I am kind of stuck in the following questions:
1.Show that if V is a complex inner-product space, then the set of
self-adjoint operators on V is not a subspace of L(V)
2.Show that if V is a real inner-product space, then the set of
self-adjoint operators on V is  a subspace of L(V)
Thanks

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Request for Question Clarification by livioflores-ga on 04 Apr 2003 14:40 PST

Hello joannehuang!!

I see this question and I will work on this; I only need that you can
wait for the answer a few hours, because I am going to the hospital to
see my father, who had a scheduled knee operation. I will go with my
notes and papers.

One more thing:
Can you tell me your definition of "self-adjoint operators" (and an
example is possible) in order to know wich way must take to solve this
problem, for example using the fact that the self-adjoint
transformation can be represented by self-adjoint or Hermitian matrix.

Thank you for your understanding.
Regards.
livioflores-ga

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Clarification of Question by joannehuang-ga on 04 Apr 2003 15:38 PST

Dear livioflores-ga :
First, I really appreciate your help! I hope your father will get well
soon.

An operator T belong to L(V) is called self-adjoint if T=T*. For
example,
if T is the operator on F^2 whose matrix( with respect to the standard
basis )is
 2  b
 3  7
then T is self-adjoint if and only if b =3 (because M(T)=M(T*) if and
only if b=3; recall that M(T*) is the conjugate transpose of M(T).
You should verify that the sum of two self-adjoint operators is
selg-adjoint and that the product of a real scalar and self-adjoint
operator is self-adjoint.

I am not sure is this what you need. Thanks again!

Hello joannehuang!!!

Letīs work!!

Question 1.
"Show that if V is a complex inner-product space, then the set of
self-adjoint operators on V is not a subspace of L(V)."


If we have a self-adjoint operator T belong to L(V), then T = T* and
in consequence M(T) = M(T*), where M(T*) is the conjugate transpose of
M(T).
Note that M(T) and M(T*) are both suqere matrices.

Remember: 
The conjugate matrix of T is the matrix  obtained by taking the
complex conjugate of each element of T.
The transpose matrix of T is the matrix obtained by replacing all
elements
T(i,j) with T(j,i).  

Then if we note the elements of M(T) as T(i,j) and the elements of
M(T*) as
T*(i,j), we have the following relationship between the elements of
the both matrices:
T*(i,j) = conjugate(T(j,i)) = T(i,j)     [1]

In particular, for the elements of the diagonal of T (the elements
T(i,i)) we have:
T*(i,i) = conjugate (T(i,i))
Because V is a complex space all the elements of T are complex
numbers, in particular if we have:
T(i,i) = a + i.b  
Then
conjugate(T(i,i)) = a - i.b 
Then by [1], for a diagonal element of a self-adjoint matrix:
a + i.b = a - i.b ==>  b = 0 ,

The conclusion is: 
The diagonal elements of a complex self-adjoint matrix (Hermitian
matrix) are pure real elements.

Now it is easy to prove that if V is a complex inner-product space,
then the set of self-adjoint operators on V is not a subspace of L(V).
In effect, if T is a self-adjoint operator on V, the matrix M(T) is a
self-adjoint matrix and it must have pure real diagonal elements.
If we multiply the matrix M(T) by the complex scalar i the diagonal
elements of the matrix iM(T) are not pure real in general that means
the operator i.T is not a self-joint operator in general (when the
diagonal elements are not all zero), then the set of self-adjoint
operators on V is not a subspace of L(V).

-----------------------------------------------------------

Question 2.
"Show that if V is a real inner-product space, then the set of
self-adjoint operators on V is  a subspace of L(V)."


Following the same reasoning of the question 1, we have the following
relationship between the elements of M(T) and M(T*):

T*(i,j) = conjugate(T(j,i)) = T(i,j)               [1]

Because V is a real space the elements T(I.j) are real numbers, then
they satisfy  conjugate(x)=x  for all number x in the set of the real
numbers.
Then we obtain from [1]:
T*(i,j) = T(j,i) = T(i,j)

This means that if a real squere matrix M is self-adjoint, then M is
symmetric.

It is easy to see that if a real matrix M is symmetric, then M is
self-adjoint:
M*(i,j) = conjugate(M(j,i)) 
        = M(j,i)           (because M is real)
        = M(i,j)           (because M is symmetric)

Now we know that self-adjoint matrix and symmetric matrix are
equivalent definition for real square matrices.

Now we only must prove that the set of the symmetric matrices is a
subspace of L(V):
If A and B are symmetrics then A(i,j) = A(j,i) and B(i,j) = B(j,i) ;
If C = A + B, then 
C(i,j) = A(i,j) + B(i,j)
       = A(j,i) + B(j,i)    (because A and B are symmetrics)
       = C(j,i) ;

Then C is a symmetric matriz and in consequence C is a self-adjoint
matrix.

If c is a real number and C = c.A ; then:
c.A(i,j) = c.A(j,i)

Then C is a symmetric matrix, and in consequence C is a self-adjoint
matrix.

Then if V is a real inner-product space and T , U are self-adjoint
operators on V, the matrix c.M(U) is self-adjoint, then the matrix M(T
+ c.U) is self-adjoint, in consequence the operator (T + c.U) is a
self-adjoint operator.

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I hope this helps you. Please tell me how this answer works, and if
you need a clarification feel free to post a request for it.
Also if you think that this is not the answer that you need, let me
know that.


It is always a pleasure to answer questions for you.
Best Regards.
livioflores-ga

That's very helpful. Thank you!