Ok. ah We are talking about eigenvector and
eigenvalues and some properties of the eigenvalues.
Today, we will discuss the more on the eigen
some properties of eigenvalues, and then we
will go for the ah define the diagonalizable
of a matrix. So, let us discuss another properties
of eigenvalue .
So, let A be an n cross n real symmetric matrix
symmetric matrix. Then ah suppose, suppose
lambda 1 and lambda 2, which are not equal,
lambda 1 and lambda 2 are two eigenvalues
of A, which are distinct ah to which are district
that means, lambda 1 is not equal to lambda
2. Consider two eigenvalues lambda 1, lambda
2, ah which are not district. Then we want
to know their ah corresponding eigenvector,
their orthogonal, so that is the theorem.
So, then ah then the corresponding then the
corresponding eigenvectors are orthogonal,
that means suppose ah X 1 is the eigenvector
corresponding to lambda 1 and X 2 is the eigenvector
corresponding to lambda 2, then this theorem
is telling that X 1 transpose X 2 is equal
to 0 . So, this is the this the this is the
property that two vectors will be orthogonal
ok.
We have to prove this, we are going to prove
this. This is the theorem . This is the theorem
or result, and we are going to prove this
theorem. So, how to prove this, so we will
just ah use the property of ah will just do
use the definition of eigenvector and eigenvalue,
and then we will take the transpose, and then
the this will this will be ok.
So, let us start with so, since X 1 is a eigenvector
of ah corresponding to lambda 1 . So, AX 1
equal to lambda 1 X 1, this is equation 1.
And X 2 is the eigenvector corresponding to
lambda 2 . So, AX 2 equal to lambda 2 X 2,
so this is equation 2 ok. So, we have these
two equation. So, now from 1, we we are going
to take the transpose on this . So, AX 1 transpose
equal to lambda 1 X 1 transpose, so from 1
.
And then this will give us this implies X
1 transpose A transpose is equal to lambda
1 transpose lambda 1 transpose, where lambda
1 is a ah just a scalar, so X 1 transpose
. And a A is a symmetric matrix real symmetric
matrix, so A transpose is same as A. So, this
implies X 1 transpose A is equal to lambda
1 X 1 transpose sorry X 1 transpose ok. So,
now we want to ah bring X 2 over here, so
AX 2 is equal to lambda 1 ah X 1 transpose
X 2 .
So, now ah we can take this, this is associativity
property of the matrix multiplications lambda
1 X 1 transpose X 2 . So, this implies now
this we can use the 2, so this will give us
lambda 2 X 2 ok. So, lambda 2 we can take
outside, so lambda 2 X 1 transpose X 2 is
equal to lambda 1 X 1 transpose X 1 transpose
X 2 . So, now we can bring this, this side,
so lambda 2 minus lambda 1 X 1 transpose X
2 equal to 0, but lambda 1 is not equal to
lambda 2. So, this is non-zero, so these has
to be 0 . So, this implies X 1 transpose X
2 is equal to 0, so that means, they are ah
orthogonal. So, X 1 and X 2 are orthogonal
.
So, if we have a ah real symmetric matrix,
then we know ah then we know that the eigenvalues
are all real. Then if you take two distinct
eigenvalues, then the corresponding eigenvectors
are orthogonal. So, this is the this theorem
is telling that. Now, I have more result on
the eigenvalue say if we have a orthogonal
matrix ah real orthogonal matrix, then if
lambda is eigenvalue, then we will show that
1 by lambda will be also a eigenvalue of A,
so that is another theorem . ah
So, A is an the square matrix A is an real
orthogonal matrix real orthogonal matrix,
so ortho[gonal] so that means, real means
the all elements are coming from real field,
so here our field is the real number. An orthogonal
means a A transpose equal to A transpose A
is equal to identity element . So, this is
the orthogonal property of a matrix. Now,
this is given. Now, since it is orthogonal,
so that means, it is a non-singular matrix,
this implies A is non-singular that means,
determinant of A is not equal to 0 . Now,
once det A is not 0, then ah then we know
that all eigenvalues are non-zero.
So, so if we take a eigenvalue, let ah lambda
be an eigenvalue of A, then lambda is non-zero,
lambda is not equal to 0. Because, if lambda
is becoming 0, then we know that product of
the eigenvalue is ah det A ah is determinant
of A of the form, so then the determinant
of A must be 0. So, this result we have seen.
So, then for any eigenvalue is not equal to
0. So, once it is non-zero, then we can think
for ah 1 by that means, 1 by lambda exist
in the real field ok. And we are going to
show these ah lambda is a eigenvalue of this,
then 1 by lambda is also an eigenvalue of
A, so that is this theorem is telling all
about. So, we have to prove this .
So, how to prove this? So, let us just so,
a a A is a real orthogonal matrix, and lambda
is an eigenvalue eigenvalue of A, then we
want to show that 1 by lambda is also an eigenvalue
of A. So, how to show this. So, since lambda
is an eigenvalue that means, the characteristic
equations will satisfy the ah root ah one
of the root of the characteristic equation
is lambda, so that means, lambda must satisfy
this det of A minus lambda I n is equal to
0 .
Now, A is orthogonal, we can use this property
over here. So, this implies det of A minus
lambda of A transpose A equal to 0 ok. Now,
we can take the determinant of det of A, we
can take common, so this is I minus lambda
of ah lambda of A transpose is equal to 0.
Now, this we can take the determinant of A
into determinant of I n lambda of A transpose
equal to 0 ok.
Now, we want to take this this side, so this
minus 1 to the power n will come, and we want
to take 1 by lambda . So, this will give us
det of A into 1 by lambda to the power n minus
1 to the power n det of ah A transpose ah
A transpose minus 1 by lambda I, this is to
be 0 . Now, this implies these are all non-zero,
so this implies det of A transpose minus 1
by lambda I n this is to be 0 .
Now, we can take the transpose of this that
is also 0. So, transpose of this is also 0
. So, this will give us so, this means this
implies det of A minus if we take the transpose
on that 1 by this is to be 0 ok. So, this
implies so, this is this imply 1 by lambda
is satisfying the characteristics equation,
this implies 1 by lambda is an eigenvalue
of A. So, this is the proof .
If lambda is an eigenvalue of A for a orthogonal
real orthogonal matrix, then 1 by lambda is
also an eigenvalue of A ok. so this is these
are some more ah properties on this, we can
have more properties in the lecture note.
So, now we will move to the ah ah definition
of a diagonalizable ah of a matrix, when we
say matrix is diagonalizable. So, for that,
we need to define the the concept of singular
matrix . So, when we say two square matrix
of same size n cross n are similar, so that
is the definition.
So, similar, so we say two matrix, this is
n cross n is similar to B matrix, this is
also n cross n. We have two square matrix,
this symbol we can use for similar, is not
kind it is not necessarily ah row reduced
or column reduce echelon form, but similar
has a definition . So, A is ah a A is the
matrix A is similar to a matrix B, which is
of same size over the same field that is important
over the same field. If there exist a non-singular
matrix over the same field, such that such
that B is of B is equal to P inverse A P . This
is the definition, B is equal to A P inverse
A P, P is a non-singular matrix, so p inverse
will exist. So, then we say A is similar to
B . This is the definition of ah this relation
this is a relation between the mat[rix] square
mat[rix].
If we have ah if we have set of if if we considered
the set of all possible square matrix of same
size n cross n, this is the matrix set of
matrix of size m cross n, then we take any
two matrix from this set, and we say they
are related with this relation similar if
and only if ah yeah if and only, one can obtained
from other by just doing this operation P
inverse A p ok. Now, if A similar to B, then
B is also similar to ah A that is this relation
is ah an and also A is similar to A, because
ah in that case P is the identity matrix.
So, this is a this relation is a ah reflexive,
because A is similar to A, and this relation
is a symmetric.
Because, if ah because if if A is similar
to B, so reflex, this is symmetric. The reflexive
is is the reflexive means A is similar to
A, because we can write A to B identity matrix
inverse A identity matrix, and this is our
P ok. Now, ah symmetric, symmetric means 
symmetric means if A is similar to B, then
we need to show B is similar to A. So, A is
similar to B implies B should be written as
P inverse A P . And from here we can write,
we can ah write this as we can P B P inverse,
so if you take this is to be A .
Now, if you take ah then A this implies A
is written as ah Q inverse B Q, where Q is
P inverse; P is non-singular, so Q is also
non-singular. So, this implies B is similar
to A . So, A is if A is similar to B, then
B is similar to A. This is the symmetric property
of the of this relation . Now, we will ah
ah talk about their eigenvalue. Suppose, two
matrices they are similar, then what about
their eigenvalues, we have seen this result
.
So, suppose we have two matrix A and B, which
is similar, so the that means, either one
of them can be written as from P inverse A
P . Now, we have seen they have the same characteristics
polynomial characteristics equation, so that
will give us the same eigenvalue, so they
have same eigenvalue. So, how to check that.
So, characteristics equation of A is nothing
but determinant of ah A minus x I n equal
to 0 or if if lambda is eigenvalue of A, then
we have determinant of A minus lambda I n
equal to 0 . Then we have to show the lambda
is also an eigenvalue of ah B.
So, for that, we can just take this ah consider
determinant of B minus lambda I n . So, if
we can show this determinant is 0, then lambda
is also an eigenvalue of B. So, this determinant
we can just write det of B is nothing but
P inverse A P, and this lambda I n we can
write P inverse P . So, this is nothing but
det of P inverse into det of A minus lambda
of I n into det of P. And the det of P, P
inverse will get ah give us the one, so this
is nothing but det of A minus lambda of I
n. So, lambda as so, since this is 0, this
is also 0, in fact we have the same characteristic
equations.
This result we have seen earlier, so the that
means, if lambda is the eigenvalue of A, then
lambda will be an eigenvalue of B. So, they
have same set of same set of eigenvalues A
and B, if they are similar ok. But, the converse
is not true that means, if the we have two
matrix, which having same set of eigenvalues,
they may not be similar ok. We have to ah
we have to justify that by taking an example
converse is not true.
So, so result this result is ah if A, B are
similar, this implies A, B have same eigenvalues,
this just now we have seen, but the converse
is not true converse is not true, so that
means, if we have a two matrix, so we have
to take a counterexample. We have to take
two matrix, which are having same eigenvalue
that means, if you have two matrix having
same eigenvalue eigenvalues, it does not mean
that this does not imply that they are similar,
this does not imply that they are similar.
So, how to prove this to I mean this is a
counter example, we have to sho[w] we have
to get take a counter example on this . So,
let us have two matrix A, B, where the eigenvalues
are same, but they are not similar .
So, I just take two one example on . So, suppose
A is ah identity matrix 1 0 0 1, and B is
1 2 0 1 ok. So, what is the characteris[tic]
how to get 
the eigenvalue of this I have to get the eigenvalue
of this, we have to take the characteristic
equations, which is determinant of ah 1 minus
x 0 0 1 minus x this is to be 0 . So, this
gives us ah 1 minus x square is equal to 0,
so that means, ah x is x is 1 . So, x is 1
with multiplicity two algebraic multiplicity.
So, we have a eigenvalue 1 with now, now how
to get the eigenvalue of B, so to get the
eigenvalue of B, we need to find the ah again
we need to find the characteristics polynomial
of B, so lambda B x, which is 1 minus lambda
2 0 1 minus sorry 1 x. So, this is the equation.
So, this will give us the same equation 1
minus x square square equal to 1 . So, they
have the same characteristic equations, and
they have the same eigenvalues. Now, the question
is is the A, B are similar. So, to be similar,
we need to have a ah square matrix like ah
yeah we need to have a non-singular matrix
P, such that B should be written as some A
ah P transpose A B.
So, we need to have a so, if A b are similar,
this implies then there has to exist a square
mat[rix] non-singular matrix P, such that
one of them should be written as ah either
B or A, one of them should be written as P
inverse A P, where P is a non-singular matrix.
Now, we have to show that we have to check
whether such a matrix P non-singular matrix
exist. Now, what is this quantity, so the
that means, B should be written as P inverse
A P . Now, A is identity and P is non-singular,
so this is nothing but P inverse identity
P, so this is nothing but identity again,
so this is 1 0 0 1. But, B is not 1 0 0 1,
B is 1 2 2 0, so the B this is not, so the
that means, A is not similar to B. So, this
is the ah this is the one example, where two
matrix is having same eigenvalue, but they
are not similar ok. Now, we will define the
ah diagonalizable of a matrix .
So, ah we say 
diagonalizable . So, suppose we have a square
matrix A of size any size n by n, now we say
this is diagonalizable, if it is similar to
a diagonal matrix that is the definition . So,
A is A is said to be diagonalizable, if and
only if ah if and only if ah A is similar
to a diagonal matrix D, which is a diagonal
matrix like we have some d 1, d 2 diagonal
elements are non-zero, others elements are
0 . So, then we say A is diagonalizable. So,
if it is similar to a diagonal matrix, then
we call A is diagonalizable. This is the definition
ok. So, similar we know ok. So, we will we
will just ah continue this in the next class.
Thank you.
