roots of f of lambda.
So let us look at a few, let us look at an
example, wherein we will do the computing
of eigenvalues and eigenvectors and look at
some of the implications. So an example, so
let us consider the matrix 
A which is given by 1, 1, 0, 1. Or we denote
this 0, 1, so it will be 1, 1. This is it.
This is the right expression of example I
would like to consider. So let us try to compute
the eigenvalues of A. In order to do that,
let us compute the characteristic polynomial
of lambda. So characteristic polynomial, we
have just noticed that the characteristic
polynomial, the rules of the characteristic
polynomial are precisely all the eigenvalues.
Polynomial of A, this is given by the determinant
of A minus lambda times I, here in this case
I 2. And that will be the determinant of the
matrix minus lambda 1, 1, 1 minus lambda.
And this tells us this is lambda times 1 minus
lambda minus 1, which is equal to lambda square
minus lambda minus 1.
So what are the eigenvalues? Eigenvalues are
all the roots of this particular characteristic
polynomial. Hence, eigenvalues are solutions
of lambda square minus lambda minus 1 is equal
to 0. This implies that lambda 1 is equal
to 1 plus square root of 5 by 2 and lambda
2 is equal to 1 minus square root of 5 by
2 are the eigenvalues of A. Let us now evaluate
the eigenvectors corresponding to lambda 1
and lambda 2.
So to compute the eigenvectors corresponding
to each, let us look at the null space corresponding
to A minus lambda 1 times identity and A minus
lambda 2 times identity. So let us now consider
A minus lambda 1 times identity. This is just
matrix is 0, 1, 1, 1 minus 1 minus square
root of 5 by 2 times 1, 0, 0, 1. And this
is 1 (plus) or 1 minus square root of 5 by
2, minus of this, 1 then there is a 1 and
there is 1 and there is 1 minus 1 minus square
root of 5 by 2, which is 1 plus square root
of 5 by 2.
So A minus lambda 1 I something in the null
space will be x, y equals to 0, implies minus
of 1 minus square root of 5 by 2 times x plus
y is equal to 0, x plus 1 plus square root
of 5 by 2 times y is equal to 0. So if you
observe carefully, if you multiply minus of
1 minus square root of 5 by 2 to the second
equation, we get back the first equation.
That is not surprising because determinant
of a minus lambda 1 I was not non-zero, it
was 0. Therefore, it does not have full rank.
So typical solution will be of this type.
Let us write it as, so v 1 equal to 1 and
or maybe, 1 plus minus of 1 plus square root
of 5 by 2 and 1 is an eigenvector of lambda
1. I leave it you to evaluate that v 2 equal
to minus of 1 minus square root of 5 by 2,
1 is an eigenvector. So lambda 1, let us see
what was lambda 1, so lambda 1 was 1 plus,
so this is let us put lambda 2 here. And this
is just going to be v 2 corresponding to lambda
2, and v 1 equal to this will be an eigenvector
of lambda 1. Lambda 1 if you recall is 1 plus
square root of 5 by 2.
So if you consider the basis, so consisting
of the vectors v1 and v2, then our matrix
for A will turn out to be a diagonal matrix.
That is something which we had proved in the
last video. So hence for the basis beta or
rather beta prime given by v 1, v 2 are the
matrix L A corresponding to beta prime will
just turn out to be equal to lambda 1, 0,
0, lambda 2. But if you recall this was nothing
but or we had A which is 0, 1, 1, 1 was Q,
let us call this D. D Q inverse where D was
D is as given above and Q is the matrix given
by v 1, v 2.
Let us apply all this to real life example.
So the example here deals with what is known
as Fibonacci's rabbit. This deals with the
growth of the population of rabbits. Let us
look at an example. So let there be x pairs
of juvenile rabbits and y pairs of adult rabbits.
Let there be x pairs of juvenile rabbits and
y pairs of adult rabbits, which we will capture
in the vector x, y. Captured in the vector
x, y. Then let us assume that every year the
population growth is given by left multiplication
by our matrix A.
Assume that the population growth is given
by y and x plus y. But what is this y and
x plus y? Is nothing but multiplication of
x, y by the matrix 0, 1, 1, 1 with this matrix
x, y. So suppose we started off with a sample
or suppose the world started off with one
pair of juvenile rabbits and no adults. So
v 0 v equal to 1, 0. So we would like to ask
the following question. After n years, after
n years what is the population of rabbits?
Juvenile or and adult rabbits.
And the answer is a straightforward answer.
It is A to the power n times v 0. We know
that A to the power n times v 0. The first
row captures the number of juveniles and the
second row captures the number of the adults
after n years as our answer. But then if you
start trying to calculate A to the power n,
in fact it is a tedious process just for this
2 cross 2 matrix to start computing A to the
power n, for n very very large.
What we will do is we will use the knowledge
that we have developed, we have noted that
you know that A is equal to QDQ inverse. So
if you recall we did mention that diagonalization
was a technique which was or it is a factorization
method which was beneficial for computing
the powers. So you notice A square is just
QDQ inverse QDQ inverse. Q inverse gives the
identity and this is assumed to be QD square
Q inverse.
Similarly, A cube will be just QD square Q
inverse QDQ inverse, which again by the same
argument gives the QD cube Q inverse. By induction
you may show that A to the power n is QD to
the power n Q inverse. But we know what D
to the power n is, D is a very special matrix.
D is a diagonal matrix consisting of the eigenvalues
in its diagonal. So we know but D is equal
to lambda 1. So let me not write the big expression.
Let me just write diag lambda 1, lambda 2.
If you recall lambda 1, lambda 2 were the
eigenvalues 1 plus square root of 5 by 2.
So let me write this, where lambda 1 was 1
plus square root of 5 by 2 and lambda 2 was
1 minus square root of 5 by 2. Where lambda
1 is equal to 1 plus square root of 5 by 2
and lambda 2 is equal to 1 minus square root
of 5 by 2. So what is D to the power of n?
That is just going to be diagonal of lambda
1 to the power n, lambda 2 to the power n.
We have also computed what our Q is. You recall
Q was just, Q is v 1, v 2. Where v 1 is minus
of 1 minus square root of 5 by 2, minus of
1 minus square root of 5 by 2 and this is
going to be minus of 1 plus square root of
5 by 2. And there was a 1 here. So this was
our Q. So by using elementary matrices and
reducing it to the epsilon form and parallely
doing the same operations to the identity,
one may check that I they will not do the
calculations. This is just going to be 1 by
root 5 times 1, minus of 1 minus square root
of 5 by 2, 1 plus square root of 5 by 2 and
1 and minus 1, this is Q inverse.
Now let us calculate, so hence A to the power
n v 0 is just QDQ inverse v 0. So in particular,
Q inverse v 0 that is just v 0, if you recall
is 1, 0. That is just going to be 1 by root
5 times. Maybe let me write it like this.
This is going to be 1 by root 5, minus of
1 by root 5. And if you multiply this to A
to the power n, this is going to be A to the
power n, I am sorry, D to the power n, Q inverse
v naught is just the lambda 1 to the power
n, 0, 0, lambda 2 to the power n times 1 by
root 5, and 1 minus of 1 by root 5. So let
me maintain the root 5 maybe now, root 5.
So now about Q, you notice, what was Q? Q
was given by this matrix. And this we will
write it down again, Q is minus of 1, Q is
minus of 1 minus root 5 by 2, 1, minus of
1 plus root 5 by 2 and 1. And if you notice
carefully, 1 plus root 5 by 2, 1 minus root
5 by 2 times 1 plus root 5 by 2 this is equal
to minus 1. This is going to be 1 minus 4
by 4 which is minus 1. 1 minus 5 by 4 which
is minus 1. And therefore, minus of 1 minus
5 by 2 is just inverse of 1 plus root 5 by
2. This is, remember this is lambda 1 and
this is lambda 2. This is just going to be
equal to lambda 2 inverse. So this is going
to be lambda 1 inverse, lambda 2 inverse 1,
1. This is lambda 2 and this is lambda 1.
Yes, if you notice carefully, that is what
our lambda 1 and lambda 2 were. Lambda 1 is
1 plus root 5 by 2 and lambda 2 is 1 minus
root 5 by 2. So let us now apply this to D
to the power n Q inverse. So this, and write
the explicitly, this is going to be lambda
1 to the power root 5, minus of lambda 2 to
the power root, lambda 1 to the power n by
root 5, lambda 2 to the power n by root 5.
So now what is QDQ inverse v naught? This
is just going to be equal to lambda 1 inverse
lambda 2 inverse 1, 1. And there is a lambda
1 to the power n by root 5 and lambda 2 to
the power n by root 5. This just turns out
to be equal to lambda 1 to the power n minus
1 plus lambda 2 to the power n minus 1 by
root 5 and lambda 1 by root 5 times lambda
1 to the power n plus lambda 2 to the power
n. There is something wrong, let me just check
something.
, there is a minus which I have very conveniently
forgotten. There will be minus here and this
is going to be minus. So yes, this is going
to be A to the power n applied to v naught.
So the population of juvenile rabbits after
n years is going to be, so the population
of pairs of (juvenile) juvenile rabbits after
n years, this will be equal to lambda 1 to
the power n minus 1 minus lambda 2 to the
power of n minus 1 by root 5. And of adult
rabbits this will be just lambda 1 to the
power n minus lambda 2 to the power n by root
5. If you notice, our A is matrix consisting
of 0’s and 1’s. v 0 is also a vector consisting
of 0’s and 1’s. So if you look at any
power that will also have integer entries.
And if you multiply it to v naught, that will
give you integers. So effectively this lambda
1 to the power n minus 1 minus lambda 2 to
the power n minus 1 by square root of 5 are
integers despite the fact that lambda 1 is
an irrational number which is 1 minus square
root of, 1 plus square root of 5 by 2. And
lambda 2 being another irrational number which
is 1 minus square root of 5 by 2. And this
also gives us that lambda 1 to the power n,
lambda 1 to the power n minus 1 and lambda
1 to the power n plus 1 all are related. So
this gives us a very clean way of calculating
all the populations after any n number of
years.
So the numbers that get popped up here are
something like say F0 equal to 0, F1 is equal
to 1, F2 is equal to 1, F3 is equal to 2,
F4 is equal to 3 and so on, are called Fibonacci
numbers. So notice that when I write F0, F1,
F2 and so on; F0, F1 was our v 0, v 1 is F1,
F2 and so on. This is going to be our, this
will capture the population of rabbits after
n years. And these numbers are called Fibonacci
numbers and they are very special and they
come up in many natural phenomenon, very natural
in fact.
So now with examples, let us discuss what
is the impact of studying vector spaces over
complex spaces. So I would like to point out
that working with vector spaces just over
the field of scalars being real numbers is
sometimes quite restrictive.
For example, let us consider this particular
case, consider the matrix, the linear transformation
corresponding to the matrix A which is given
by say 0, minus 1, 1, 0. This is in some sense
the rotation by 90 degree. So if you consider
this particular matrix, then you will notice
that the characteristic polynomial, this is
given by lambda square plus 1. And lambda
square plus 1 equal to 0 does not have any
roots in real numbers. And therefore, we do
not have eigenvalues over the field of scalars
when it is real numbers. However, if the theory
over to be developed over the field of scalars
being complex numbers. Then lambda square
plus 1 equal to 0 would have roots i and minus
i, the complex numbers i and minus i.
And we would have been able to compute the
corresponding eigenvectors in C2 instead of
R2. And we would have had a much richer theory.
So it is quite important to note at this juncture
that it is many times far more beneficial
to study vector spaces over more general fields,
more general fields of scalars. In particular,
it is far more powerful to study vector spaces
over field of complex numbers. Which however,
let us not deal with in this particular course,
which I will leave you for another course
in the future.
