Ok, we are finding the geometric and algebraic
multiplicity of a matrix. So, we have seen
the matrix is ah ah 1 1 1 minus 1 minus 1
minus 1 0 0 ah 0 0 1. And we have seen the
roots of this matrix, we we got the ah characteristics
equation like this. And if we equate the roots,
we are getting the two distinct eigenvalues.
So, this 0 is coming twice, so that is why
ah 0 is ah eigenvalue 0 is having a algebra
multiplicity 2. And eigenvalue 1 is having
algebra multiplicity 1. Now, we will talk
about geometric multiplicity. So, for that
we need to find the set of eigenvectors corresponding
to this eigenvalue.
So, let us try with the first one with the
eigenvalue 0. And I need to find the eigenvectors
the set corresponding to the eigenvalue 0.
So, eigenvectors for the eigenvalue lambda
1 equal to 0. So, how to get eigenvectors
corresponding to the eigenvalue 0? So, for
that we need to get the all the non-zeros
so this is X all the non-zero vector X such
that this is true. Now, lambda is 0 so that
means ah this is 0. So, we have to find the
all the non-zero vector X for which X equal
to 0.
So, what is A, A is nothing but 1 1 1, ah
then the minus 1 minus 1 0 minus 1 and 0 0
1, so these into x 1, x 2, x 3, this is the
0 vectors. So, if you simplify this, we will
be getting ah like x the equation likes x
1 plus x 2 plus x 3 n equal to 0, then minus
x 1 minus x 2 minus x 3 equal to 0, and then
x 3 equal to 0. So, this will give us x 1
plus x 2 equal to 0, and x 1 ah yeah if you
put this to be 0, so we have x 1 plus we have
basically two equations, and x 1 plus x 2
equal to 0. So, you have option for x 3, so
if you choose say x 3 is 0, we have no option
for x 3 sorry.
So, now what are the values, now we can choose
x 1 to be K or x 2 to be K, so this will give
us x 1 equal to minus x 2, which is say minus
K, which is x 2 be K, and x 3 is 0. So, these
are the this is the set. Now, this will corresponding
to the vector like ah ah so x 1 x 2 x 3 if
you the write it in a column way, this is
nothing but if you take the K common , so
minus 1 1 0. So, this is the set of all possible
eigenvectors corresponding to the eigenvalue
0.
So, now this is ah this set along with the
0. So, this set if K could be 0, all these
are all real numbers. So, this is the real
set. So, the you know this is a vector space,
and dimension of this space is called geometric
multiplicity of this. So, this is generating
by the all vector, so dimension of this is
1. So, the geometric multiplicity of the eigenvalue
0 is 1. Geometric multiplicity of the eigenvalue
0 is equal to 1. Whereas, algebraic multiplicity
of 0 is 2, because 0 is coming twice in the
root of the characteristics equations.
Now, we can similarly we can find the ah which
is ah this which is 1, and which is less than
2, which is the algebraic multiplicity of
the this. Now, we try to find the ah geometric
multiplicity of the eigen ah value lambda
2, which is nothing but 1. So, to get this
we need to ah solve the homogene yeah will
need to solve the equation A X equal to lambda
1 lambda 2 X lambda 2 X. So, this will give
us ah lambda, lambda 2 is 1 we can just write
1 into X. So, if X is X is x 1, x 2, x 3,
so this will give us ah 1 minus 1 1 1, then
minus 1 minus 1 minus 1 minus 1, then 0 0
1 minus 1, x 1 x 2 x 3.
This is the corresponding ah A minus lambda
X A minus lambda i 0 0 0. So, this will give
us the solution like, so you this is 0, this
is 0. So, this will give us solution like
x 2 plus x 3 equal to 0, and so this means
x 2 equal to minus x 3. And if you solve it,
we will be get a x 1 to be 2 x 2 minus x 3.
So, if you just choose x 3, ah x 3 equal to
be K, then x 2 will become minus K, and x
1 is becoming 2 of minus ah yeah so x this
is minus. So, 2 K minus K, so this is also
K.
We have this x 1, x 2, x 3 is nothing but
of this form. So, x 1 x 2 x 3 we take K common
which is ah 1 minus 1 1. So, this is generating
the so this is the corresponding ah this is
the all possible eigenvector. So, every eigenvector
is of this form. So, V of lambda 2 is nothing
but so the dimension so this this this vector
is generating the group, ah generating the
field.
So, the dimension of lambda 2 is equal to
1, which is same as this is the geometric
multiplicity. So, geometric multipli[city]
which is the geometric multiplicity of of
lambda 2, which is same as algebraic multiplicity
here. Geometric multiplicity is less than
equal to algebraic less than equal to, so
here the equalities occurring here ok. So,
these are the geometric and algebraic multiplicity
of ah geometric multiplicity of a eigenvalue
ok.
So, now we will talk about some properties
of eigenvalues. Say, for example, if we have
a real symmetric matrix ok, we if we have
a real symmetric matrix, then the eigenvalues
are all real. And if we have a skew symmetric
matrix, we have we want to see eigenvalues
are either 0 or they are completely imaginary.
So, let us try first one. So, the theorem
is telling the eigenvalue of a real symmetric
matrix. We know the form symmetric matrix,
symmetric transpose of ah A is A are all real.
So, eigenvalue of all real eigenvalue of a
real symmetric matrix.
Suppose, you have a real symmetric matrix
a i j n cross n square matrix, where a i j's
are real, so ah that means, if you take the
conjugate of this, and that is same as this,
this is conjugate. And it is symmetric means
transpose if you take the transpose of this,
this is same as A. This is an example of symmetric
matrix. If we have 1 say minus 2 3, we have
diagonal we have nothing to do, but if the
value over here is same as value over here.
So, if you have a say 4 over here, this has
to be 4. So, value over here is same as value
over here. If it is minus 5, you have minus
5 over here. If you have a value over here
say 9, this has to be 9, so that means, a
i j equal to a j i. So, this is the for all
i and j, then the matrix is called symmetric
matrix.
So, now this theorem is telling if you have
a symmetric matrix, and if the all coefficients
are ah real that means, if the conjugate of
a is same as this. So, conjugate means, if
you have a complex number, a plus i b ah a
plus i b, then the conjugate of this is defined
as a minus i b. But, if b is 0, then it is
called real real ah purely real, then a ah
conjugate is same as this. And if a is 0,
it is called purely imaginary, so that is
the thing. So, now we have to so conjugate
matrix means, we take the conjugate matrix
we defined as this where we take the conjugate
of each of these numbers.
So, you are going to prove this that real
symmetric matrix in ah real eigenvalues only.
So, how to prove this? So, for this we taken,
ah so suppose lambda is eigenvalue of the
matrix A so that means, A x there is a non-zero
vector X such that A X equal to lambda X or
we can say X 1. Some non-zero vector X 1 not
equal to 0 vector. So, what we do, we take
the transpose of this now before transpose
yeah, we take the conjugate. So, conjugate
must be same, and then we take the transpose.
So, if we take the conjugate, then it is nothing
but 
A conjugate X 1 conjugate transpose equal
to ok lambda conjugate X 1 conjugate transpose,
then there is a scalar quantity ok.
So, now this will give us so transpose will
ah reverse this, so this is x 1 conjugate
transpose transpose, and then lambda bar x
1 conjugate transpose. So, we have to simplify
further ok. So, ah so now this ah this we
can write as this ah X 1 conjugate A X 1 equal
to lambda bar, we just ah taking a both X
1 both side, lambda bar X 1 transpose conjugate
X 1. So, if we further simplify this, we take
this transpose we have to be carefully this
transpose and T ah bar and transpose. So,
this is A X 1, because which is associativity
property is there bar X 1 bar transpose X
1 ok. So, let us just ah so up to this is
fine.
Now, this is again lambda of X 1, so the lambda
this is again ah lambda X 1 bar transpose
X 1 equal to lambda bar X 1 bar transpose
X 1. Now, this implies lambda equal to lambda
bar, because if you take this this side, lambda
minus lambda bar X 1 transpose X 1 equal to
0. Now, this is X 1 is non-zero, so that that
means, X 1 bar transpose X 1 is not equal
to 0. So, this means this has to be 0, lambda
is equal to lambda bar. So, this means lambda
is ah this means lambda is real purely real,
no complex part, no imaginary part in it.
So, this is the ah this is one of the theorem.
If the eigenvalues are if the matrix is symmetric
real symmetric matrix, then the corresponding
i all the eigenvalues are real.
Now, this is true for Hamilton matrix also.
So, we will define the Hamilton matrix sorry
Hermitian matrix. So, when we say matrix is
Hermitian matrix, if ah A bar transpose transpose
equal to A. This is the definition of Hermitian
matrix, so then this is called ah then then
the matrix is called Hermitian matrix. Now,
the way we proving the last just now same
technique will go. And we can say the all
the eigenvalues are all the eigenvalues of
Hermitian matrix are real all eigenvalues
of Hermitian matrix are real ok.
So, now we talk about ah real skew symmetric
matrix. And we will see how their eigenvalue
are A is n by n matrix skew symmetric matrix
of the coefficient of real, it is real skew
symmetric matrix. Skew symmetric matrix means
I will I will define that symmetric matrix.
Then the eigenvalues, eigenvalues are either
purely imaginary or 0 and the eigenvalues
of a are purely imaginary means there is no
real part. So, like 2 plus 3i it is not purely
imaginary. So, this part should not be there,
so that is the purely imaginary.
So, eigenvalues are purely imaginary or 0
ok. So, ah so this is how to so what is skew
symmetric matrix, so skew symmetric matrix
is A transpose equal to minus A that is the
skew symmetric matrix definition that means,
if A is a i j n cross n, then a i j must be
equal to minus a j i. And this true for all
i and j starting from 1 to n, so that is why
if the diagonal element, where i is j is equal
to i, diagonal event a i i equal to minus
a i i, so that means, 2 a i equal to 0. So,
a i equal to 0. So, for any skew symmetric
matrix the diagonal elements are 0. And off
diagonal element ah if we have a so all the
diagonals elements will be 0's over here,
and off diagonal element so the this is 2
this has to be minus 2. If this is minus 4,
this has to be plus 4, so that is the definition
of skew symmetric matrix.
And it is a real skew symmetric matrix the
that means, ah this a i j's a real number.
So, a i j's conjugate if you take it is a
i j so that means, ah ah this conjugate of
so conjugate of sorry conjugate matrix is
same as this. So, this is the definition of
real skew symmetric matrix. So, if we have
a real skew symmetric matrix minus A, yes
if we have a real skew symmetric matrix, then
we can show that ah the eigenvalues are either
0 or 1 ah sorry eigenvalues are either imaginarily
or pure. So, similar way we can argue that.
Now, we will talk about orthogonal matrix.
So, this prove we are not doing is this will
be in the note lecture note. So, now we talk
about eigenvalues corresponding to the to
a orthogonal matrix. So, ah so suppose A is
orthogonal matrix orthogonal matrix. So, orthogonal
matrix means A to A transpose equal to identity,
so that means A A transpose is an inverse
of A. So, if we have a real orthogonal matrix,
then the theorem is telling the eigenvalues
values of A ah has unit modulus. Unit modulus
means mode of that eigenvalue that means,
if lambda is the eigenvalue, it is a complex
number maybe mod of this is 1 that is the
unit modulus had units mod modulus ok.
So, this we have to proof. So, how to proof
this to prove this suppose ah yeah suppose
lambda 1 is eigenvalue. So, and corresponding
eigenvector is A X 1. So, we have this while
lambda, where X 1 is not equal to 0 and lambda
1 this. So, what we do we take the transpose
of this, and it is ah which is the conjugate,
then we will take the transpose of this. Then
this is nothing but A bar transpose ah A bar
X 1 bar transpose, then this similar thing
we did in the last proof. So, it is is this
this is orthogonal matrix real orthogonal
matrix. So, we have to use that property also
here to achieve that ok.
So, A bar transpose so if you take the transpose,
it will give us X bar transpose and A bar.
A bar is ah this is a real matrix so this
A, so this is nothing but A transpose equal
to so X 1 is there, so yeah X 1 is already
there equal to ah lambda bar X 1 bar transpose
ok. Now, we multiply both side by A X 1. So,
it is mode of it so we multiply both side
by A X 1 transpose this A X 1 equal to lambda
bar X 1 transpose A X 1 ok.
So, now we take this together A transpose
A X 1 equal to ah this is A X 1 is nothing
but lambda X 1. So, transpose lambda X 1,
so this lambda X 1 what we can do yeah. So,
this is identity matrix, because this is the
property of orthogonal matrix. So, this will
give us X 1 transpose X 1, this identity matrix.
And this will give us ah lambda lambda 1 we
take this side, so ah lambda sorry lambda
1 these are lambda or lambda 1 any way lambda
is lambda. So, so lambda bar lambda ah X 1
bar transpose X 1 ok.
So, this will give us X 1 bar X 1 into 1 minus
lambda bar lambda equal to 0. Now, this is
non-zero quantity. So, this implies lambda
lambda bar equal to 1. Now, this this implies
mod of lambda equal to 1. So, if lambda is
the complex number, so modulus is 1. So, it
has unit modulus ok, so this has unit modulus.
So, we will talk about more on this. ah In
the next class, we will define ah diagonalization
of the matrix. So, those we will will will
will we will discuss in the next class.
Thank you .
