Welcome viewers, this lecture is in continuation
to my earlier lecture on Eigen Values
and Eigen Vectors.
.
In this lectures, we will be discussing Eigen
values and Eigen vectors, of some special
matrices. And we will be discussing similar
matrices. I assume that viewers have all
ready gone through my 1st lecture on Eigen
values and Eigen vectors and characteristic
equation.
..
We start with Eigen values of some special
matrices. We have all ready been discussing
matrices, which are real matrices. That means,
the elements of the matrices are real
numbers, however in many situations the elements
maybe complex numbers. And we say
the matrices are complex matrices. Let us
consider a vector V in R n, consisting of
components X 1 X 2 X n. So, V is equal to
X 1 X 2 X n, it is transpose is vector in
R n.
Then, the length of the vector V is defines
as X 1 square plus X 2 square plus X n square.
Now, length as we all know is a positive number.
And it is 0, only when the individual
components X 1 X 2 X n are 0. So, you can
verify this when X 1 X 2 X n are 0, then the
length of V is 0. And, we defined length of
V as V transpose V, which is the vector X
1
X 2 X n. The row vector multiplied by the
column vector X 1 X 2 X n transpose.
This is also denoted by inner product of X
and X. If we extend this concept to complex
numbers and let us apply to complex vector
1 comma i transpose. Then, one can notice
that its length will be 0. So, this vector
is not 0 vector, but its length is 0. If we
apply this
definition of length which we have already
developed for real matrices. So, we need to
redefine this length in the relation to complex
matrices. So, let us redefine it, that length
V square is equal to V conjugate transpose
times V.
V conjugate transpose means, we first take
the conjugate of the numbers. These numbers
are complex numbers. So, conjugate means the
complex conjugates, and then multiplied
by the vector X 1 X 2 X n transpose. And if
you perform this multiplication, then this
.comes out to be X 1 bar X 1 plus X 2 bar
X 2 and X n bar X n. And we define it as the
inner product of X n and X. And if you have
used this definition, this revised definition
for length. Then, 1 plus i transpose square
comes out to be 1 minus i. Because, a minus
i
is a conjugate of i multiplied by 1 i it is
transpose.
And, if you multiply it, it is 1 and minus
i square which is also 1, so the length is
2. So,
length square is equal to 2. And this is not
0, when this vector is not 0. So, we know
that
this number is now a real number. And this
has to be because length is a positive real
number. So, we extend the definition of length
in this particular manner.
.
And accordingly, the inner product of two
vectors x and y is defined as x bar transpose
y.
One may notice that, x bar transpose y is
not the same as y bar transpose x. This is
in
contrast to what we have in real numbers.
When length from x to y is the same as from
y
2 x. Similarly c x, the length of c x is equal
to c bar x, it is not c it is c bar. So, if
we have
a complex number, we have a complex that then
c of x is c bar x. Of course, when we
apply to real number and c happens to be real
number, then c bar and c are equal.
Then, we write down, we combine the conjugate
and transpose. And we write it as star
denoting the conjugate transpose. And accordingly,
we can say that inner product x, y is
equal to x star y. So, with this notation,
we have introduced x star y as the inner product
of x and y. The vectors x and y are orthogonal,
when x star y is equal to 0. So, this is true
for this is the definition we apply for complex
vectors x and y.
.Now, let us illustrate these concepts with
the help of an example. So, find length and
inner product of the vector x, which is 1
comma 1 plus i transpose. And y is equal to
1
minus i comma 2 transpose. So, length of vector
x will be equal to 1 comma 1 plus i
square 1 comma 1 plus i, it is length will
be equal to under root of this product. And
if
you work it out, it is 1 into 1 plus1 minus
i into 1 plus i, that gives me under root
3.
.
While the length of the second vector 1 plus
i 2 is equal to 1 minus i 2. That is the
conjugate of this number multiplied by 1 plus
i 2. And then it is transpose we take the
multiplication of these two matrices. In fact,
these two vectors and this comes out to be
under root 6. And if we have to calculate
the inner product of x and y, it is defined
as x
bar transpose y.
Then, x bar transpose is 1 comma 1 minus i.
And y is 1 plus i comma 2, that is given to
us. And if you multiply it, it comes out to
be 1 into 1 plus i plus 1 minus i into and
that
simplifies to 3 minus i. Now, this is about
vector. Now, we go to a special matrix is
a
Hermitian matrix, it happens to be a complex
matrix and in fact a square matrix. So, we
say a complex square matrix A is Hermitian.
If A is equal to it is transposed conjugate.
That means, A is equal to transposed conjugate.
So, we have first take the conjugate and then
transpose. In fact, it does not make much
difference. Whether we first take the transpose,
or we take the conjugates. And we
denote it by A star. So, if A is equal to
A star, then the matrix which is of complex
.elements is a Hermitian matrix. Now, if we
have a square matrix A, then a typical
element in this is a i j. Then according to
this definition A is Hermitian. If a i j is
equal to
a bar j i, this bar denotes the conjugate.
And this index denotes that it is being transposed.
So, a i j is equal to a j i bar, if this
condition is satisfied, then the matrix is
Hermitian. However is j is equal to I, then
one
can notice that, a i i is equal to a bar i
i. That means, the diagonal elements have
the
property, that the element is equal to it
is conjugate. And this will happen only when
the
number is a real number. That means, if we
have a Hermitian matrix, then diagonal
elements will always be real.
.
And from this one can also derive, that when
A is a real matrix, then A bar is equal to
A.
The real matrix, so conjugate will be the
number itself. So, A bar is equal to A. So,
if we
consider this A star, since A bar is equal
to A. So, this A star will actually reduce
to A
transpose. So, then A happens to be a real
matrix, then A is equal to A transpose for
matrix to be Hermitian.
And this means, a real Hermitian matrix is
a symmetric matrix. So, if the matrix happens
to be real matrix, then it is nothing but
a symmetric matrix. Let us take an example.
We
have complex matrix, the diagonal elements
are 1 and 2. And 1 plus i and 1 minus i are
non-diagonal elements. So, let us calculate
it is conjugate. So, conjugate of 1 is 1,
.conjugate of 2 is 2. Conjugate of 1 plus
i is 1 minus i, and conjugate of 1 minus i
is 1
plus i.
And this means, if you take the transpose,
then A star is equal to A. Or when you take
the transpose, then this 1 plus i will go
here. And this 1 minus i will come here. So,
this
A bar transpose is nothing but A. So, this
matrix is a Hermitian matrix. Then, the result
is
that X star X is always real, whatever be
x. So, let us consider x is a plus i b comma
c
plus i d. So, these are complex elements of
a 2 by 2 matrix. I prove this result for 2
by 2,
in fact, this can be done for n th order vector.
So, x star x is equal to a minus i b into
c minus i d, that is the x star. And this
is the
column vector, so x star will become the row
vector. And the corresponding elements
will be the conjugate. So, a plus i b will
become a minus i b. And c plus i d will become
c minus i d multiplied by a plus i b and c
plus i d transposed. And when you multiply
it,
it is a minus i b multiplied by a plus i b,
that gives me a square plus b square. And
the
second element here is then multiplied by
this, gives me c square plus d square. So,
a b c
d being real, because a plus i b and c plus
id being complex. So, this is a real number,
so
x star x is always a real number.
.
Then, if A is Hermitian matrix, then x star
A x is real number. So, this is another result,
let us try to prove this. Let A is a Hermitian
matrix, we have prove that it, this number
is
real number. This number is a real number,
then it is conjugate is a real number. And
in
.fact, x star A x is nothing but a 1 by 1
matrix. Because, x star is a column vector
and x is
a column vector and x star is a row vector.
So, this product will be 1 by 1 matrix. So,
if it
is transpose conjugate is a real number, then
that means, that this is a real number.
So, to prove x star A x is equal to x star
A x transpose. So, we are proving this result
by
proving this property, x star A x is equal
to x star A x star. For this purpose, we will
evaluate x star A x star, this can be evaluated
using these properties. We know that, A B
transpose is equal to B transpose A transpose.
That means, when you take the transpose
of a product the order will change. So, this
means A B star is equal to B star A star.
Now, this result we will be using to establish
this result. So, I start with the right hand
side, x star A x star. So that means, this
is the first matrix and this is the second
matrix.
So, second matrix will be started first. And
this first matrix will be star next. So, it
is A x
star multiplied by x star star. Now, A x star
again I am using this property. So, it
becomes x star A star and when I take conjugate
twice. Because, conjugate is a operation
which will be negated, if we do it twice.
So, x star star will become x.
And then this is nothing but x star A x, because
this multiplication is associative. So,
these brackets does not have any meaning.
So, we have proved that x star A x star is
equal to x star A x. And this is possible,
only when this number is a real number. So,
we
have proved that if A is a Hermitian matrix,
then x star A x is real. Now, on the basis
of
this result, we will prove that the Eigen
values of Hermitian matrices are real. So,
let us
see how we prove this result.
..
So, let us take lambda as an Eigen value of
Hermitian matrix A. Then we can write A x
is equal to lambda x. X being the Eigenvector
corresponding to Eigen value lambda and
A is Hermitian. We pre-multiply this equation
by x star. So, we will have x star A x is
equal to lambda x star x. Now, we have already
proved the left hand side is real number
in my earlier result. And also I have proved
that x star x is a real number. So, left hand
side is a real number right hand side is a
real number. So, this lambda cannot be
complex, because if this is complex. Then,
this has to be complex, so lambda is real.
..
And that is how is say that Hermitian matrix
matrices, will always have real Eigen
values. We illustrate this with an example.
So, let us consider an a matrix A consisting
of
0 minus i, in the first row i 0 in the second
row. So, it is a 2 by 2 Hermitian matrix.
So,
we see what are its Eigen values to calculate
the Eigen value of this matrix. I have to
compute lambda i minus A. So, determinant
of lambda i minus A equal to 0 is the
characteristic equation for this given matrix.
And for this lambda i minus A is to be computed.
So, lambda into i, that is lambda A is
not contributing here in the first element.
So, it is simply lambda. Then, the second
is
lambda i. So, no contribution from i in this
element. So, it is minus A, so it is i here
then
this i will become minus i and then we have
this lambda. Now, if you evaluate this
determinant it is lambda square minus. And
this minus will make it plus, so it is plus
i
square. And plus i square becomes minus 1.
So, determinant lambda i minus A is lambda
square minus 1 equal to 0. And, this is a
second order equation in lambda, which can
be easily solved. And we get the Eigen
values of given matrix are real and they are
1 and minus 1. So, we have established a
result and we have verified this in this example.
..
Now, the determinant is product of Eigen values.
This is the result, which we have
established in earlier lectures. So, here
we have Hermitian matrix, which has real Eigen
values. And if we combine these two results,
we can say that the determinant of
Hermitian matrix is real, because all the
Eigen values are real. So, there product will
also
be real and product of Eigen values is determinant.
So, the determinant of Hermitian
matrix is real.
Now, this is an important result it says that,
a real symmetric matrix has real Eigen
values. Now, this result can easily be derived
from what we have done so far. We have
Hermitian matrix having real values, and a
real Hermitian matrices, real symmetric
matrix. So, if we use that result, we can
easily arrive to the result that a real symmetric
matrix has real Eigen values.
However, if you consider this matrix A it
is symmetric, because this is a mirror image
of
this. So, symmetric matrix you can calculate
its Eigen values. Determinant lambda i
minus A is equal to lambda and this also lambda.
Here, it is minus i and minus i
simplified it is lambda square plus 1 equal
to 0. Therefore, this characteristic equation
lambda square plus 1 equal to 0, gives me
Eigen values as i and minus I.
And that means, this real symmetric. This
symmetric matrix gives me Eigen values as
complex numbers i and minus i, where we have
gone wrong. This is a symmetric matrix,
but not a real symmetric matrix. So, the result
which we have stated is related to real
.symmetric matrices. And the result is that
a real symmetric matrix has real Eigen values.
The symmetric matrix may have other Eigen
values.
.
Now, we come to another concept Skew Hermitian
Matrix. A complex square matrix A
is Skew Hermitian if minus A is equal to it
is transpose conjugate. That means, A is
equal minus of A star. And when is a real
matrix, then we know A bar is equal to A.
And
this reduces to A star is equal to A transpose.
And that means, A is equal to minus A star
means minus A transpose.
So, if we have a real matrix then it will
be Skew Hermitian. If A is equal to minus
A
transpose or we say that a real Skew Hermitian
matrix is a Skew symmetric matrix,
because this is the definition of a Skew symmetric
matrix. So, if a matrix is real and it is
Skew Hermitian, then that matrices is going
to be a Skew symmetric matrix.
..
You may notice that, if A is Skew Hermitian
then i times A is Hermitian. Let us, take
an
example we have been given a matrix A as 0
1 plus i minus 1 plus i 0. Let us see whether
what is its conjugate A bar. So, these diagonal
elements will not be effected. But, this 1
plus i will become 1 minus i, when we take
the conjugate. And minus 1 plus i it is
conjugate will be minus 1 minus i.
So, A star will be the transpose of this.
So, this row becomes this column. And this
row
becomes this column. And from here, if we
compare A and A star, one may notice that
A
star is nothing but minus times A. This is
0 and 0 does not make much difference. But,
this element is negative of this. And this
element is negative of this and hence, A star
is
equal to minus A. Or we can say that, this
matrix is Skew Hermitian matrix.
So, given this Skew Hermitian matrix. We will
see that, if we multiply this matrix by A
this matrix by i. Then, what we have a resultant
matrix as Hermitian matrix. So, let us do
it here, the matrix i A is Hermitian. So,
we multiply this matrix A by i. Let us call
this
matrix as B, 0 multiplied by i this remains
as it is. So, diagonal elements are not affected,
but this element will become minus 1. Why
because, i in to i is minus 1 and 1 into i
is i.
And this element will become minus i minus
1 and 0 remain as such. So, this B as i times
A we calculate its conjugate. So, conjugate
of minus 1 plus i is minus 1 minus i and
conjugate of this is minus 1. And this minus
i will become plus i conjugate, means the
imaginary part will be negated. And then B
star B transposed conjugate or conjugate
.transpose is equal to the transpose of this
matrix. That is row becomes column and this
row become this column and from here. If you
compare B and B star this is nothing but
B is equal to B star. And that proves, that
B is Hermitian matrix. So, A given to be Skew
Hermitian i times A is Hermitian.
.
Then, every Eigen value of Skew Hermitian
matrix is pure imaginary. We have proved a
result related to Hermitian matrix. And we
found that Hermitian matrix has real Eigen
values. But, if the matrix happens to be Skew
Hermitian then its Eigen values will be
pure imaginary. So, to prove this let K is
Skew Hermitian matrix. Then, i K is Hermitian
matrix, let lambda is an Eigen value of real
Hermitian matrix. That is lambda is a real
Eigen value of i K, then Eigen values of K
are pure imaginary. So, this what we get from
this result that if K is Hermitian. Then,
i K will be if K is Skew Hermitian then i
K will
be Hermitian.
..
Let us check this, in this example the given
matrix is Skew Hermitian. It is Eigen values
are i and minus i. Therefore, this Skew Hermitian
matrix A has pure imaginary Eigen
values.
.
Now, we were talking about Skew Hermitian
matrices. And Hermitian matrices we are
talking about their Eigen values. Hermitian
matrices have real Eigen values, while Skew
Hermitian matrices have pure Eigen values
purely imaginary Eigen values. Now, we will
.talk about orthogonal matrices. And we see
that an orthogonal matrix, has Eigen values
1
or minus 1. So, let us try to prove this result.
It is being given that A is orthogonal matrix.
Then, from the definition of orthogonal
matrices, we know that AA transpose is identity
or A transpose is nothing but it is
inverse A inverse. So, let us say lambda is
an Eigen value of this matrix A. Then, A X
is
equal to lambda X. And also A inverse y is
equal to 1 upon lambda 1 upon lambda times
y. And that means, A transpose and A have
same Eigen values. And that means, lambda
is equal to 1 upon lambda or lambda square
is equal to 1. And that gives me lambda is
equal to plus minus 1. So, if A is orthogonal
matrix, then its Eigen values are 1 or minus
1.
.
So, let us see how they look like on this
plane. If I have a real axis here, and imaginary
axis here, then all the Eigen values of symmetric
matrices of real symmetric matrices
will lie on the real axis. And the Eigen values
of Skew symmetric matrices, will lie on
the imaginary axis. While, this is a unit
circle and then all the Eigen values of orthogonal
matrices will lie on this unit circle.
..
Now, we come to similar matrices. Let A and
B be two square matrices of order n. Then,
A is said to be similar to matrix B. If there
exist an invertible matrix P of order n, such
that A can be represented as P inverse B P.
The following elementary properties can be
easily established with respect to similar
matrices.
.
The first is that A is similar to A. So, if
A is similar to A means there should exist
some
matrix P. Such that, A is represented as A
times such that A is P inverse A P. So, in
this
.case, if we take P inverse as equal to identity
which is the same as P. Then, we can prove
easily that A is similar A.
Secondly, if B is similar to A, then A is
similar to B and this can we very easily
established. And on the basis of this one
can say the matrices A and B are similar.
You
do not have to say B is similar to A or A
is similar to B, we can simply say that the
matrices A and B are similar.
Thirdly, if A is similar to B and B is similar
to a third matrix C. Then, A is similar to
C.
Now, one can see that the first property is
reflexive property. Second is symmetric and
third is transitive. And on this basis one
can say that, the relationship. That is similar
A is
similar to B is an equivalence relation. And
one can also see that the identity matrix
of
order n that is I n is similar to itself,
because in that case P is I itself.
.
Now, if the matrices A and B are similar.
Then, we can prove the determinant of A is
equal to the determinant B. Or we can say
similar matrices have same determinants. Not
only this, but the trace of A and the trace
of B are also equal. Further, if A m is similar
to
B m for if A and B are similar. Then, A m
and B m will also be similar for any positive
integer m. And finally, if A is similar to
B and if A is invertible, then A inverse is
similar
to B inverse. These are some of the properties
of similar matrices, let us prove them 1 by
1.
..
So, to prove the first property, let us say
A and B are similar. Then, this statement
means
that there exist a matrix P which is nonsingular
such that A is equal to P inverse B P.
And from here, we can get determinant of both
the sides. So, determinant A is equal to
determinant of P inverse B P. And we know
the property of determinants that, this
determinant of product is equal to product
determinants. So, determinant P inverse B
P is
equal to determinant P inverse multiplied
by determinant B into determinant P.
Determinants happens to be real number so
they can we they can change the order. And
that means, we can bring this determinant
P here. So, this determinant P inverse into
determinant P into determinant B. And one
can combine these two, it is determinant P
inverse P into determinant B. And this is
nothing but determinant of i which is 1. So,
we
can say determinant A is equal to determinant
B. That is the first property, which we
have listed.
The second is trace of A is equal to the trace
of B. So, we can write down trace of A is
equal trace inverse trace of B inverse B P,
because B is similar to A. So, there exist
such
B there exist such P. So, to prove this result
we will use this result with respect to
product and trace, where we have just earlier
established. That trace of A B is equal to
trace of B A.
So, we can write down this trace of P inverse
B P as trace of first matrix into second
matrix. So, this means trace of P into trace
of P inverse B. So, order has been changed.
.And that means, this can also be using this
product is associate. So, we can write trace
of
P in to P inverse B and P into P inverses
identity. So, identity into B is B itself,
so trace
of A is equal to trace B. So, second result
is established.
.
To prove the third result, let us consider
A is equal to P inverse B P. We will prove
this
result by induction. So, we will let us see
what happens to A square. So, if we write
down if we multiply A with A. Then, it is
A square which is equal to P inverse B P A,
which is equal to P inverse B P into P inverse
B P. I have written this is first A and this
is
second A. So, we multiply it this is multiplication
is associate. So, you can change the
brackets.
And that is why I write down this product.
As P inverse B into P into P inverse into
B P
and that gives me P inverse B this is I and
B P. So, it is P inverse B I B P and then
one
can write down this product as P inverse B
square P. So, there we have written A is equal
to P inverse B square P. And from the definition
of similar matrices, one can say that A
square is similar to B square.
So, now let us assume that A k minus 1 is
similar to B k minus 1. Now, we write A k
minus 1 as P inverse B k minus 1 in to P.
Because, they are similar matrices and then
A k
is written A k minus 1 in to A. So, A k minus
1 is this which is given to us and A is
similar to B. So, I write it as P inverse
B P. If you simplify, then this expression
comes
.out to be P inverse B k P and that proves
that A k is similar to B k. So, by induction
this
result is true for any value of positive m,
that proves the result.
.
Then, the fourth according to fourth property
if A is similar to B, then B can be written
as P inverse AP. It is given that A is invertible,
so determinant A is not 0 that is, what we
have proved earlier, and; that means, determinant
of B is equal to determinant P inverse
A P which is equal to determinant of A. Since,
determinant B is not 0, so B is also
invertible.
Once B is invertible, then we consider A into
A inverse A is similar to B, so I write it
as
P inverse BP into A inverse. Now, A into A
inverse is identity, so this is simplified
to P
inverse B P A inverse or we can write it as
P into I if I premultiply this expression
by P.
So, it is P into I is equal to P times P inverse
B P A inverse or that from here we can get
A k as A k minus 1 into A which is equal to
A k minus 1 is replaced by this P minus 1
B
k minus 1 P into P inverse B P. And, if you
simplify this expression this comes out to
P
inverse B k P and this proves that A k is
similar to B k
Now, by induction the result is true for m
positive, . now, to
prove the fourth result if A is similar to
B, then B is expressed as P inverse A P. It
is
given that A is invertible; that means, determinant
A is not 0, then determinant of B
determinant of A are equal. So, determinant
B is equal determinant P inverse A P which
.is determinant A and this means determinant
B is not 0 and; that means, B is invertible
also.
Then, A times A inverse is equal to A is similar
to this expression A similar to B, so we
write down A as P inverse BP multiplied by
A inverse, now this is nothing but identity
on the left hand side. And then P inverse
we rearrange this term P inverse is take out
B P
A inverse and; that means, if i multiply this
by P then P I is equal to P into P inverse
into
B P A inverse and; that means, P is equal
to B P A inverse or B inverse P is equal to
P A
inverse from here. And finally, P inverse
B inverse P is equal A inverse and; that means,
A inverse is similar to B inverse. So, this
is what we have to prove in the fourth property,
that A inverse and B inverse are similar.
.
Now, these properties are helpful in checking
whether A and B are similar, like if I have
given these 2 by 2 simple matrices. Then one
can very easily check that they are similar
or not. We can first calculate the trace of
this matrix A it is 4 and trace of B is 4,
but this
result is not conclusive. So, we check the
determinant, determinant of A is 3 plus 2
and
determinant of B is 4 minus 2, so determinant
A is not equal to determinant B although
trace A is equal trace B and this proves that
the two matrices are not similar.
..
In this example, I have been given 3 by 3
matrices trace A is equal to trace B, let
us
check trace A is 2plus 1 plus 1 the sum of
diagonal elements it is 4 and trace B is sum
of
diagonal elements as 3. So, trace A and trace
B are not equal and one can conclude that
the matrices are not similar.
.
In this example, I have been given 2 by 2
matrices there trace 2 and 2 4 and 2 and 2
4.
So, trace A is equal to trace B one can calculate
their determinants this determinant is 4
.and this determinant is 4, so there determinants
are equal. Can be conclude that A and B
are similar.
So, my claim is that these two matrices are
not similar, although there determinants and
traces are equal. Now, to show this result
let me have an a square matrix P which is
taken
a general matrix consisting of elements a
b c d such that one can express A in terms
of
the matrix B as P inverse B P. So, let us
say there exist some P for which this is possible,
so if we can find out a b c d in such a manner
making P invertible, then we say that A
and B are similar. If we cannot then of course,
A and B will be not be similar.
.
So to do this, we write down A is equal to
P inverse B P or this can be simplified to
PA
is equal to B P. So, we first calculate, what
is P A, I have assumed P as a b c d a b is
given to me as this.
So, let us multiplier them, so it is 2 a plus
b and 2 b and this multiplication will give
me
this row and P A is this 2 by 2 matrix. Similarly,
I calculate B P, so B is given to me as
this matrix and a b c d. I have assumed as
this get this product and this product comes
out to be this, so if A and B are similar,
then P A should be equal to B P as I have
shown
here.
So, let us try to prove, let us try to show
that this matrix is equal to this matrix,
so we
have 2 a plus b into 2 a plus b 2 b 2 c plus
d 2 d is equal to 2 a 2 b 2 c 2 d. So, if
there are
.these are equal element wise then the result
is proved. So, this gives me four equation
2 a
plus b is equal to 2 a this means 2 b is equal
to 2 b this means 2 c plus d is equal to 2
c
and this means 2 d is equal to 2 d, so these
are four equations, since from here, this
are
identically satisfied, but from here.
.
And from this, one can show that d is equal
to 0 and b is equal to 0; that means the
matrix P will be of the form a c 0 0 and that
simply means that determinant of P,
whatever be the values of a and c will be
0 and that proves that P is not invertible
for any
combination of a and c and; that means, we
cannot find a matrix P which is invertible,
such that A can be expressed as P inverse
B P or finally we can say that A cannot be
similar to B.
So in this example, we have proved that two
matrices which have same determinant
same trace, but still there they cannot be
similar. So, we it is not conclusive that
determinant A is equal to determinant B trace
A is equal to trace B then the matrices will
be similar. Now, similar matrices share many
other properties we have seen that they
have same trace and determinants, now I will
show that they have same Eigen values
..
So let us I have this theorem, let A and B
be two similar matrices and if lambda is an
Eigen value of A, then lambda is also an Eigen
value B. So, to prove this result it is it
is
being given to us that A and B are similar,
so we can say that there exists nonsingular
matrix P. So, that A is equal to P inverse
B P by the definition of similar matrices.
Let
lambda be an Eigen value A; that means, there
are nonzero vectors X such that A X is
equal to lambda X, so its Eigen its characteristic
equation will be lambda I minus A into
X is equal to 0.
And, lambda I minus A, let us say it has the
matrix A has Eigen values lambda 1 lambda
2 lambda n. Then, lambda I minus A is equal
to lambda minus lambda 1 lambda minus
lambda 2 into lambda minus lambda n equal
to 0, because we have assumed that lambda
one lambda two lambda n are the roots of this
characteristic equation. Now, since A and
B are similar, so we can say P inverse B P
X is equal to lambda X, so I have written
the
value of A in this equation.
..
And, this gives me lambda I minus P inverse
B P is equal to lambda P inverse I P minus
P inverse B P is equal to 0, so this lambda
I, I have written in this manner. So, lambda
P
inverse I P minus P inverse B P equal to 0.
And, that gives me that P inverse lambda I
minus BP its determinant equal to 0 and this
simply means that determinant P inverse
into determinant of lambda I minus B into
determinant of P and that gives me
determinant of lambda I minus B. So, I have
started with lambda I minus A and this
comes out be determinant of lambda I minus
B.
And; that means the characteristic equation
for A and for B they are same, so we can
write down determinant of lambda I minus B
as lambda minus lambda 1 lambda minus
lambda 2 into lambda minus lambda n. And,
since lambda 1 lambda 2 lambda are n are
other roots of this characteristic equation
lambda I minus A equal to 0, so; that means,
they are roots of this equation determinant
lambda I minus B also.
..
And; that means, the characteristics polynomial
of A and B are same and therefore, the
Eigen values of A and B are also same. So,
that proves the result and this means that
if
lambda is an Eigen value of A, then it is
also an Eigen value for B, now although the
Eigen values of similar matrices are same
their Eigen vectors corresponding to given
Eigen values maybe different. So, this is
an observation that Eigen values maybe same
we have proved in the form of a theorem that
the Eigen values of similar matrices are
same, but the Eigen vectors need not be the
same. In fact, we say that similar matrices
can be thought of as describing same linear
transformation, but with respect to different
bases.
..
Let me illustrate, this with this diagram,
let us say we have a linear transformation
in fact
linear operator from V to V, V may be R n.
Then, any vector X under this operator will
map to L x. Let us, say there are two bases
for V the s and t, so x s denotes the
coordinates of this vector x with respect
to basis s. And x s, so this is L x the x
will map
to Lx., so Lx with respect to basis s can
be written as B times x s. So, this vector
x s will
map to this vectors L of x s. So, B is the
transformation matrix, so x s is written as
B
times x s.
Now, his vectors the coordinates of this vector
x in s is x s, but with respect to y the
relationship between with respect to be another
basis t, the relationship will be x s is
equal to t times x t, so these are transformation
matrix which takes the vectors of which
takes the vectors I mean coordinates x s to
x s x t. So, this vector then maps to this
P of x
t is A times A of P x t. And then you can
again comeback from here you can come back
to this, so transformation P inverse is to
be applied matrix P inverse is to be multiplied
to
get this matrix. So; that means, it is B P
is P inverse A P x t, so we can say it is
the same
vector, but in a different bases.
..
Now, we consider different we consider an
example in which T is a transformation from
R 2 to R 2 defined by this transformation.
We have to find the matrix representation
for
the transformation related with the standard
basis, and then we will also find the matrix
with respect to the basis 2 1 and 1 1 and
we will show that the two matrices are similar.
This will illustrate what we have discuss
just now in a diagrammatical representation
So, we start with the vector 1 0 and 0 1 as
the base vector in the standard basis, so
1 0
will map to 1 3 under this transformation.
One can check x 1 is equal to one and x 2
is
equal to 0, so this comes out to be 1 and
the second element will map to 3. Similarly,
T 0
1 will be 6 0. Then, the vector 1 3 will be
represented as a linear combination of the
base
vectors 1 0 and 0 1; that means, alpha is
equal to 1 and beta is equal to 3, so these
are the
coordinates of this vector. Similarly, 6 4
will be represented as alpha times 1 0 plus
beta
times 0 1 and that gives me alpha is equal
to 6 beta is equal to 4.
Now, matrix of transformation with respect
to standard basis will now be written as 1
3 6
and 4. So, this was the first column vector
and this is the second column vector which
we
have obtained earlier. And accordingly, if
we consider the second basis as 2 minus 1
and
1 1, then the transformation y is equal to
A x will give me T is equal to 2 minus 1 is
equal to this matrix A. And then 2 minus 1
and this gives me 2 and minus 6 is 4, 3 into
4
multiplied by 2 minus 1 gives me 2, so T 2
0 1 will become 4 minus 2. Similarly, T 1
1
can be expressed as this matrix A multiplied
by 1 1 and that give me 7 7, so
representation of minus 4 2 and 7 7.
.
.With respect to second basis comes out to
be a linear combination of this, so minus
4
comma 2 is equal to alpha times, the base
vector 2 minus 1 plus beta times 1 comma 1.
And then one can is simplify this equation,
so it is 2 alpha plus beta is equal to minus
4
and minus alpha plus beta is equal to 2. So,
when you solve these two equations the
coefficient alpha comes out to be minus 2
and beta comes out to be 0. Similarly, if
we
write down 7 7 as a linear combination of
base vectors, then again we will have system
of equations 2 alpha plus beta is equal to
7 minus alpha plus beta is equal to 7 solve
them, we will get alpha is equal to 0 and
beta is equal to 7.
.
And; that means, the matrix with respect to
second basis comes out to be minus 2 0 and
0
7. So, we have now we have two different matrices
with respect to two different basis
and one can note that determinant A is equal
to determinant B and trace A is equal to
trace B for these two matrices. Now, the two
matrices A and B may be similar, so simply
that these two results will not mean that
A and B will be similar we can say that A
and B
may be similar.
..
Now, for this let us consider, there exists
a nonsingular transformation P which is a
b c d.
So, that we can write down A is equal to P
inverse B P, and then P A is equal to B P
and
from here we can get AB into the first matrix
is equal to second matrix multiplied by P.
And from here, if we simplify we will have
left hand side as a plus 3 b 6 a plus 2 b
6 c
plus 3 d and 6 c plus 4 d and the right hand
side will have minus 2 a minus 2 b 7 c and
7
d. and from here.
..
If we solve for a and b, then we will have
a plus 3 b is equal to minus 2 a, 6 a plus
4 b is
equal to minus 2 b, c plus 3 d is equal to
7 c and lastly 6 c plus 4 d is equal to 7
d and
from here one can get that 3 times a plus
b is equal to 0 from these two equations.
And that means, if you solve these four equations
in four unknowns we will get a is
equal to 1 b is equal to minus 1 c is 1 and
d is equal to 2,.so this is one such combination.
And since, determinant P is not 0; that means,
P is nonsingular, so inverse exist. So, we
can say that a and b are similar, because
we could be able to find out such a matrix
a b c
d which is nonsingular and which gives as
A is equal to P inverse B P. So, one can notice
from this system that the matrix B is a diagonal
matrix.
..
Now on the basis of this, we can say that
it is easy to find the inverse determinant
and
Eigen values for a diagonal matrix, because
if it is a diagonal matrix the determinant
will
be simply product Eigen value. Then, diagonal
elements are the Eigen values, so if and
the inverse will be simply the inverse of
diagonal elements, so if the matrix happens
to
be a diagonal matrix then number of things
can be simplified.
So, algebra will be simpler if we can perform
operations on a diagonal matrix similar to
the given matrix. So, that is how we say that,
finding a similar matrix which is a diagonal
matrix we simplify the algebra. So, finding
a diagonal matrix similar to given matrix
is
discussed in the next lecture on diagonalization,
so with this remark I close this lecture.
..
And to summarize, what I have done in this
lecture, I have given the basic concepts
needed for diagonalization. I have discussed
Eigen values of real symmetric matrices and
similar matrices and with this we are ready
to discuss the diagonalization that will be
the
content of my next lecture.
Thank you.
.
