Welcome back and this is a lecture number
47, we will continue our discussion on Eigenvalues
and Eigenvectors.
In particular today we will focus more on
evaluation of the eigenvalues and some good
worked out problems will be discussed.
So, let us start with this problem here find
eigenvalues and eigenvectors of this matrix,
again a very simple matrix we have taken 2
square root minus 2 and then square root 2
and again here square root 2 and this one
there. So, the characteristic equation we
have to write down as a first step to find
the eigenvalues; that means, the determinant
of this A minus lambda I is equal to 0 that
is the characteristic equation we have.
So, here the lambda will be just subtracted
from the diagonal entries otherwise this is
just the matrix a. So, 2 minus lambda and
the square root lambda again here the same
element is square root 2 and then 1 will be
1 minus lambda is equal to 0 and now we can
simplify this easily. So, we can multiply
2 minus lambda 2 this 1 minus lambda and then
this minus this 2 so that can be simplified
to give that minus 2.
So, this is the 2 minus lambda and then we
have 1 minus lambda and minus 2. So, what
do we get here? The 2 and then this will be
minus 2 and here minus 2, so 3 minus 3 lambda
and plus this lambda square and then we have
minus lambda minus 2, so this 2 gets cancelled
and we have exactly this characteristic equation
which is lambda times this lambda minus 3.
So, here we have this characteristic equation
which tells that there are 2 eigenvalues lambda
is equal to 0 and lambda is equal to 3.
So, the eigenvalues here are lambda is equal
to 0 and lambda is equal to 3. So, now, we
will compute the eigenvector corresponding
to lambda 1, the first eigenvalue that is
0 here. So, we have to solve the system of
linear equation A minus lambda x is equal
to 0. So, the lambda is 0, so this is simply
this a so we are solving basically this A
x is equal to 0 or we are trying to get the
null space of this A x is equal to 0 and the
generator of the null space or the basis of
the any basis of the null space will be the
eigenvector.
So, here getting to this A here we have to
square root 2 and square root 2 and 1 here
and then the vector x 1 x 2 is equal to 0.
So, we need to solve this equation, so how
to get this one? Let us just take a while
here, so this 2 and square root 2 that is
the first row which we can keep as it is.
The second one is square root 2 and 1, so
we want to make this 0, so we can divide the
first row for instance by a square root 2
and this will be 0 then and here 1 1, so that
would be also 0.
So, this is the rho reduce a echelon form
of the given matrix. So, we have basically
this first equation which says that 2 x 1
minus square root 2 x 2 is equal to 0 and
we have one free variable which we can take
whichever we want, so let us take this x 2
is a free variable because as per our notations
here this is the first the p both here. So,
this x 1 is not free and then we can take
as x 2 free, but any case in any case we can
take any one of them and the other one will
depend on the given chosen number. So, here
the x 2 we can choose and then we can get
the x 1.
So, out of these many possibilities we can
just pick one vector that we have done here
for instance. The x 2 was taken as the free
variable alpha and then x 1 is coming 1 over
square root 2 with the minus sign or any multiple
of this we can for instance x square or 2.
So, we can also take the vector here the square
root 2 with minus sign and the square root.
So, here when we multiply 2 square root 2
to both, so this will become 1.
So, this vector also we can take minus 1 and
square root 2 or we can take 1 and minus square
root 2. So, we can multiply by any number
here that will be the characteristic a vector
here or the eigenvector of the given equation.
So, here the one characteristic vector or
the eigenvector we got corresponding to lambda
1 is equal to 0.
Similarly, we can also look for the eigenvector
corresponding to lambda 2 is equal to 3. So,
again we have to solve the system of equation
a minus lambda I x is equal to 0 this time
and then, so again this lambda will be subtracted
from this matrix A.
So, that will be the matrix here for the coefficient
matrix for the system of equation and again
we have to get the solution of this one which
we can we can again choose one of because
the second row can be made to 0 when we multiply
the first equation by square root 2 and then
subtract from 2, so the second row will be
0.
So, the corresponding to this the row reduced
form will be minus 1 and square root 2 0 0
and from here now we can choose this x 2 variable
s as alpha and then here the x 1 from this
first equation minus x 1 plus square root
2 x 2 is 0; that means, the x 1 is square
root 2 x 2, x 2 is alpha. So, we have taken
here x 1 as square root 2 alpha which we have
written in this form the solution x 1 is square
root 2 alpha and x 2 is alpha. So, this is
another vector here any multiple of this is
square root 2 1 we can take as an eigenvector.
So, in this case the eigenvectors corresponding
to the 0 was this minus 1 and square root
2 and here we have this is square root 2 and
1. Again we can note that these eigen vectors
are linearly independent and as I said in
the previous lecture that we will also proof
formally this result that corresponding to
distinct eigenvalues we have the eigenvectors,
the set is linearly dependent the set of eigenvectors
is linearly independent or the eigenvectors
corresponding to distinct eigenvalues are
always linearly independent. So we can check
here also that these 2 vectors are linearly
independent.
Another problem this is 3 by 3 matrix, so
3 minus 2 0 and so on. So, here again we want
to find the eigenvalues and eigenvectors of
this given matrix which is 3 by 3 now. So,
we have to again follow these steps that first
we have to write down the characteristic equation
from there we can get the eigenvalues. So,
the characteristic equation will be determinant
of this a minus lambda I, so we have to subtract
this lambda from all the diagonal entries
which is 3 3 and 5 there.
So, that is the characteristic equation of
this matrix and then we have to evaluate this
determinant which is not so difficult. So,
we have this 3 minus lambda here and that
will be multiplied now with this 3 minus lambda
into this 5 minus lambda and minus 0, so this
is done and then we have this minus minus
plus 2 here. So, this will be then minus 2
times this 5 minus lambda and then the rest
will be 0.
So, here and then again, so this is the value
of this determinant which should be 0, then
this 5 minus lambda we can take as common
from both the terms. So, here we will get
this 9 and then we have also minus 3 lambda
minus 3 lambda, so minus 6 lambda plus lambda
square, so minus 6 lambda plus lambda square.
So, there is 3 minus lambda whole square,
so and then we have here minus this 4 term
2 2 the 4 here. So, this is the characteristic
equation here with again we can factorize
the second term, so your 5 minus lambda and
this is lambda square minus 6 lambda and this
is 5.
So, we got here this 5 minus lambda this factor
and then this is lambda minus 5 and lambda
minus 1, so is equal to 0. So, that is the
characteristic equation from there we can
get the roots of the equation, so we have
this 5 minus lambda or lambda minus 5 this
whole square and thus we have this lambda
minus 1 the another factor. So, the characteristic
equations suggest now that we have basically
2 distinct eigenvalues the one is lambda is
equal to 1 the other 1 is 5 which is repeated
two times.
So, eigenvalues are now 1 and 5, these are
the 2 eigenvalues and we need to compute the
eigenvectors corresponding to each.
So, when we take this lambda 1 is equal to
1 what do we get this system of equation a
minus lambda I x is equal to 0 and our a was
this here 3 minus 2 0 and minus 2 3 0 0 0
5. So, here a minus this lambda I and lambda
is 1 now, so 1 will be subtracted from the
diagonal and that will be our matrix here
with this 2 minus 2 0 and this is again 2
here the diagonal entries is reduced by 1.
So, we have this matrix which we want to now
solve the system here and we need to get this
reduced to echelon form, so here 2 minus 2
and 0 here we want to make it 0, so we can
just add the rule number 1.
So, this is 0 and this will become 0 and this
is 0, so here 0 0 4, this is the one step
of the row reduced form and now we can interchange
these 2 rows. So, we will get 2 minus 2 0
0 0 and this 4 and the last row will be the
0 row. So, we have this, so we have the pivot
here in the first column, we have pivot in
the third column the second, so corresponding
to x 2 here we can choose as a free variable
and then x 1 and x 3 will be computed based
on this free variable.
So, here the second equation; however, suggests
that the x 3 is 0 because this third second
equation is now 4 times the x 3 is 0. So,
which tells us that x 3 is anyway 0 always
and then x 2 is a free variable. So, we can
take any alpha and then from this equation
number 1 which says the 2 x 1 is equal to
2 x 2, so this x 2 is free variables. So,
here x 1 is also alpha because x 1 and x 2
both are equal. So, we have here x 1 alpha
x 2 alpha and x 3 is 0 in all the cases, so
what do we get?
We get this as the solution here of this equation
that x 1 x 2 x 3 is the alpha times the 1
1 0 because x 1 was alpha x 2 was also alpha
and the third component x 3 is coming to be
0. So, in this case the solution of this system
is alpha times 1 1 0 and that is exactly the
generator of the null space of this matrix.
So, we can pick any vector, any nonzero vector
from this null space. So, we can take 4 instances
1 1 0 or any multiple of it that will be the
eigenvector corresponding to this eigenvalue
lambda 1 is equal to 1.
Moving to the other one, so we have this lambda
2 is equal to lambda 3 the 2 eigenvalues are
equal there which is 5 and then if we set
up the system of equations for this given
A which will be in this case because the 5
will be subtracted from the diagonal entry,
so we will get minus 2 here minus 2 and the
last entry the diagonal entry will be 0. So,
we have this system and this is interesting
now to see that we can reduce to this row
reduce echelon form very easily. So, when
we subtract row number 1 from row number 2,
so we will get 0 0 and 0 here again this is
0.
So, this is the row reduced echelon form corresponding
to this matrix which suggests now we have
here this pivot element and that is the only
pivot element, so we can choose x 2 and we
can choose x 3 whatever we like. So, here
let us take this alpha 1 and here if we take
this l alpha 2, then what will happen from
this equation 1 we have the minus this x 1
and minus this x 2 is equal to 0 or minus
2 x 1 minus 2 x 2 is equal to 0, meaning this
x 1 is minus x 2 that is coming from this
equation number so this x 1 is x 2 means this
minus alpha 1.
So, we have now the vector here the x 1 x
2 and x 3 with alpha 1 and alpha 2. So, alpha
1 with x 1 we have minus 1 and x 2 also we
have a component here this is 0 with alpha
2, again we have in x 1 component there is
no alpha 2 in x 2 or so there is no alpha
2 and here we have this 1. So, this is the
solution of this system of equation x 1 x
2 x 3 is alpha 1 times minus 1 1 0 and alpha
2 times this 0 0 1. Having this, so, what
we have exactly this is written here, so x
2 we have chosen this free variable x 3 also
free variable and x 1 is coming as this minus
alpha 1 which is we have evaluated and this
is the solution now which we have already
written.
So, here we have these 2 generators minus
1 1 0 and 4 alpha to the 0 0 1. So, what do
you observe in this case? That we are getting
this is either 1 is satisfying the given differential
equation the given system of the question,
so here minus 1 1 0 satisfies the given equation,
also 0 0 1 is satisfying the given equation.
And these two are also linearly independent
which we can see from the structure itself,
any linear combination like alpha 1 plus alpha
2 is equal to if we set to 0 the alpha 1 alpha
2 has to be 0.
So, that says easy check for the linear independency
in for this case. So, what we are getting
corresponding to this lambda is equal to 5
which was the repeated eigenvalue and we are
also getting here the 2 linearly independent
vectors which satisfy this A minus lambda
I x is equal to 0. So, basically corresponding
to lambda is equal to 5 we are getting 2 eigenvectors
or rather to say 2 linearly independent eigenvectors,
we are getting in this case.
Now, if we look into another problem here
the eigenspace of this matrix that is 2 1
and 0 0, then we have this is actually 4 by
4, so little bigger matrix, but the calculations
are, so in this case if we set up the characteristic
equation again A minus lambda I x is equal
to 0. So, the determinant of this will give
the characteristic equation and what is the
determinant here?
So, 2 minus lambda 2 minus lambda and so the
determinant is this 2 minus lambda 1 0 0 and
here also 0 this 2 minus lambda 1 and 0 and
to minus lambda and then we have no the second
component is 0, then third one is 2 minus
lambda and this 1 and 0 0 0 at this 2 minus
lambda. What is interesting now that if we
compute this determinant, so this and then
the rest determinant again we will take, so
this is 0 here. So, again this, so this is
coming that 2 minus lambda power cube nothing
else.
So, in this case this determinant is nothing,
but these 2 minus yeah the determinant is
here when we get we will solve for this determinant
we are getting this 2 minus lambda cube; so,
2 minus lambda sorry for because there are
4 entries here, so we will get just the product
of the diagonal entries for such matrices.
So, we get this 2 minus lambda power 4 is
equal to 0, so this is another special case
where all the eigenvalues, all these 4 eigenvalues
are same and that is the 2 or the value is
2. So, if we compute now because we want to
compute the eigenspace means the eigenvectors
set of all eigenvectors and when we include
0 or so that is called the eigenspace corresponding
to the given eigenvalue.
So, here the eigenvalue is 2 which is repeated
4 times, so to get the eigenvector. So, we
have to set up this a minus lambda x is equal
to 0 this system of equation corresponding
to this lambda is equal to 2. So, what do
we get now corresponding to this one because
this 2 will be subtracted from the diagonal
entries and our matrix there was having this
2 in the diagonal entries. So, this diagonal
entries will become 0 now. So, we got this
as our system of equation and here we do not
have to do anything to get the reduced form
because we can clearly see this structure
here this is already the reduced form the
reduced echelon form.
So, this is the pivot element and this is
another pivot element this is also a pivot
element. So, how many pivot elements are there
this x 2 is the dependent variable, x 3 will
be the dependent variable, x 4 will be the
dependent variable and here we have this x
1 which we call the free variable. So, our
free variable is this x 1 which is corresponding
to this column number 1 because we do not
have pivot in this column. So, we can choose
this x 1 as a free variable, so x 1 is not
taken as alpha and now we can get the other
variables.
So, from this equation number third, from
the third equation what we are getting, so
we are getting this x 4 is equal to, so no
dependency on alpha indeed. From the second
equation we are getting this x 3 is equal
to 0 and from the first equation we are getting
x 2 is equal to 0. So, that is the solution
here x 1 we can choose any vector alpha as
alpha we have taken here, and now if we write
down the solution x 1 x 2 x 3 and x 4.
So, in terms of the alpha only the first component
has alpha all others are 0, so this is the
solution of this system of equation; that
means, here we have only one free variable
at though the 2 was repeated several times
the 2 was repeated 4 times in this particular
example, but what we are getting here we are
getting only 1 linearly independent eigenvector.
Whereas, in the previous example the eigenvalue
was repeated two times and we were also getting
two linearly independent solution of this
equation, but now though the 2 was repeated
4 times, but we are getting only 1 eigenvector
corresponding to this 4 times repeated eigenvalue
2.
So, what we want to discuss now with this
example that that anything is possible just
based on this eigenvalue whether it’s repeated
several times we cannot claim anything about
I mean directly looking at the eigenvalue,
we cannot claim that how many linearly independent
eigenvectors we will get as this is the case
here the eigenvalue was repeated 4 times,
but we are getting only 1 linearly independent
eigenvector and that is this 1 0 0 in this
case.
So, and in some cases if it is repeated for
example, 3 times you may get 3 linearly independent
eigenvectors also, so the other way around
is also possible ok. So, thus a basis of this
eigen space is this 1 the three 0’s here
the transpose is used here, so that is the
basis for the eigenspace or an any vector
from this eigenspace other than 0 is the eigenvector
right. So, here this is also the basis for
the null space of this matrix here 1 0 0 0
and this the transpose to show that this is
a column vector ok.
So, the another example where we will look
for the eigenvalues and the eigenvectors of
this matrix A which is given by this 1 1 and
minus 2 and 3. So, here if we compute this
determinant of a minus lambda I is equal to
0, so what will happen? So, we have 1 minus
lambda and then 3 minus lambda minus 2 and
1 remain, intact. So here if we compute this,
so we have 1 minus lambda and multiplied by
this 3 minus lambda and minus minus this becomes
2 is equal to 0. So, let us do the calculation
so 3 and then we have 4 lambdas also we have
minus 4 lambda and then this lambda square
plus this 2 is equal to 0.
So, we are getting this equation lambda square
minus this 4 lambda and plus this 3 plus 2
this 5 is equal to 0. So, we are getting this
lambda minus 5 and lambda plus so the plus.
So, once more here, so we have the 3 and then
we have a minus 3 lambda also minus lambda.
So, will be minus 4 lambda here the minus
minus, so lambda square and then plus 2.
So, we have this lambda square term as it
is and minus 4 lambda and plus 5 is equal
to 0. So, we get this roots here for this
equation, so we have lambda square then minus
5 times lambda and plus this lambda plus 5
is equal to 0, so this minus 4 lambda we have
written as minus 5 lambda plus lambda. So,
here this lambda and this lambda minus 5 and
plus this 1 and lambda minus 5 lambda plus
5. Now this is not working, so let us just
check here that the equation is lambda square
minus this 4 lambda n plus 5 exactly and the
roots here because it does not have the real
roots.
So, that is the point here we are not getting
any factorization in this case we are not
getting this factorization here lambda plus
5 is coming and then we cannot take common.
So, the factorization was not possible because
this equation does not have a real root. So,
when we solve this equation for this lambda
it is a quadratic equation. So, we can easily
do that. So, minus b; that means, 4 b square
minus 4 ac and then we have here 2 a, so divided
by 2.
So, this is 4 plus minus and then here also
4, so this will becoming 2 i and then we have
a 2 there. So, this is a 2 plus minus I is
the eigenvalue here, so there are 2 eigenvalues
what is interesting in this example that we
are getting 2 complex values; 2 complex value
for the eigenvalues though the matrix here
was real matrix. So, we had the entries 1
1 minus 2 and 3, but we are getting here 2
complex values for the eigenvalues in this
case.
So, that again shows that the matrix entries
may be real we can have a real matrix, but
still we may get the complex roots of this
characteristic equation meaning the eigenvalues
may be complex. So, a real matrix may have
complex eigenvalues that is the remark and
then eigenvectors corresponding to now this
2 plus i the first eigenvalue. So, we have
1 eigen value 2 plus i another one is 2 minus
i and we will see later on that these eigenvalues
will always appear in conjugate form as the
root always appears there and not only that
we will have something more interesting for
the eigenvectors corresponding to these conjugate
eigenvalues.
So, if we take this 2 plus i and again the
same system of equation this a minus lambda
I x is equal to 0 we need to solve. So, that
will be this 1 minus lambda, 3 minus lambda
here and x 1 x 2 is equal to 0. So, this equation
and now for lambda 1 this 2 plus i we need
to substitute for this lambda.
And what we will get, so corresponding to
this we have now this A now the coefficient
matrix of this eigenvector equation. And now
we will make this transformation that this
R 2 will be now R 2 minus R 1 multiplied by
this 1 minus i. So, if you multiply this equation
by 1 minus i and subtract from this equation
number 2, what we will get? We will get the
reduced form of the echelon form and this
will be precisely 0 here.
So, when we multiply by this 1 minus i this
will be minus 2 and this will become 1 minus
i. So, when we subtract this is exactly what
happening here the 0 and 0 we are getting
the 0 row, so this is already the reduced
row reduced echelon form and then what we
we have here? We have this first equation
which is 1 plus i times this x 1 and is equal
to x 2 this is the first equation and we can
take any arbitrary value for x 2 and we can
get the corresponding x 1 or we can get any
arbitrary value for x 1, we can get the x
2 because there is only one equation.
So, one variable we have to choose arbitrarily.
So, by doing, so if we take this x 2 1 plus
I then we are getting x 1 is equal to 1. So,
that is the one possible eigenvector we are
getting here. So, the eigenvector corresponding
to this lambda 1 is nothing, but 1 and 1 plus
i that is the eigenvector corresponding to
this eigenvalue 2 plus i.
The same calculation we can repeat for the
other root which was 2 minus i and again what
we will get here this is our matrix now for
this system of equation and by using this
transformation here that are the row transformation
that R 2, we will now get R 2 minus when we
multiply 2 R 1 by 1 plus i and again the same
situation here, so we can get when we multiply
2 1 plus i, in this case we will be getting
again this minus 2 there and here 1 plus i
1 when we subtract, the second row will become
0.
So, we have this row reduced form by this
transformation and again we have this equation
1 minus i x 1 is equal to x 2 from where we
can just set this x 2 is equal to 1 minus
i and we can get this x 1 as 1. So, a eigenvector
corresponding to this lambda 1 is now this
1 and the 1 minus i, so we have 1 and 1 minus
i as the eigenvector. So, if A is a real matrix
this is the result not related to this example
the general result what we have now because
what we observed in this example, the earlier
eigenvalue was 2 plus i and the corresponding
eigenvector was 1 plus i and now we have this
2 minus i and the eigenvector here what we
get 1 minus i.
So, the eigenvector or one of the eigenvectors
here we are getting exactly the conjugate
of what we have got for the eigenvector corresponding
to its conjugate there. So, that is a nice
result here in general we have this then A
is a real matrix and has complex eigenvalues.
Then the conjugate ; then the conjugate this
lambda bar, so lambda is an eigenvalue, then
the conjugate will be also the eigenvalue
and, but that is natural which is coming from
this characteristic equation. So, the conjugate
will be always there as the root.
So, here and what is interesting? The interesting
is that this we have from this Ax is equal
to lambda x if we take the conjugate both
the side what we will get because a is a real
matrix, so A will remain as it is and this
will be the conjugate here for x and the right
hand side the conjugate of lambda and the
conjugate of x. So, again we have this equation,
the eigenvalues eigenvector equation is satisfied
for this x x bar, the conjugate of x and the
right hand side is just the lambda bar that
is the eigenvalue now and x bar.
So, what this tells now that this x bar here
the conjugate of x will be the eigenvector
corresponding to this conjugate of lambda
bar. So, indeed we do not have to compute
this since we know this result from this calculation
that once we have eigenvector corresponding
to one complex value of this lambda and its
conjugate will be will be the eigenvector
of the other conjugate eigenvalue. So, here
that is the interesting result we have about
the conjugate roots or about the conjugate
eigenvalues complex conjugate here.
So, in this case the eigenvectors corresponding
to distinct eigenvalues are linearly independent
this is what we have observed at least for
the calculations we had in numerical calculations
and also what we have observed here that a
real matrix may have a complex eigenvalues.
So, that is interesting here and both the
eigenvalues and eigenvectors are correct complex
conjugate pair.
So, the once we have the complex value as
the eigenvalue its conjugate will be there
and not only that the eigenvectors will be
also the conjugate pairs. So, that is interesting
result we have seen and that is true in general
not for the example we have just shown.
So, these are the references used. Thank you
for your attention.
