Hello, welcome to another module in this massive
open online course. So, we are looking at
examples ah; on matrices and also convex sets.
So, let us continue a discussion, let us look
at another example related to Positive Definite
Matrices alright. So, let us consider continue
our discussion or continue looking at examples
related to matrices and convex sets . So,
ah this example number 3 what you have to
show, that is A is a PD matrix that is positive
definite, if A is positive definite this implies
A inverse is also positive definite ok .
Want to show that if A is positive definite
A inverse is also positive definite, this
can be shown as follows. If A, A can be written
expressed as we already seen this U lambda
U Hermitian where ah U is a unitary matrix
satisfies U U Hermitian equals U Hermitian
U equals identity. And lambda is a diagonal
matrix of eigenvalues, and also further we
have seen that the eigenvalues of any positive
definite matrix have to be greater than 0.
The eigenvalues of a positive semi definite
matrix are greater than equal to 0; the eigenvalues
of a positive definite matrix they have to
be strictly greater than 0, it cannot have
any eigenvalues equal to 0.
And therefore, now if you look at A inverse
its rather easy to see A inverse equals U
lambda U H U lambda U Hermitian inverse, which
is basically U Hermitian inverse times lambda
inverse times U inverse. But we have seen
that U is a unitary matrix which implies U
Hermitian U equals U U Hermitian equals identity,
ah well this implies U equals U Hermitian.
So, what this implies is that if you look
at you Hermitian inverse that is U itself
because, U Hermitian to use identity times
lambda inverse into U inverse is U Hermitian
because again U U Hermitian equals identity.
And therefore, this is again has the same
structure except you can see with eigenvalues
1 over lambda 1 1 over lambda 2 1 over lambda
n times U Hermitian.
And ah therefore, now you can see if lambda
i is greater than 0, this also implies that
1 over lambda i is greater than 0. So, eigenvalues
of A inverse or also greater than 0 in effect
A inverse can be expressed as U lambda inverse
U Hermitian. And therefore, it has and it
has positive eigenvalues alright and therefore,
it is also a positive definite matrix.
And you can also check this as follows for
instance if you consider Z bar Transpose for
any vect real vector if you consider Z bar
Transpose ah A inverse Z bar I can now write
this as now since this is a real vector I
can write this as Z bar Hermitian inverse
we have seen is U lambda inverse U Hermitian
Z bar. Now if u set U Hermitian Z bar if you
set this is equal to Z tilde then this will
become Z tilde Hermitian lambda inverse Z
tilde.
Which is equal to Z tilde 1 Z tilde 2 so on,
up to Z tilde n into the product 1 over lambda
1 1 over lambda 2 so on, 1 over lambda n times
Z tilde 1, Z tilde 2 up to Z tilde n. And
if you look at this, this is nothing but well
this will be Z tilde conjugate since this
is the Hermitian of the vector ok. So, we
are setting ah U Hermitian Z bar 
Z tilde. So, this will be summation over i
equals 1 to n over lambda i 
Z tilde i conjugate into Z tilde i that is
magnitude Z tilde i square.
Now, 1 over lambda i this is greater than
0, magnitude Z tilde i square this is greater
than 0. And therefore, ah this implies ah
that ah Z bar transpose A inverse Z bar is
also greater than 0 for each for all Z bar
for all vectors Z bar, and this implies that
A inverse is a 
this implies that A inverse is a positive
definite matrix. So, A is a positive definite
matrix A inverse is also positive definite
matrix. In fact, the eigenvalues of a inverse
are the inverse that is if lambda i is an
eigenvalue of A then eigenvalues of corresponding
eigenvalue of A inverse is 1 over lambda i.
Eigenvalues of A are strictly greater than
0 if A positive definite matrix and similarly
the eigenvalues of A inverse or also strictly
greater than 0, if strictly greater than 0
correct.
Since lambda is greater than 0 1 over lambda
is also greater than 0 ok alright, ah; let
us continue our discussion let us look at
another problem problem number example number
4. Ah What we want to show is that if A, B
ah these are 2 invertible invertible n cross
n matrices then, A B, B A have the same A
B and B A have the same eigenvalues we want
to show this property that A B eigenvalues
of A B are equal to eigenvalues of B A well.
We start with the characteristic polynomial
remember to compute the eigenvalues of any
matrix in this case the eigenvalues of the
matrix A B. So, we start with the characteristic
polynomial 
of A B , that is obtained by nothing but,
that is basically obtained by looking at the
determinant of A B minus lambda i remember
the eigenvalues are computed as the roots
of the characteristic polynomial. The characteristic
polynomial of A matrix A is A minus lambda
i you want to look at the characteristic polynomial
of the matrix A B therefore, that will be
the determinant of A B minus lambda i.
So, this is equal to I can write this as determinant
of A B A A inverse minus lambda A A inverse
because remember A is an invertible matrix
matrix both A and B are invertible matrices
so, A A inverse equals identity. So, I can
always write this as A B determinant of A
B A inverse minus lambda A A inverse. Now
I can extract the A on the right so, this
will become B A minus lambda I times A inverse
this is determinant of A times B A minus lambda
I times A inverse. The determinant of A matrix
product is the product of the determinants
that is determinant of A times determinant
of B A minus lambda I times determinant of
A inverse. Determinant of A inverse is basically
1 over the determinant of A because A times
A inverse is identity.
So, this is B A minus lambda I into 1 over
the determinant of A determinant of A times
1 over determinant of A these cancels. So,
this becomes the determinant of A minus lambda
I. Therefore, we have this interesting property
that is A B determinant of A B minus lambda
I equals determinant of B A minus lambda I
implies characteristic polynomials of B A
this implies the characteristic polynomial,
the characteristic polynomial 
of A B which is determinant of A B minus lambda
I. This equals the characteristic polynomial
of B A ok .
The characteristic polynomial of A B equals
the characteristic polynomial of B A this
implies the roots are equal roots are equal
or identical and this implies. So, characteristic
polynomials are equal implies the roots are
identical. And this implies therefore, eigenvalues
of A B equal eigenvalues of this implies eigenvalues
of A B equals 
this implies that eigenvalues of A B equal
eigenvalues of B A ok.
And in fact, you can see, if you write A B
equals U lambda U inverse remember this eigenvalue
decomposition I can write these are eigenvalues
are equal I can write B A equals some V times
lambda times V inverse because remember their
eigen values are equal. So, that diagonal
matrix of eigenvalues will be the 
equal or identical similar eigenvalues they
have the same eigenvalues.
This is the same eigenvalues the diagonal
matrices lambda right, the diagonal matrix
of eigenvalues will be the same for both A
B and B A in their eigenvalue decomposition
ok. So now, this implies that B A equals V
lambda V inverse so, this implies V times
or V inverse times B A into V equals lambda.
Now substitute lambda in the first one this
implies A B equals U times lamda, but lamda
is V inverse B A into U inverse V and this
is nothing, but U V inverse let us call this
as U tilde B A. If U inverse U V inverse is
U tilde then U inverse B becomes U tilde inverse.
So, I can write the matrix A B as some matrix
U tilde I can write this as you tilde times
B A times U tilde inverse such matrices are
said to be similar matrices. So, A B implies
A B is similar to B A. In general C similar
to D if there exist M such that C equals M
inverse D M. So, if there exists a matrix
M, such that you can write C equals M inverse
D into M, then the matrices then the matrices
C and D are said to be similar matrices. So,
in this case you can see these 2 matrices
A B and B A in fact, which of the same eigenvalues
alright these are similar matrices ok alright.
Let us now look at another interesting property
that is the eigenvalues of unitary matrix.
So, this is our example number 5 again another
simple property, what can we say about the
eigenvalues of a unitary matrix, now let U
be a unitary matrix . 
Remember the unitary matrix is defined by
the property U Hermitian U equals U U Hermitian
equals identity. Ah This is the property of
the unitary matrix now let, x bar be the eigen
vector 
and lambda equals corresponding eigenvalue.
Now this implies, what this implies is that,
U x bar equals lambda times x bar correct,
which implies now you can multiply U x bar
Hermitian U x bar that will be equal to lambda
x bar Hermitian. Because U x bar equals lambda
x bar U x bar Hermitian equals lambda x bar
Hermitian multiplied by lambda x bar.
And this implies x bar Hermitian U Hermitian
U x bar equals lambda Hermitian, but lambda
is a number so lambda Hermitian is simply
lambda conjugate x bar Hermitian lambda into
x bar. Now U Hermitian U is identity because,
U is unitary matrix that, leaves x bar Hermitian
x bar which is remember norm of x bar square
this is equal to lambda conjugate lambda,
that is magnitude lambda square times x bar
Hermitian x bar, which is again once again
norm x bar square which implies cancelling
the norm x bar square on both side. This implies
magnitude lambda square equals 1 which means
magnitude lambda equals 1 .
So, this implies basically that eigen so,
this basically shows a very interesting property
eigenvalues of unitary matrix , eigenvalues
of a unitary matrix have unit magnitude that
is the interesting property that this shows
alright. And now similarly, if you consider
the determinant of the unit the magnitude
of the determinant let us consider the magnitude
of the determinant. Remember we have seen
that the determinant is nothing, but the product
of the eigenvalues so, the magnitude of the
product of the eigenvalues, which is nothing,
but the product of the magnitudes of the eigenvalues.
And each eigenvalue is unit magnitude so this
is equal to the product of ones which is one
which shows this ancillary property or you
can also think of this as an axiom that the
determinant of a unitary matrix is identity.
All the eigenvalues of a unitary matrix are
magnitude 1 and the determinant of unitary
matrix has the magnitude of the magnitude
of the determinant of a unitary matrix is
1 as well ok alright. Let us continue a discussion
let us start with another example ah let us
consider the norm relation between the 1 norm
of a vector we want to show that the 1 norm
of a vector x bar is less than or equal to
square root of n times the 2 norm.
Now if x bar is a vector with elements x 1,
x 2 up to x n. Now remember the 1 norm, this
is simply the sum of the magnitudes, the magnitude
of magnitude x 1 plus magnitude x 2, magnitude
x n and the 2 norm is the square root of magnitude
x 1 square plus magnitude x 2 square plus
so on, plus magnitude x n square . Now to
show the property about what we will do is,
we will consider two different vectors will
construct 2 vectors u bar and components the
elements of u bar are magnitude x 1 magnitude
x 2.
So, I am constructing two different vectors
magnitude x 1 magnitude x 2 so on, magnitude
x n and v bar is a vector n dimensional vector
of all ones. Now what I going to do is, I
am going to apply the koshish squads inequality,
remember we have seen the koshish squads inequality
which states that the inner product square
u bar v bar that is less than u bar, that
is norm u bar square into norm b bar square.
This implies that if you look at u bar transpose
v bar square, that is less than or equal to
norm u bar square, norm v bar square. And
this also implies that u bar transpose v bar
is less than or equal to norm u bar into norm
v bar we know this property.
Now all we have to do is substitute the definition
from the definition above substitute u bar
and v bar, you can see u bar transpose v bar
is nothing but, magnitude x 1 plus magnitude
x 2 so on, up to magnitude x n. Which is basically
norm x bar of 1 and norm u bar that is the
2 norm remember the 2 norm u bar is square
root of magnitude x 1 square plus so on magnitude
x n square which is nothing, but the 2 norm
of x bar. And finally, the 2 norm l 2 norm
of v bar is square root of this is 1 plus
1 plus 1 n times this is nothing, but square
root of n. And now using this property using
1 substituting all these in 1, this basically
yields norm x bar 1 that is u transpose v
bar less than or equal to norm v bar that
is square root of n into norm u bar that is
a 2 norm of x bar.
So, this is an interesting property that we
have ok. So, this is the property or the relation
you can say characterizes the relation between
the 1 norm and the 2 norm. In fact you can
also show something between the relation between
the 2 norm the infinity norm. You can also
show ah; that ah the infinity norm that is
if you look at the 2 norm, now this is equal
to well we have seen magnitude x 1 square
plus magnitude x 2 square plus magnitude x
n square, this is a sum of the squares of
the magnitude is all the elements.
Now this is greater than or equal to you simply
take the maximum of 
the maximum of the magnitude correct. This
is the sum of the squares of the magnitude
of all the elements, which is greater than
equal to the square of simply the magnitude
of the maximum of these elements, which is
equal to now you take the square root all
you are left with is the maximum of magnitude
x i which is nothing, but the l infinity norm
ok.
So, therefore, this shows that ah so, this
shows that ah basically your this thing is
greater than equal to the a l 2 norm is great.
So, this basically shows that your l 2 norm
is greater than or equal to the l infinity
norm ok alright. So, let us stop here and
we will continue with other aspects in the
subsequent modules.
Thank you very much .
