See 
in the last lecture we had discussed the properties
of unitary operators on a finite dimensional
inner product space in todays lecture I will
discuss the normal operator, the case of normal
operators and prove what is called as the
spectral theorem for normal operators on a
finite dimensional inner product space. We
must compare this spectral theorem for a normal
operator with the one that I proved a couple
of lectures ago for the self adjoint operator
ok. Will make this comparison just before
the statement or perhaps after the statement
of the spectral theorem for a normal operator.
Let’s look at some properties that we will
require for a normal operator. So first of
all what is a normal operator? Let me we have
seen this before let me quickly recall the
definition and prove some properties of normal
operators. So definition let V be an inner
product space the finite dimensional 
inner product space, let T be your linear
operator on V be a normal operator I am sorry,
I want to define a normal operator so let
me say I want to define a normal operator,
an operator T element of L V is called normal
if is called normal if T-T star equals T star
T. T star at the adjoint operator of T. Ok
that is the definition.
We have seen this before let’s look at a
couple of examples one for the real inner
product space and another one for the complex
inner product space. Just a couple of examples,
we will define the linear transformations
through matrices, so the first one let’s
take A, actually A theta to be the rotation
operator, we have come across this operator
before so I am writing down the matrix first
and then I will define the operator through
this matrix.
A theta is Cos theta Sin theta minus Sin theta
Cos theta, for theta lying in 0 to Pie, define
a linear transformation I will call it T theta
from R2 to R2 by the formula that T theta
of x is equals A theta x, x in R2 for a fixed
theta. Then you can verify that T theta is
a normal operator that is see this a real
inner product space so instead of star we
consider the transpose T. So this equation
T-T star equal to T star T in this case is
T-T transpose equals T transpose T that is
T is a normal operator ok so this is an example
of a normal operator on a real inner product
space let me give an example for a complex
inner product space.
Example 2, again I take a matrix and then
define the linear transformation through the
matrix, say 1 i, define let’s call it S
perhaps S from C2 to C2 by S of x equals A
x, x belongs to C the I leave it as an exercise
for you to verify that S is normal that is
S-S star equals S star S. if I remember right
it turns out to be identity so these are this
is a unitary operator this is an orthogonal
operator ok, so these are some of the examples
of normal operators. Let’s now look at some
properties a couple of properties really for
a normal operator that will be required in
the rest of the discussion.
So the first property is the following, this
first property that I am going to write down
should remind you of a result for a self adjoint
operator ok. So the first one is let me call
it as a theorem let T be a normal operator
on a finite dimensional inner product space
V. We don’t really need the space to be
finite dimensional but still I will assume
that V is finite dimensional. Let x be an
eigenvector of T corresponding to the 
eigenvalue lambda of T then we will show that
x is an eigenvector 
of T star corresponding to 
the eigenvalue lambda bar.
We will require this result in later discussion
ok so let’s prove this result a I told you,
you, you must make a comparison of this result
with the case of self adjoint operators. In
the case of self adjoint operators we have
shown that the eigenvalues of operator are
real numbers we have also shown that eigenvectors
corresponding to distinct eigenvalues of a
self adjoint operator are orthogonal ok. Let
me include the second part, also eigenvectors
so we have a all that I am saying is that
we have a similar result for a normal operator
also.
Also eigenvectors corresponding to distinct
eigenvalues 
of T are not just linearly independent they
are orthogonal also are orthogonal. Ok so
let’s see the proof of this result, T is
given to be a normal operator let us now define
an operator U by the formula T minus lambda
i, lambda is any complex number in particular
we take lambda to be the number the eigenvalue
that we started for example. So take U to
be this then it is easy to see that U is normal
I leave that part for you to verify then u-U
star equals U star U.
Ok so let’s now look at so U is normal let’s
now look at norm 
of U star x the whole square this by definition
inner product U star x with itself I use a
property of the adjoint operator this operator
U star when it comes to this argument it goes
as U so this is x U-U star x I will now use
the equation U-U star equals U star U so I
can write this as x, U star U x again I bring
this to this argument it will comes as U so
that is U x inner product of U x with itself
which is now norm of U x square.
So in particular what this means is that norm
of U star x equal to zero if and only if norm
of U x equal to zero. Now look at U star,
what is U star? U is T minus lambda i then
U star is T star minus lambda bar i, so zero
equals norm of I am writing T star minus lambda
bar i of x if and only if norm of T minus
lambda i x equals zero. So I have just written
U star here and U here from this it follows
that if lambda ok look at this equation. If
lambda is eigenvalue x is a corresponding
eigenvector for the transformation T for the
normal operator T then it follows that x is
an eigenvector the same x is an eigenvector
for T star corresponding to the eigenvalue
lambda bar ok.
Now this is what we wanted to show, that is
the first part. Let’s show the second part,
second part is to show that eigenvectors corresponding
to distinct eigenvalues of a normal matrix
R orthogonal. Again the this part of the proof
should remind you of what we did for the case
of a self adjoint operator so let me go through
this quickly. Now second part let T x equals
lambda x and T y equals Mu y where I know
that Mu and lambda are distinct, I must show
that x is orthogonal to y. So I must show
that inner product x y is zero.
So let’s start with consider lambda times
inner product x y I can bring this lambda
to the first argument lambda x, y and then
substitute lambda x equal to T x, this is
T x, y apply the definition of the adjoint
that is x T star y now use the previous result
y is an eigenvalue sorry y is an eigenvector
corresponding to the eigenvalue Mu of the
operator T then we have seen that it must
follow that y is an eigenvector corresponding
to the eigenvalue Mu bar for the operator
T star that is t star y equals Mu bar y. So
this is using the first part and Mu bar is
in the second argument it comes out with a
conjugate and so conjugate double conjugate
Mu times x y.
Now the result follows, if lambda is not equal
to Mu then inner product x y must be zero
and so x and y are orthogonal ok. So x is
perpendicular to y that is what we wanted
to show. Let’s go back and see every unitary
operator is a normal operator ok but a unitary
operator the entry is atleast when you treat
the columns or rows of unitary operator they
are very much tractable in the following sense.
If U is a unitary operator then without writing
on the board let me just explain this, if
U is a unitary operator then it satisfies
U-U star equals U star U equals identity ok.
What this means is the following (just) let’s
look at the equation U-U star equal to identity,
it means that if you look at the inner product
of the first row of U into the inner product
of the first column of U star ok that is the
inner product of the first row of U with itself
then by looking at the right hand side it
follows that this is equal to one whereas
the inner product of the first row with the
inner product of the second column of U star,
third column of U star, etc the last column
of U star, these inner products are all zero.
This is the same as saying that the inner
product of the first row with the second row
the third row etc with the nth row that inner
product is zero, is that clear? So every row
of the matrix U if U is unitary has the property
that every row if you take the dot product
of the row with itself then it is a vector
norm 1 whereas the dot product of the vector
with every other row is zero. This means what?
The rows of a unitary matrix form an orthonormal
basis for C n. A similar argument can be given
by looking at U star U equal to the identity
to make the following statement that the columns
of a unitary matrix form an orthonormal basis
for C n.
A similar statement can be given for the case
of a orthogonal matrix that is a real unitary
operator. So for an orthogonal matrix the
rows and columns form an orthonormal basis
for R n for instance. But the case of a normal
operators much more general, the product is
not equal to identity it not so it not need
be invertible a normal operator need not be
invertible. So the statement about the entries
of a normal operator is not easy to make but
we will look at a particular case that is
the essence of the next theorem that is if
you know that an operator a normal operator
T has the property that the matrix of T relative
to a particular orthonormal basis is triangular
that is upper triangular or lower triangular
then we can say something.
Then it follows that the matrix must be a
diagonal matrix ok, so this is what we will
show, so let me make this statement precise.
So this is what I want to prove I will state
this as a theorem. Let T belong to L V where
V is the finite dimensional inner product
space. Suppose that there exists an orthonormal
basis I will call it script B orthonormal
basis B of V such that the matrix of T relative
to this basis equals A is let us say upper
triangular, till now I have not made any assumption
on T, T us just any operator with the property
that there exists an orthonormal basis with
a extra property that the matrix of T relative
to the orthonormal basis is upper triangular
or lower triangular it does not matter.
Then T is then we can characterize normality
of T, necessary sufficient condition for T
to be normal is that A is diagonal ok, when
T is normal if and only if A is diagonal.
As I told you this say something less about
the matrix A that is you need to make some
assumption on the matrix of T to an orthonormal
basis but nevertheless this will be useful
for us in proving the spectral theorem for
a normal operator ok, let’s see the proof
of this. So proof, let’s perhaps prove the
easy part here. Suppose that A is diagonal
by the way all the spaces are complex spaces.
Suppose that A is diagonal then any two diagonal
operators commute in two diagonal matrices
commute so I can say this much that A-A star
equals A star A this is always true any two
see is diagonal A star is also diagonal so
this happens this means what? Thus, what is
A? A is matrix of T relative to B and A star
is the matrix of A star is conjugate of the
matrix of T relative to B but then we know
we have proved this before that T B star is
T star B so I have this equation and on the
right hand side I will write A star A that
is T star B into the matrix of T relative
to B have this equation and there is a property
that we have proved for the product you can
replace that by the composition, composition
I have not used the circles I will simply
say that T P star relative to B is T star
T relative to B, now this a calculation that
we have done a number of times before.
What follows from this is that the operators
must be the same that is action of the operators
so I have two operators let us say S and T
the action of S on a basis equal to the action
of T on that basis then we have seen that
this defines S uniquely so from this it follows
that T-T star equals T star T. So is it clear?
So T is normal if A is diagonal then we have
shown the T is normal, let’s prove the converse
part, that is not straight forward. The converse
part is to show that if T is normal then A
is diagonal will do will consider one column
of the matrix A at a time and show that all
elements except the principle diagonal element
must be zero.
Ok so let T be normal and what is given I
am just and I am calling that as a matrix
A right, what is given is the following, A
equal to the matrix of T relative to B is
upper triangular so let me write the matrix
the form of the matrix it is upper triangular
so it is something like this, a 11, a 12 etc
a1 n this entry must be zero a 22, a23 etc
a2 n this entry must be zero this must be
zero etc let me write the last row 0, etc
0 this is a n n. So this is form of the matrix
A if T is normal I must now show that all
these entries are also zero, this is the principle
diagonal primarily we must show that all these
entries are zero ok so will do it one row
at a time ok.
Now let’s look at the relationship between
the entries of A and the matrix of T relative
to B. In other words I will just use the formula
for this equation, so this says T of U j equals
summation i equals 1 to n A I j Ui this is
true for all j, 1 less or equal to j, less
and or equal to n. Now this equation we have
encountered a number of times earlier. If
A is a matrix of T relative to B then this
is what this equation must hold. Where see
we are making the standard assumption that
what are these U j’s? U j’s form a basis
that is U1, U2 etc U n this is an orthonormal
basis where B is an orthonormal basis for
V, this is standard terminology.
Ok let’s now look at T U1, that is j equals
1, we have T U1 to be so j equals 1 so it
is ai1 u1 so T u1 is let me there just write
this a11 u1 plus a21 u2 plus etc an 1 u n
but see (a21) a23 etc a21 a31 etc an 1 they
are all zero so this is just a11 u1 ok, T
u1 equals a11 u1 I will appeal to one of the
properties that we have proved just now. By
the way what this means is that u1 is an eigenvector
corresponding to the eigenvalue a11 for the
operator T what follows is that u1 must also
be an eigenvector for the conjugate a11 bar
corresponding to T star so let me write that
equation.
So let me say that a11 bar u1 equals T star
of u1 and then look at T star of u1, what
is the formula for t star of u1? Similar to
what I wrote down for T U j I will write down
a similar equation I will write down an equation
for T star U j and then look at the particular
cases. Also T star of U j is summation i equals
1 to n this time it is a j i bar T star U
j it is a j i bar U y so where I have made
use of the fact that the matrix of the conjugate
of the matrix of T relative to B is the matrix
of the conjugate T star relative to the same
basis B.
Ok so I have this again 1 less and or equal
to j less and or equal to n in particular
look at j equals 1 and so let me go back to
this equation a11 bar u1 equals T star u1
which can now be computed as j is 1 so it
is a11 bar u1 plus a12 bar u2 plus etc plus
a1 n is that ok j is 1 and i varies from 1
to n. So a1 n bar u n. Now a11 bar u1 that
can be cancelled look at what remains, what
remains is a12 bar u2 etc p plus a1 bar u1
equal to 0.
Now u1 u2 etc u n form an orthonormal basis
in particular they are linearly independent
so subset is also linearly independent so
u2 etc u n is a linearly independent subset
so this is an equation like let us say alpha
to u2 etc plus alpha n Un equal to 0 so each
of the scalar must be zero so what follows
is that a12 bar so see the conjugate are zero
so the complex number also must be zero so
a12 equals a13 etc equals a1 n all this must
be zero. Observe what this are, this are precisely
the first row entries except the first entry.
So the first row entries have been shown to
be zero, what we have shown is that a12, a13
etc a1 n they are all zero in particular they
ok (in particular) see will appeal to induction
but we need to make one observation, in particular
a12 0 so if you look at the formula for T
u j when j equals 2 what happens? The formula
for T u j when j equals 2 gives me a22 u2
and that is like this equation I am sorry
yeah that is like this equation T u1 equals
a11 u1 so once I have a guarantee for that
then we can appeal to induction, that is I
will say that in particular what follows is
that T u2 equals a22 u2, because a12 is zero
and so I am looking at the second column,
all entries below a22 are already zero because
this is upper triangular matrix.
Now what follows is that u2 is an eigenvector
corresponding to the eigenvalue a22 for the
operator T so u22 will be an eigenvector corresponding
to eigenvalue a22 bar for the operator T star
I proceed as before, there are only finitely
many columns so it follows that a is diagonal.
So I will simply say by induction it follows
that a is diagonal ok, now this an important
intermediate result in order for us to prove
the spectral theorem for a normal operator
ok. Now before going to the spectral theorem
we will prove what is called as a Shur Triangularization
theorem ok Shur Triangularization theorem.
So let me write this theorem, this is called
Shur Triangularization theorem, the statement
is the following. Let V be a finite dimensional
inner complex inner product space and T be
any operator. Now this result is true for
any operator there are no conditions on T
normality is there is no such condition. So
what is the statement? Finite dimensional
complex inner product space T is a linear
operator on V then it says that you can do
a little less than diagonalization then there
exists an orthonormal basis 
script B of the vector space V such that the
matrix of T relative to B is upper triangular
ok.
For any operator T on a complex inner product
space what is important to observe is that
the eigenvalues belong to the underline field
because C is an algebraically closed field
ok. So let’s understand that there is no
counter part of this result for the case of
a real inner product space. However if you
know that T is an operator on a real inner
product space with the property that all the
eigenvalues of the operator T are real then
we can write down a similar statement ok.
So let’s understand the importance of the
fact that this is a underline field is a is
the complex field means it is an algebraically
closed field.
Ok will show that it is upper triangular,
the proof we be by induction ok. Proof by
induction on the dimension let’s call this
n the case n equal to 1 there is nothing n
equal to 1 it is trivial that is when n equal
to 1 what is means is that we want to show
that there is an orthonormal basis such that
T B is upper triangular there is just one
entry so 1 cross 1 matrix, now what is that
matrix, see this is where we are using the
fact this a complex inner product space there
is an eigenvalue for the operator T. In other
words let me say that there exists x not equal
to zero such that T x equals lambda x for
some complex number lambda.
We can call x1 as x by norm x, the Euclidean,
x is not zero so this is well defined then
it follows that T x1 equals lambda x1 with
norm x1 V in 1 you take the basis consisting
of just this single vector then the matrix
of T relative to B is the number lambda which
is trivially upper triangular. Let’s assume
that the result is true for finite dimensional
inner product spaces of dimension n minus
1 and then prove it for spaces of dimension
n. So in that it was the induction hypothesis
is made will prove this result for all finite
dimensional inner product spaces of dimension
n.
So let’s take dimension of V to be n, let
us now start with ok suppose dimension V is
n see we made a statement that the operator
T has an eigenvalue a similar statement holds
because underline field is complex a similar
statement can be made for T star so let me
exploide that. Let x not equal to 0 be such
that this time T star x equals lambda x for
some complex number lambda. Existence of an
eigenvalue and so there is by definition a
non-zero vector x satisfying this equation.
Let us now define W to be span of this vector
x and take its perpendicular. So W span x
perpendicular, then what follows is that then
obverse that x belongs to W perpendicular.
So what it means is that, T star of W perpendicular
is contained in W perpendicular that the space
W perpendicular is invariant under T star
because x is an eigenvector ok it is a one
dimensional W perpendicular is a one dimensional
subspace spanned by the single vector x ok.
Now T star W perpendicular is contained din
W perpendicular we know that this is the same
as saying that T of W is contained in W, this
one of the result that we proved earlier.
If T star is if W perpendicular is invariant
under T star then W perpendicular, perpendicular
must be invariant under T star-star that is
W is invariant under T.
So T W contained in so it makes sense to so
S from W to W defined by so I define a new
operator S this is defined as the restriction
of S to the restriction of S I am sorry the
restriction of T to W this is well defined.
So this operator is well defined S from W
to W is well defined. See all that I need
to know is that if you take an element W in
W whether S of W belongs to W ok but by the
definition S of W is T of that little W but
T of W is contained in W so this S is well
defined. This is another construction that
we have used when we discuss the self adjoint
operator case. So S is well defined, what
is the dimension of W? Dimension of W is n
minus 1, see it is orthogonal compliment of
a single vector so dimension W is n minus
1, I have an operator S on a finite dimensional
inner product space of dimension n minus 1
so the induction hypothesis is applicable
that is there exists an orthonormal basis
so let me say there exists an orthonormal
basis let me write down the elements as u1,
u2 etc maybe x1, x2 etc x n minus 1 right,
the space is of dimension n minus 1. Orthonormal
basis of W such that the matrix of S relative
to this x1 etc x n minus 1 is upper triangular.
Ok so I have this by the induction hypothesis.
Let just write this full and see what it says
ok but before that let’s I will make use
of this and then define a new vector x n as
the first vector x that we started with divided
by its norm. The vector x that we stared with
need not be of norm 1 so I will take that
divide it by its norm and call it x n then
it is clear by the construction that the vector
x1, x2 etc x n minus 1 which have been obtained
before and the vector x n that we have defined
newly through this vector x is an orthonormal
basis this is an orthonormal basis for V that
is we are writing V as W plus W perpendicular
and then for W this is the basis orthonormal
basis for W perpendicular x2 etc x n minus
1 is an orthonormal basis.
So come we know that when you combined this
two it give rise to an orthonormal basis for
V all that we need to do is to locate the
matrix of T relative to this basis script
B, which means what? You need to you take
T apply it on x1 write it as a linear combination
of x1, x2 etc x n minus 1 x n etc finally
take x n look at the image of T look at the
image of x n under T write it as a linear
combination of x1, x2, etc x n. Now T x1,
x2 etc x n minus 1 belong to W and the action
of T is like the action of S and for S I have
an orthonormal basis I am sorry the form of
the matrix of S relative to x1, x n minus
1 is upper triangular ok so let’s make use
of that.
So all that I am saying is that the matrix
of T relative to this basis B is of the form
let’s say a11 it is upper triangular for
the first a12 etc a1 n minus 1 then it is
upper triangular this entry is zero a22 etc
a2 n minus 1 etc all this entries are zero
a n minus 1 n minus 1 and then look at what
happens so this purely comes from the matrix
of S relative to x1, etc x n minus 1, what
is T of x n? T of x n, see x n belongs to
W I am sorry x n belong to W perpendicular
and so I have I make use of this invariant
subspace thing. So what follows is that the
last entry will be 0, 0 etc lambda. Ok just
check this steps, now it is now clear that
this is a required form.
And so the matrix so if T is normal I am sorry
if T is an operator on a finite dimensional
complex inner product space then the matrix
of T relative to B is upper triangular, this
is the so called Shur Triangularization theorem.
Let’s now move on to prove the spectral
theorem for a normal operator. Let T be a
normal operator on a finite dimensional complex
inner product space V then there exists a
basis script B of V such that the matrix of
T relative to B is diagonal. That is every
diagonal you can say this is the same as saying
that this is a matrix version of the statement
that if you have a complex normal matrix on
if you have a complex normal matrix then it
can be diagonalized by a unitary matrix.
There is also another way of stating this
which is that if T is a normal operator on
a complex finite dimensional inner product
space V then there is a basis orthonormal
basis B of V which has the property that each
vector in the basis is an eigenvector for
T we have seen this before, we have seen the
statement before so let me not emphasize ok,
how does the proof go? It makes use of the
previous two results the Shur Triangularization
theorem and the fact that if the matrix of
a linear transformation corresponding to an
orthonormal basis is triangular then the operator
is normal if and only if the matrix is diagonal
ok.
So let me say by Shur’s theorem, the Shur’s
theorem is applicable for any operator by
Shur’s theorem the matrix of I will simply
say there exists a basis such that the matrix
of t relative to B let us call it as A is
upper triangular. So that is Shur’s theorem.
Now A is upper triangular appeal to one of
the results the earlier result that we proved,
since T is normal 
it follows that A is diagonal which is what
we wanted to show that is there exists an
orthonormal basis B of the vector finite dimensional
vector space inner product space V with a
property that the matrix of T relative to
this basis is a diagonal matrix ok.
Now you must compare the statement with the
statement for a self adjoint operator. See
I will leave it as an exercise for you to
write down the matrix versions, the matrix
version is something that I told you immediately
after writing down this theorem so that is
anyway there. Now you must compare this with
the spectral theorem for a self adjoint operator
where the underline inner product space was
not assume to be complex ok.
To summarize this is what we have. If you
have a self adjoint operator on a finite dimensional
inner product space it doesn’t matter whether
it is a real inner product space or a complex
inner product space it always has the property
that there is an orthonormal basis for the
space V which has the which also satisfy the
additional condition that each vector of the
orthonormal basis is an eigenvector for the
operator T.
For the case of complex no for the case of
normal operators I have already made the statement
that if you have a normal operator on a real
space then it is not necessarily diagonalizable
that is if T satisfies T into T transpose
equal to T transpose into T then doesn’t
necessarily follow that T is diagonalizable
the example that was given was that of the
rotation operator which we also see which
we have also seen in the beginning of today’s
lecture. But if it is a complex inner product
space and T is a normal operator then there
is an orthonormal basis with a property that
each vector is an eigenvector ok let me stop.
