.)
So, welcome back and this is lecture number
48 and today we will talk about the properties
of Eigenvalues and Eigenvectors.
So, here the first property so if we have
a lambda as an eigenvalue of A and then x
be its corresponding eigenvector; then, this
alpha A has eigenvalue lambda A and the corresponding
eigenvector is x. So, what is this property
that if lambda is the eigenvalue of A and
x is its corresponding eigenvector in that
case the alpha times a alpha is some constant
some scalar quantity from the set of real
numbers.
For instance so here the alpha A will have
the eigenvalue alpha lambda. So, this alpha
will be multiplied simply to lambda and the
corresponding eigenvector will remain the
same that is x this property we can easily
verify. So, here Ax is equal to lambda x that
is the relation we have for eigenvalues and
eigenvectors.
So, if you multiply by alpha here both the
sides so alpha A and this will be alpha and
then this alpha a we are treating as matrix.
So, alpha will be multiplied to each of the
entries of A and here alpha times lambda.
So, with this relation what do we see that
this alpha A has the eigenvalue lambda alpha
lambda and x as the eigenvector from this
relation the matrix times x should be some
scalar into x.
So, that is from here we conclude that this
alpha lambda is the eigenvalue of this matrix
alpha A with the eigenvector same as before
that is x. So, another property of the eigenvalues
eigenvectors we have if A power m, so we are
multiplying this m times so A power m has
eigenvalues lambda power m and the corresponding
eigenvector is x again for any positive integer
m. So, if we have for instance A square so
its eigenvalue will be just the lambda square
with the same eigenvector as the A has.
So, this also we can easily see because we
have this Ax is equal to lambda x and now
we can multiply for example, just to see this
result for A square. So, we multiply here
A both the sides and the right hand side here
this lambda it is a constant here we can bring
to the outside. And then we have this Ax there
and the Ax we can replace again by lambda
x by this relation Ax is equal to lambda x.
So, if you replace your Ax by lambda x then
what will happen we have this lambda into
lambda x that is lambda square x.
So, what relation we have now here A square
x that was the multiplication of this A with
A. So, A square x is equal to lambda square
x which tells that this square the matrix
a square has the eigenvalue here lambda square
and lambda was the eigenvalue of A. So, with
this relation we can easily get the eigenvalues
of the A square or A cube or A power any integer
m. Because it will just the power will go
to the eigenvalue and the remaining the eigenvector
will be also the same which was the eigenvector
of this lambda for matrix A.
Now the next property so here we have seen
this lambda square is the eigenvalue of A.
Now, the two eigenvectors of a square matrix
A corresponding to two distinct eigenvalues
of A are linearly independent. So, what we
will prove here that two eigenvectors corresponding
to two distinct eigenvalues are always linearly
independent. This observation we already have
seen for numerical examples, but we can prove
here for more general matrix for any matrix
we can prove this result theoretically.
So, what do we consider now so let us say
x 1 and x 2 be two eigenvectors of A corresponding
to two distinct eigenvalues lambda 1 and lambda
2. So, this lambda 1 and lambda 2 these are
two distincts eigenvalues we have assumed
and their corresponding eigenvectors are denoted
by the x 1 is corresponding to lambda 1 and
x 2 is the corresponding to this eigenvalue
lambda 2.
So, with then meaning is that we have this
relation they satisfy this relation that Ax
1 is equal to lambda 1 times x 1 and Ax 2
is equal to lambda 2 x 2 because they are
the pair of this eigenvalues eigenvectors.
And now what actually we want to show that
this x 1 and x 2 are linearly independent.
So, for that you will consider this linear
combination c 1, x 1 plus c 2 x 2 is equal
to 0 and. So, this is the 0 vector here the
right hand side and then we will show that
this will imply that this is true this linear
combination is 0 this is true only when x
c 1 is 0 and c 2 is 0, that shows that this
x 1 and x 2 are linearly independent.
So, to do so we will consider here the c 1
and this x 1. So, we have multiplied basically
by the matrix this A by the given matrix A.
So, we have c 1, Ax 1 because c 1 was a constant,
so we have taken out here. So, c 1 times Ax
1 and then c 2 times Ax 2 that is so c 1 times
Ax 1 and the c 2 times Ax 2 this is the relation
we got from this equation by just multiplying
this to A.
Well, so the next having so here now we have
Ax 1 and we have Ax 2 right there which we
can replace by lambda 1 x 1 and the lambda
2 x 2. So, we have basically these two equations
now, one is the c 1 x 1 plus c 2 x 2 is equal
to 0 and the another equation we have here
c 1 lambda 1 x 1 plus c 2 lambda 2 x 2. Indeed
this unknown we can consider as c 1, x 1 and
the another one c 2 x 2.
So, here also we have c 1 x 1 and here also
we have c 2 x 2. So, these two forms a system
of a linear equation; system of this linear
equation and with unknown here the unknowns
are the c 1 x 1 and c 2 x 2. So, these are
the two linear equations or they form the
system of linear equation with unknowns so
c 1 x 1 and c 2 x 2.
With this so we have two equations now this
c 1 x 1 plus c 2 x 2 is equal to 0 and with
this lambda 1 c 1 x 1 plus lambda 2 c 2 x
2 is equal to 0. So, we want to solve now
for this c 1 x 1 and c 2 x 2. So, what we
do we multiply this equation number 1 here.
So, we multiply this equation by lambda 1.
So, if you multiply here then we have lambda
1 c 1 x 1 and lambda 1 c 2 x 2 and now with
these two equations we can subtract this equation
number 2 from this equation 1.
And then what we will get this lambda 2 minus
this lambda 1 because this term will cancel
out. So, we will get lambda 2 minus lambda
1 with this c 2 x 2 is equal to 0. And now
what we see that this equation will imply
simply that c 2 equal to is equal to 0 because
this lambda 2 minus lambda 1 cannot be 0.
Because we have two distinct eigenvalues and
this x 2 is the eigenvector which again is
a non-zero vector. So, this equation implies
that c 2 must be 0 as this lambda 1 minus
lambda 2, or lambda 2 minus lambda 1 is not
0 and x 2 is also not 0. And then from this
equation number 1 again here c 1 x 1 plus
c 2 x 2 if we substitute this c 2 equal to
0.
So, this term will be will be 0 and then we
have this relation that c 1 x 1 is equal to
0 and again with the same argument because
this x 1 cannot be 0. So, here again this
implies that c one is equal to 0, since this
x 1 is not equal to 0. So, with these we have
our now the c 1 0 and c 2 0 and that was the
aim to show that in this linear combination
c 1 x 1 plus c 2 x 2 equal to 0 is possible
when this c 1 is 0 and c 2 is 0, meaning that
these eigenvectors are linearly independent.
So, the eigenvectors here x 1 and x 2 both
are linearly independent.
So, this was the case when we have considered
two distinct eigenvalues, but we can also
generalize this case for more eigenvalues
for instance here, we have eigenvalues x 1,
x 2, x 3, x r are corresponding to our distinct
eigenvalues here. So, these are the eigenvectors
corresponding to these eigenvalues lambda
1 lambda, 2 lambda respectively and in that
case also we can use the similar trick to
prove that these eigenvectors are linearly
independent. So, we have this very nice result
that corresponding to distinct eigenvalue
the eigenvectors are linearly independent.
So, another result we have they said if x
is an eigenvector of A corresponding to the
eigenvalue lambda, then this kx is also the
eigenvector corresponding to the same eigenvalue
lambda. So, this we have also seen before
that for the given eigenvector you can multiply
by any constant and that will also remain
the eigenvector and this is what we will see
here, and more formally theoretically that
this is true for any matrix.
So, here Ax is equal to lambda x that is a
relation, so it tells us that lambda is one
of the eigenvalue and the corresponding eigenvector
is x and then this k time, so we multiplied
the equation by k here both the sides and
then we have k times this Ax is equal to k
times this lambda x and then we can combine
this like A into this kx and is equal to lambda
times this kx there.
So, we have this relation that A some vector
here is equal to lambda the same vector kx
which tells us that this kx is the eigenvector
again, if the x was the eigenvector the k
times x is also the eigenvector for any this
k and nonzero scalar. Here if x is the eigenvector
of the matrix A, then x cannot correspond
to more than one eigenvalue of A. So, another
important result that this eigenvector is
like a unique.
So, if you have an eigenvector corresponding
to let us say the lambda, then this x cannot
correspond to any other eigenvalue, so it
is a unique in that sense. So, here if we
assume that for a given matrix here we have
Ax is equal to lambda 1 x, this is our assumption
and we also assume that this Ax is equal to
lambda 2 x, meaning we have assume that this
x the eigenvector x corresponds to two eigenvalues;
that means, the lambda 1 and lambda 2.
So, these two eigenvalues correspond to the
same in vector x this is our assumption and
we will see now that this is not possible.
So, having this relation we have actually
the lambda 1 x is equal to lambda 2 x because
they have the same value here of vector Ax;
Ax, so they both are same. So, lambda 1 x
is equal to lambda 2 x which tells here the
lambda 1 minus lambda 2 times x is equal to
0 and this x is a where eigenvector, so it
cannot be 0.
So, naturally we should have here that lambda
1 minus lambda 2 is equal to 0. So, lambda
1 minus lambda 2 is equal to 0 meaning say
lambda 1 is equal to lambda 2 since this x
is a eigenvector and so our assumption here
is this somehow says now if we have taken
that there were two eigenvalues. So, these
two eigenvalues have to be the same eigenvalues
you cannot have 2 distinct eigenvalues which
can correspond to the same eigenvector.
So, here A minus KI has eigenvalue lambda
minus k and the corresponding eigenvector
is x, so another result which says that this
A minus KI so if we subtract this k from the
diagonal entries here. So, the eigenvalues
will be lambda minus k of this new matrix
and the corresponding eigenvector will remain
as x. So, to see this again we start with
this standard result on the eigenvalues eigenvector;
that means, Ax is equal to lambda x.
Having this Ax is equal to lambda x we cannot
subtract this kx from both the sides, so here
from lambda x we have subtracted this kx and
here also the same thing the k x, so this
is nothing, but the kx because this ix is
just simply x. So, here also we have kx, here
also we have kx both the sides we have subtracted
this kx here for x we have written ix because
we have also the matrix together, so it will
be easy now to combine.
So, having this now we can take this x common
from this left hand side, so A minus this
KI into x and is equal to here also this lambda
minus k into x. So, with this relation tells
that if this A minus KI has the eigenvalue
lambda minus k, so this A minus KI vector
your matrix sorry is has the eigenvalue A
lambda minus k with the same eigenvector x
as before.
So, this is another result that if we have
this new matrix which is just the A minus
KI, then we know about the eigenvalues from
the eigenvalues of A. And this A inverse again
important here that if it exists of course,
then only we are talking about this result.
So, if A inverse exists for a matrix, then
this A inverse will have eigenvalue here or
eigenvalues one over lambda. So, if we have
lambda 1, lambda 2 eigenvalues for instance
of A, then A inverse will have 1 over lambda
1, 1 over lambda 2 as an eigenvalue.
So, here the A inverse will have eigenvalue
1 over lambda and the corresponding eigenvector
will be x, so eigenvector will not change
only the eigenvalue will change for this inverse
matrix. To see this result we have this Ax
is equal to lambda x and if you multiply by
A inverse both the sides, so we have A inverse
into Ax the right hand side also we have A
inverse into this lambda x. So, we have multiplied
both the sides by this A inverse and then
what we have this A inverse x is equal to,
so A inverse x from this side what we have
here? A inverse x.
So, here we have A inverse x and this lambda
it is a constant term we can take to the left
hand side, where this A inverse A n is just
the identity matrix and identity matrix with
this x will give us x here and this lambda
goes to this left side, so that we will get
1 over lambda. And this A inverse x remains
here, so what relation we have now, that A
inverse x is equal to 1 over lambda times
x. That means, this 1 over lambda here, 1
over lambda is the eigenvalue of this A inverse
matrix, so A inverse x is equal to 1 over
lambda times x.
So, A and A transpose have the same eigenvalues,
A and A transpose have same eigenvalues and
which we can again easily see because the
determinant of a minus lambda I that is the
characteristic equation which actually gives
the eigenvalues. So, here this characteristic
polynomial which is A minus lambda I we know
the property of the determinant, that the
determinant of this matrix A minus lambda
I will be the same as the determinant of a
minus lambda I transpose.
So, the transpose does not change the determinant
of a matrix, so that property we have used
here that the determinant of A minus lambda
I is equal to determinant of this a transpose
of that matrix A minus lambda I. Now the property
of the transpose says here that A minus lambda
A transpose will be A transpose minus lambda
and I transpose which is again I.
So, here this is equal to the determinant
of A transpose minus lambda I and that source
itself there the determinant of this. So,
this characteristic polynomial here A minus
lambda I same as the characteristic polynomial
of A transpose minus lambda I. And this relation
says that we have the same characteristic
equation for A and A transpose; that means,
they will lead to the same eigenvalues.
So, here the result is that A and A transpose
have the same will have the same eigenvalue.
So, A and A transpose have same eigenvalue,
so that is another important result which
easily we can find out with the help of this
determinant property.
Next theorem, so here the characteristic roots
I mean the eigenvalues so sometimes we also
call the characteristic roots. So, the eigenvalues
of Hermitian matrix are real. So, what is
the Hermitian matrix? So, we know that A is
Hermitian when A star is equal to A meaning
the conjugate transpose, A star means here
that we are taking the transport and also
we are taking the complex conjugate of the
matrix A.
So, this complex conjugate of this transpose
is equal to A, then we call that A is Hermitian
matrix. So, what is this result that for Hermitian
matrices the roots are real because what we
have also seen that though the matrix having
all real entries, but we can get the characteristic
roots as complex number we have seen in previous
lectures one of the examples where we had
a very simple 2 by 2 matrix with real entries.
And its characteristic polynomial or I mean
the characteristic roots the eigenvalues were
non real so the complex.
So, here we have at least the results for
the Hermitian matrix that all the characteristic
roots are real in this case. So, if lambda
be a characteristic root of A and lambda the
corresponding eigenvector and then we will
show that this lambda has to be real, how?
So, we have this Ax is equal to lambda x that
is the property of the relation of the eigenvalues
eigenvector again.
We multiply here by this x star term, so what
is x star again the transpose of x and its
complex conjugate. So, we have multiplied
by this vector both the sides and then this
lambda is a constant term, so we can always
take into the front here, so this is x star
x. So, we have x star Ax is equal to lambda
x star x that is a one relation and now we
take the conjugate transpose both the sides
of this equation here x star Ax is equal to
lambda x star x.
So, what do we get here? X star Ax complex
conjugate and here also we take this conjugate
transpose again this x star and then we have
the properties here that this will be x star
and A star and again x star there. So, I mean
the x star or star that will be x and here
also we will have the same scenario the x
star and then the x star star will become
x and this lambda will have its conjugate
there lambda bar.
So, we have this relation and we have also
this relation x star Ax is equal to lambda
x star x, we have this relation x star Ax
is equal to lambda bar x star x. Just by taking
the complex conjugate from this equation we
got this equation and now we have these two
equations here whose left hand side will be
the same because A star is A, so here A here
also this A star is A.
So, with these two equations what we can conclude
that this right hand side should be equal
to 0; that means, lambda x star x is equal
to lambda bar x star x, the reason is that
this A star and this A are the same here,
they are the same. So, naturally the right
hand side will be also the same here and we
have lambda x star x is equal to lambda bar
x star x and this one now what it tells? That
this lambda, so we can bring to the left hand
side. So, lambda minus this lambda bar x star
x is equal to 0 and x is the eigenvector,
so that cannot be 0 x star x cannot be 0.
So, here the lambda must be equal to lambda
bar because this quantity cannot be 0, so
this has to be 0. So, here what we have seen
that the lambda is equal to lambda bar and
that is what we want to see here that the
lambdas are real. So, if lambda is the eigenvalue
of Hermitian matrix, then the lambda is equal
to lambda bar meaning it is a real number,
it cannot be a complex number ok.
So, another similarly we can prove these following
results which have the similar lines of the
proof which we have just done, that the eigenvalues
of this real symmetric matrix are also real
that is what we can also do and the eigenvalues
of real a skew symmetric matrix. So, here
we have; that means, this a transfer is minus
of the A. So, here this real a skew symmetric
matrix are purely imaginary.
So, this is also interesting here that eigenvalues
of such matrices are either purely imaginary
or 0 that is the two possibilities, which
again if you follow the earlier proof we can
also do this one and the eigenvalues of the
a skew Hermitian matrix. So, for Hermitian
matrices we have seen, but now there is a
skew Hermitian matrix; that means, this A
star is equal to minus A.
So, for those cases the eigenvalues are purely
imaginary or 0 again. So, these are the consequence
of the earlier proof which we can easily see
here.
Now, another important result that the eigenvalues
of unitary matrix are of unit modulus, so
this also we can prove in general that if
you have this unitary matrix; that means,
this A star A is equal to is equal to identity
matrix, so such matrices are called the unitary
matrix.
So, if we have unitary matrix then we will
prove now the eigenvalues the modulus of the
eigenvalues is 1; that means, if you consider
here Ax is equal to lambda x and then taking
the complex conjugate here again Ax star is
equal to lambda x star what we will get so
this again we will use this property.
So, x star and the A star is equal to this
will be lambda bar x star and from this A
star Ax is equal to lambda x and from this
equation x star you know A star is equal to
lambda bar x star. We will now continue the
product here we will take the product, so
here x star A star the product with this Ax
these two these two vectors.
So, x star A star multiplied by this vector
Ax is equal to this lambda bar x star and
multiplied by this lambda x. So, we have done
just the product here of these two and then
what we see here x star and this associativity,
so we can use this aced a together and then
x here lambda bar this lambda is a constant,
so we can easily take out.
And then we have x star x there and then we
can take this common because this A star A
is equal to I. So, here we have the identity
matrix meaning this term is nothing, but this
term here is nothing, but the x star and x.
So, we have x star x here also we have x star
x. So, we take common this x star x and we
get 1 minus this lambda bar lambda and that
is equal to 0 and with this we got this result
that this lambda bar lambda is equal to 1
or lambda bar lambda is nothing, but the absolute
value of lambda square. So, this absolute
value of lambda square is equal to 1 because
this cannot be 0. So, this has to be 0 which
tells us this lambda bar square is equal to
0.
So, meaning this we got that this absolute
value of lambda has to be 1. So this unitary
matrix the eigenvalues of the unitary matrix
are of unit modulus. Same results we can also
use for the orthogonal matrices because they
are also having the same property a transpose
A is equal to I. So, for orthogonal matrices
also now we can prove thus the similar all
absolutely all same steps here and we can
again prove the there eigenvalues are also
of unit modulus.
The location of the eigenvalues now what we
have just seen in previous slide. So, if you
have a skew Hermitian matrix their eigenvalues
are imaginary, purely imaginary here the unitary
matrix they lie on this modulus 1 and for
the Hermitian matrix or the symmetric matrix
the values are sitting on the real axis, so
meaning they are the real numbers.
So, here for a skew Hermitian and exclusive
metric the same thing unitary and orthogonal
we have the same result, that they are of
a unit modulus for Hermitian and symmetric
we have also the same result for both that
they are the real entries.
Getting to the conclusion, so we have seen
several properties of this eigenvalues and
eigenvectors of a matrix, we have considered
different; different types of matrices where
we can tell about whether the eigenvalues
will be real imaginary if your imaginary you
are 0.
So, here in all these properties the simple
idea was to use this Ax is equal to lambda
x and we played with this equation only to
prove all these properties. And now they can
be used now without doing all these numerical
calculations we can compute directly also
with the help of these properties.
So, these are the references we have used
to prepare these lectures.
Thank you for your attention.
