Welcome back, so this is lecture number 49
and we will be talking about today the geometric
and algebraic multiplicity and the similarity
of matrices.
So, what is the algebraic multiplicity? So,
algebraic multiplicity of lambda as a root
of the its a multiplicity of lambda as a root
of the characteristic equation. And the geometric
multiplicity is nothing but, the dimension
of the eigenspace of lambda; that means, the
number of linearly independent eigenvectors
corresponding to an eigenvalue lambda.
So, these are the two numbers which we will
now use for telling about the multiplicity
of this lambda, because we have seen in several
examples the characteristic, roots all the
eigenvalues were repeated. So, that we can
now quantify with the help of this algebraic
multiplicity; so, algebraic multiplicity if
for instance one root is repeated 3 times.
So, then it is algebraic multiplicity of that
particular root is 3 and the geometric multiplicity
will be the dimension of the eigenspace or
the number of linearly independent vectors
we have corresponding to that particular eigenvalue
lambda.
So, with these two classification we will
move further, but before that there is a note
here, that this geometric multiplicity is
always less than or equal to the algebraic
multiplicity, that is a important result which
one can formally prove. But, it is it requires
little more knowledge of a diagonalization
etcetera.
So, we will not prove this result now, but
we will keep in mind that this geometric multiplicity
is always less than or equal to the algebraic
multiplicity. Meaning that for example, one
a particular root; one particular eigenvalue
is repeated 3 times than the corresponding
geometric multiplicity meaning they are number
of linearly independent eigenvectors cannot
be more than 3, they have to be less than
or equal to 3.
So, first we will see with the help of many
examples that what are the situations arises
here. So, in this case we find the eigenvalues
and the eigenvectors of this matrix A which
is given as a 6 minus 2 and 2 here of minus
2 3 1 and 2 minus 1 3. So, for this eigen,
for this matrix we will compute the eigenvalue
and eigenvectors we have already computed
for several matrices in the last lecture.
So, we are on now the familiar with the computation
of the eigenvalues.
So, here first we need to write down the characteristic
equation for this given matrix which is the
determinant of this A minus lambda I, determinant
of this matrix A minus lambda I is equal to
0 and for this matrix we can compute this
determinant here. So, the determinant would
be like the 6 minus lambda minus 2 2 and then
minus 2 here 3 minus lambda and minus 1 2
minus 1 3 minus lambda. So, this lambda would
be subtracted from the diagonal entries.
And with this now we can expand this, so here
the 6 minus lambda and then we have this product
minus this, then we will take this 2 then,
minus 2. So, with this value of this determinant
here which will be coming as; when we do the
factorization of this polynomial that will
become 2 minus lambda lambda minus 2 and the
lambda minus 8. So, I skip this portion here
because in this lecture that is not important,
we have already seen for several examples
in the last lecture.
So, what we have? We have this characteristic
equation of this matrix here as this 2 minus
lambda lambda minus 2 and lambda minus 8 equal
to 0. So, what we observe now that there are
two distinct eigenvalues and one is repeated
2 times so; that means, the lambdas are the
2, 2 and 8. So, this eigenvalue 2 is repeated
2 times and this 8 is repeated 1 times, and
exactly that is what we have discussed about
this algebraic multiplicity.
So, the algebraic multiplicity of this lambda
is equal to 2 this eigenvalue 2 is because
it is repeated 2 times here. So, that I the
algebraic multiplicity of this 2 is 2 and
the algebraic multiplicity of 8 because this
is repeated only once. So, here the algebraic
multiplicity of this lambda is equal to 8
is 1. So, this is how the algebraic multiplicity
and the geometric algebraic multiplicity is
defined to define the geometric multiplicity
corresponding; to lambda is equal to 2 or
lambda is equal to 8, we need to get the eigenspace
of these vectors of these eigenvalues.
So, the eigenvector corresponding to this
lambda is equal to 5. So, remember this lambda
is equal to 8 was repeated once. Though, we
have already the result that the eigen the
geometric multiplicity cannot be more than
1 now in this case, because the algebraic
multiplicity of this lambda is equal to 8
is 1. So, without calculation of the eigenvectors
as well we can claim that the geometric multiplicity
will be 1, because definitely there will be
one linearly independent eigenvector corresponding
to this lambda is equal to 8.
So, there cannot be two linearly independent
eigenvectors, that was that result says where
we have that these algebraic multiplicity
is always bigger than the geometric multiplicity.
So, here we know though beforehand that this
there will be there cannot be two linearly
independent vectors, it has to be only one
because the algebraic multiplicity of this
lambda is equal to 8 is 1.
And it had it just must to have at least 1
eigenvectors because that is what the foundation
says so, we have already this a minus lambda
I and the determinant is equal to 0 we have
non-trivial solution always. For this equation
here A minus lambda x is equal to 0. So, there
will be definitely one linearly independent
eigenvector, but there cannot be two this
is what we will see in this case as well.
So, here this is a minus lambda I; so, this
lambda means 8 here was subtracted from the
diagonal entries of A and then we have x 1
x 2 x 3 and the right hand side this 0 vector.
So, we can reduce to this echelon form this
matrix and so, 2 minus 2 2 first row as it
is and then here we can subtract this. So
this will be 0, when you subtract will be
minus 3 and this is minus 3 and here also
we can add in the first step. So, this will
be 0, this will be minus 3 and this will be
minus 3 and then in the second step again
with the help of the second column.
So, in the first step what do we get, it is
like minus 2 minus 2 2 and then here we have
0 and then minus 3 and minus 3 here we get
when we add row 1 and row 3. So, we will get
0 well get minus 3 and minus 3. So, again
with the help of the second row, we can now
get actually get rid of this number here minus
3, but this will also become 0 together. So,
this is the situation, this is the row reduced
echelon form for the system of equations for
this matrix.
And now, we observe here that this is the
pivot element and here also we have the pivot
element. So, the first two columns have pivot
element that third column does not have pivot
element. So, that corresponds to this x 3
component of this vector and which we can
take as the free variable. So, there will
be only one free variable which was clear
from there also because the algebraic multiplicity
was one and corresponding to that we cannot
get two free variables. So, the number of
free variables tells about the number of linearly
independent eigenvectors.
So, here we cannot have two linearly independent
eigenvector. So, we know in advance that there
will be only one free variable in this case,
there cannot be two free variables. So, this
x 3 is the free variable which we can choose
again as as alpha; having that alpha we can
compute the x 1 and x 2 in terms of of alpha.
So, then we can write down this solution of
this equation x 1 x 2 x 3. So, here this x
3 was taken as alpha and x 2 comes to be from
here the minus of x 3. So, we got this minus
alpha and from this equation number 1, we
will got this two time; we will get this 2
times alpha, this vector here x 1. So, we
have this alpha not equal to 0 and alpha belongs
to this real number we can take any real number
here. So, this is for any alpha not equal
to 0 these are the eigenvectors. And basically
the dimension of this eigenspace is one or
in other words we got only one linearly independent
eigenvector which we can take for instance
this is 2 minus 1 1.
So, that is the only one eigenvector which
is linearly independent, any other eigenvector
which we get out of taking this value alpha
where they are the dependent eigenvectors
on this 2 minus 1 1. So, in this case we got
only one linearly independent eigenvector
and therefore, we say that the geometric multiplicity.
The geometric multiplicity that is the number
of linearly independent eigenvectors corresponding
to a given eigenvalue here the lambda is equal
to A. So, the geometric multiplicity of this
lambda is equal to 8 is 1. So, here that is
the number of linearly independent eigen vectors.
When we come to this eigenvalues lambda is
an; eigenvalue lambda is equal to 2 and remember
it was repeated 2 times meaning the algebraic
multiplicity of this lambda is equal to 2
was 2. So, in this case we have the possibility
that the corresponding eigenvectors the corresponding
linearly independent eigenvectors there may
be 2, I mean at most 2, but there may be 1
as well, we do not know now in advance we
have to compute them.
Because, looking at this eigenvalue we cannot
just tell how many eigenvectors will be linearly
independent corresponding to a given eigenvalue,
but what we can tell now because the multiplicity
of this 2 was 2 or the algebraic multiplicity
was 2 and we know that the geometric multiplicity
will be less than or equal to 2. So, we know
now that the number of linearly independent
eigenvectors could be 1 or it could be 2 also
now in this case.
So, like let us compute this. So, this a minus
lambda I when we subtract from this diagonal
entries this number lambda. So, we get this
equation the system of linear equation and
then by reducing to this echelon form. So,
indeed these two rows are the same. So, we
can set one of them equal to 0 immediately
and out of this first row again because it
is half of this is again when we add to this
row number 2. So, this will become 0 and similarly
row number 3, if we subtract half of the row
number 1 this will also become 0. So, this
is here the operation we have taken that the
R 2 is nothing but R 2 plus half of R 1 and
here for R 3 we have taken now the R 3 and
minus the half of R 1.
So, with this two operations; two elementary
operations we have written got this row reduced
echelon form of this system of linear equation.
And then and now we can identify that how
many linearly independent eigenvectors we
are going to have in this particular case.
So, here this is the pivot element which is
minus 2 in this case and the column number
two does not have a pivot here also we do
not have pivot. So, there is only one pivot
that is in the column number 1.
So, here x 2 and the x 3; x 2 and x 3 will
be will be free variables so, there will be
free variables now free variables. So, we
can assign any value to them; that means,
we are going to have now two linearly independent
eigenvectors because a number of free variables
decide exactly how many linearly independent
eigenvectors we will get.
So, in this case we will get two linearly
independent eigenvectors, and that is what
we write. So, here x 2 is taking alpha 1,
x 3 is taken as alpha 2 and then we have computed
this alpha 1 from; this x 1 from this equation
number 1 when we write in the vector form
as this alpha 1 we have this half and 1 0,
alpha 2 a minus half 0 1. So, we got this
two linearly independent eigenvector. So,
this one and this one, one can check where
they are linearly independent.
So, corresponding to those lambda is equal
to 2 because it was repeated this 2 times
the algebraic multiplicity was 2, this algebraic
multiplicity of this was 2 and we also got
now the geometric multiplicity. So, the geometric
multiplicity is also 2 in this case. So, we
have the algebraic multiplicity 2 and as well
as the geometric multiplicity 2; geometric
cannot be more than the algebraic one again,
but in this case we got at least the equality.
So, the geometric multiplicity of this lambda
is equal to 2 is 2, because we have two linearly
independent eigenvectors corresponding to
this lambda is equal to 2 ok.
So this example 2, where we determine the
eigenvalues and eigenvectors of this A the
matrix is given here 2 4 0 0 2 0 0 0 3 and
in this case one can compute easily the eigenvalues
will be the diagonal entries because it is
a lower triangular matrix and for the triangular
matrices, we have all the eigenvalues sitting
on the diagonals here.
So, we have this 2 2 3 there these are the
eigenvalues and the eigenvalues of a triangular
matrix are always it is a diagonal element;
so, we have these eigenvalues 2 2 3. The eigenspace
now we will compute for lambda is equal to
2. Again I mean here, the if we want to know
the algebraic multiplicity so it is 2 4 2
and the algebraic multiplicity of 3 is 1.
So, here the eigenspace you want to compute
now to get the geometric multiplicity.
So, here the eigenspace; so, A minus lambda
I so here 2 will be subtracted from the diagonal
entry. So, we will get 0 there, 0 there, 1
there. So, this is the now the matrix A minus
lambda I and x is equal to 0. So, what do
we see here, we can actually just take this
we can interchange the row and then we have
this echelon form; row reduced echelon form
the 0 we can bring to the bottom if we like.
So, we can easily convert to this echelon
form here and then we will see there will
be 2; there will be 2 pivot elements here.
So, this will be the pivot element and this
will be also the pivot element when we convert
into this echelon form. And this middle one
so, here the first column will have a pivot
and the third column has a pivot and this
x 2 is going to be the free variable. So,
this x 2 is going to be the free variable;
that means, only one free variable and we
will get only one linearly independent eigenvector.
And surprisingly here that, when we compute
this x 3 is equal to 0 that is straight away
from this equation and from this equation,
we will get this x 1 is equal to0. So, out
of this we are getting x 1 is equal to 0 also
x 3 is equal to 0 and this x 2 will be the
free variable which we can take as alpha,
and then this x 1 x 2 x 3we can write down
as alpha times 0 1 0. So, here in this case
what we observe though the algebraic multiplicity
of this lambda is equal to 2 was 2, but now
we got the geometric multiplicity as 1.
So, not surprising as I said before that for
given eigenvalues we cannot predict in advance
at how many eigenvalue vectors will be linearly
independent so, we have to compute them. What
we know from that result that algebraic multiplicity
is less than or equal to the, or the geometric
multiplicity is less than equal to the algebraic
multiplicity that the algebraic multiplicity
of this 2 was 2. So, we know that there will
be at most 2 linearly independent eigenvectors.
There cannot be three linearly eigen linearly
independent eigenvectors for example, in this
case, but we do not know whether there will
be 2 or there will be 1. So, what we have
observed? In the previous example though the
algebraic multiplicity was 2 and the geometric
multiplicity was also 2. In this case we have
the algebraic multiplicity 2, but geometric
multiplicity is just 1 in this case.
So, here the geometric multiplicity is 1,
because we have 1 linearly independent eigenvector
and the algebraic multiplicity of 2 is 2 because
this 2 was repeated 2 times.
Coming to the eigenspace of this lambda is
equal to 3. So, we know already that there
would be only one. So, in this case we know
that there will be only one free variable
definitely because the algebraic multiplicity
is 1. So, we cannot have more than 1 linearly
independent eigenvector.
So, here if we compute this a minus lambda
I x is equal to 0. So, we have this and then
when we solve this system. So, we will observe
that there are 2 pivots in this case, when
we just we can just make this to 0 and then
this will become also a pivot because this
will not be 0 in that case. So, you will have
2 pivot elements and this x 3 will be the
free variable in this case.
So, therefore, this alpha is corresponding
to x 3 and this x 2 will be 0 and x 1 will
be also 0 from this structure of the matrix.
So, we will get the solution alpha x 0 1 1
and as expected or there is only one linearly
independent eigenvector corresponding to this
lambda is equal to 3. So, the geometric multiplicity
of this lambda is equal to 3 is 1 and the
algebraic multiplicity of this lambda is equal
to 3 was also 1 in this case.
Another example we will find the dimension
of this eigenspace of this lambda is equal
to this very special matrix here 1 0 0 1 1
1 and 0 0 1. So, in this case again we need
to write the characteristic equation. So;
that means, a minus lambda I is equal to 0;
so, 1 minus lambda here, 1 minus lambda and
1 minus lambda and that determinant.
So, what we will observe in this case that
the characteristic equation is lambda minus
1 power 3 is equal to 0. So, we have these
3 roots; so, 1 1 1. That means, this algebraic
multiplicity of this lambda is equal to 1
is 3 now. So, we have an example where all
these we have the same eigenvalues, but repeated
three times. When we compute the eigenspace
here meaning we have to compute the eigenvector
so with this equation a minus lambda ix is
equal to 0. So, what will happen in this case?
That when we take this minus lambda I so,
minus 1 from the diagonal entries.
So, this diagonals will be also 0 and we have
this very a simple example and in this case,
there will be only 1 pivot. So, this first
column will have pivot and the second and
the third one will be the free variables.
So, this is going to be the pivot element
and then nothing else. So, we have the free
variable we have the free variable. So, there
are two free variables, meaning 2 linearly
independent eigenvectors.
So, here for x 2 be sine alpha 1, x 3 we sine
alpha 2 and from this equation which says
x 1 plus x 3 is equal to 0; means, x 1 is
equal to minus x 3. So, we get here x 1 is
equal to minus alpha 2. So, when writing in
this vector form we have x 1 x 2 x 3 as alpha
1. So, this component here is 0; so 0 and
then the second place will be 1 and then 0
for alpha 2 at 1 place you have minus 1 and
then 0 and x 3 is alpha 2 here so 1.
So, we have this x 1 x 2 x 3 as alpha 1 times
is 0 1 0, alpha 2 times minus 1 0 and 1. So,
there are 2 linearly independent eigenvectors
corresponding to this eigenvalue 3; so, eigenvalue
1 here, which was repeated 3 times. So, the
algebraic multiplicity of this lambda is equal
to 1 was 3 and the geometric multiplicity
of this lambda is equal to 1 is 1 oh sorry
2.
So, they are 2 linearly independent vector.
So, the dimension of this eigen space is 2
or the geometric multiplicity of this lambda
is equal to 1 is 2, in this case. So, again
though it was repeated 3 times, but we got
only this dimension as 2 not 3 and not 1,
but the possible values here could be 3, it
could be 2 as this is the case here, but it
can be 1 as well.
In this example again we will take this identity
matrix. So, very simple to evaluate so we
have, we want to find the dimension again
of the eigenspace of this A is equal to this
identity matrix. And if we write down its
characteristic equation, we will get this
lambda minus 1 power 3 is equal to 0; so,
again we have this 1 1 1 which the algebraic
multiplicity of this eigenvalue is just 3
now.
So, corresponding to this 1 so; algebraic
multiplicity is 3 and if we compute the geometric
multiplicity now that is interesting. So,
the eigenspace will be computed by this A
minus lambda I, x is equal to 0 and therefore,
when we subtract from the diagonal entries
this eigenvalue 1. So, what we will get this
0 matrix here and x 1 x 2 x 3is equal to again
the 0 matrix.
So, what do we see now in this case? That
there is no there is no pivot here; there
is no pivot here and all the variables x 1
x 2 x 3 they are the free variables. So, we
can choose, we can assign any value to x 1
x 2 x 3 they are free here and that is a very
special case which we have just seen now,
that we got the 0 matrix here as A minus lambda
I and then we have the possibility of choosing
this x 1 x 2 x 3 freely. So, whatever we like
and we have taken alpha 1 here alpha 2 and
there alpha 3 because all three are free variables.
And then this x 1 x 2 x 3 we can write down
in terms of these alpha 1, alpha 2, alpha
3 as this combination alpha 1 this 1 0 0,
alpha 2 0 1 0 and alpha 3 will be 0 0 1. So,
we have three linearly independent eigenvector
in this case corresponding to this lambda
is equal to 1. So, the algebraic multiplicity
of lambda 1 was 3 and also the geometric multiplicity
which is the dimension of the eigenspace that
is also 3 in this case. So, we have seen in
this example that, if it is repeated 3 times
it is also possible that we can get the full
dimension, here the dimension of the eigenspace
that is 3.
As many times as the lambda was repeated,
but what that result says that it cannot be
more than 3. And naturally, that is the case
here because the dimension that is the full
dimension because the elements belongs to
this R 3 and we cannot have the dimension
more than 3 in that sense also we can conclude
here.
There is a concept here the similarity of
matrices which will introduce here and we
will continue for the discussion in the next
lecture. So, an n cross n matrix B is called
similar to an n cross n matrix A, if we have
this B is equal to P inverse AP. If we have
this relation between the between the matrix
A and B, then we call this P is similar to
the matrix A or A is similar to the matrix
B. And what to we have to we this P what is
the P before some non singular matrix P, if
there exists a matrix here this P inverse
I mean, this non singular matrix P therefore,
that P inverse make sense.
So, if we have this relation between the 2
matrices here B and A that P inverse AP gives
the B the other matrix, then we call that
these two are similar. Why do we use the similar
words some of the properties we will check
today itself, that they share away many common
properties this B and A in terms of the eigenvalues,
eigenvectors and there are other considerations
as well which we will continue in the next
lecture.
So, today we will see that if B is similar
to A, then the B has the same eigenvalues
as A and if x is an eigenvector of A. Then
this y is equal to P inverse x is the eigenvector
of B corresponding to the same eigenvalue.
So, meaning if we know the eigenvalues and
eigenvector of one, we can get the eigenvalues,
eigenvectors of the other. In fact, they same
they have the same eigenvalues and the eigenvector
also will be just the P inverse x where P
we have introduced already in the similar
t definition. So, what we take that let us
say this lambda is the eigenvalue of this
matrix A and the similar to A, we have the
B matrix. So, first of a relation we have
for this A that x is the eigenvector and lambda
is the eigenvalue.
So, we have this relation Ax is equal to this
lambda x. And what we do now? We multiply
by this P inverse here. So, the right hand
side we have P inverse, P is that matrix which
we are talking about the similarity there.
So, we have P inverse e Ax and here also P
inverse. So, the lambda is constant so, we
have P inverse x there. And then what we do,
we have here the lambda P inverse x the same,
the P inverse A again we have introduced this
identity matrix.
So, here we have introduced identity matrix
which we have written as P and P inverse x
and then what we do, this we combine here
P inverse AP; P inverse AP and then we have
P inverse x. So, what do we see now, this
P inverse AP as per the definition of the
similarity that A similar to the or B is similar
to A; that means, this B we can write as P
inverse AP. So, this we have this lambda P
inverse x is equal to B times, this is B P
inverse x. So, what we observed now from this
relation, that this is the eigenvector P inverse
x and this lambda is the eigenvalue of this
B.
So, if this B is similar to A the B will have
the same eigenvalue as A because this lambda
was the eigenvalue of A and the eigenvector
will be this P inverse x. So, we can get the
eigenvector and eigenvalue of the similar
matrices if we know for one. So, lambda is
an eigenvalue of B and P inverse is the eigenvector
corresponding to the eigenvalue lambda.
Another result which actually we have seen
already in this first result A B are the square
similar matrices, then they have the same
characteristic polynomial. So, eventually
we have seen already that they have the same
eigenvalue. So, if we have the same eigenvalues
meaning they have the same characteristic
polynomial, but this is just another way of
looking at it.
So, we take this B is P inverse AP this relation
and then if we get this determinant B minus
lambda, if that is the characteristic polynomial
B minus lambda I the characteristic polynomial
of this B here is equal to the determinant
this B, we will replace by this P inverse
AP minus the same thing this P inverse P that
is the identity matrix we have introduced
here. I mean, you can see easily that, this
is nothing but the lambda I because lambda
we can take common, then we have P inverse
I P and then P inverse I is equal to nothing
but the P inverse P lambda times and this
is I. So, lambda times I so, we have again
here this is nothing but the lambda times
I only, but we have rewritten in this form
that P inverse lambda I and P.
So, here the determinant we have P inverse
let us take common from both and then we have
A here and minus this lambda I and P from
this right hand side, we can take as common.
So now, this A minus lambda I and then, this
we can use the property of this determinant
here the product of these three matrices;
that means, the determinant of P inverse determinant
of this middle one A minus lambda I and the
determinant of P.
So, here the determinant of P inverse and
determinant of P will cancel out each other,
we will get just one here and what we will
get that is the property of this determinant
P and P inverse they are just the reciprocal
and here we have determinant of A minus lambda
I.
So, what we have seen that the determinant
of this B minus lambda I is equal to determinant
of this A minus lambda I so, they have the
same characteristic polynomial. In other words,
we can say again that this A and B will have
the same eigenvalues and we have seen again
in the previous slide here the relation for
the eigenvectors as well ok.
Coming to the conclusion so, in this lecture
we have talked about the algebraic multiplicity
and that was nothing but the number of occurrence
of an eigenvalue. And we have also seen the
geometric multiplicity that was the number
of linearly independent eigenvectors associated
with that eigenvalue. And always this is the
case that geometric multiplicity is less than
equal to the algebraic multiplicity and we
have also talked about the similar matrices;
that means, B and A are called the similar
to each other they are the similar matrices,
if we have this relation that B is equal to
P inverse AP for some invertible matrix P.
So, these are the references used to prepare
these lectures and thank you for your attention.
