1. Jul 22, 2009

### evilpostingmong

1. The problem statement, all variables and given/known data
Prove or give a counterexample: the product of any two selfadjoint
operators on a finite-dimensional inner-product space is

2. Relevant equations

3. The attempt at a solution
I'd say that if we let a diagonal matrix represent T (after all, its transpose representing
T*=the matrix representing T) and multiply it by a diagonal matrix representing
the transformation S, then we'd end up with a diagonal matrix as a product.
So the product is self adjoint since all diagonal matrices are equal to their
transposes.
Another case is with an nxn matrix where all entries are equal. This matrix represets T and its
transpose is T*. Its matrix = its transpose so its self adjoint. Now multiplying it with another nxn matrix
representing S with all entries equal to each other would obviously produce a matrix with all entries
equal to each other. Or multiplying the matrix for T with a diagonal matrix would produce a diagonal matrix.

Last edited: Jul 22, 2009
2. Jul 22, 2009

### Cyosis

$$\left(\begin{matrix} 2 & 0 \\ 0 & 1 \end{matrix}\right) \left(\begin{matrix} 1 & 1 \\ 1& 1 \end{matrix}\right)$$

I must say I don't quite understand what you're trying to do. What is this transformation S, how is it relevant? Also there exist more matrices that are self adjoint than just diagonal matrices and matrices where every entry is the same. You're given two self adjoint operators on a finite-dimensional inner product space. You must prove that a product of two such operators is also self adjoint on that inner product space.

So try to answer these questions mathematically that is without using any words.

What is the definition of self adjoint?
What do you want to prove?

Last edited: Jul 22, 2009
3. Jul 22, 2009

### evilpostingmong

Yeah self adjoint means Tv=T*v so T does the same to v as T* does.
You map from V to V to make this possible. I didn't really know that there were other matrices besides
the ones I mentioned that were self adjoint. Thats why I attempted the proof the way I did
since I thought those were the only types of matrices. I mean for T=T*, the matrix for T needs to equal
its transpose for T*, thats what I thought. I just started learning this stuff last night, so I guess I jumped
in too quickly. I thought it was easy as that.

Last edited: Jul 22, 2009
4. Jul 22, 2009

### Cyosis

This doesn't really have anything to do with whether you just learned it or not. This has to do with your methodology. When you start a proof and are given a certain object, you use the definition of that object. You don't add something to it because you think that it may be true. If you want to add some extra property to an object you better prove that it is actually true before assuming it.

I will answer those two questions myself:
1) $A=A^*$
2)if $A=A^*$ and $B=B^*$ then $AB=(AB)^*$

You want to prove 2). However 2) is only true in very special cases you can see that by using the inner product. Now all we have used is what you're given we didn't invent any irrelevant stuff that only makes a proof more opaque and most likely wrong.

Its conjugate transpose. For real valued matrices this is the same as the transpose.

5. Jul 22, 2009

### evilpostingmong

Ok thanks! So it isn't always true for the product of two matrices representing
self adjoints to be self adjoint by the example you gave in post 2.

6. Jul 22, 2009

### HallsofIvy

Staff Emeritus
The problem is that (AB)*= B*A*= BA for if A and B are both self adjoint, not AB.

Saying that A is self-adjoint means that <Au, v>= <u, Av> for any u and v in the vector space. Now, <ABu, v>= <A(Bu), v>= <Bu, Av>= <u, BAv>, not <u. ABv>.

7. Jul 22, 2009

### Cyosis

Well yes one counter example is enough to discard a lemma. What worries me is that you managed to prove that it is true while it is not true.

Using the objects given by you, your proof should have lead you to,

Let A and B be self adjoint, we want to prove that AB is self adjoint as well, that is $AB=(AB)^*$.

Using the inner product:
$$<v,(AB)^*w>=<(AB)v,w>=<Bv,A^*w>=<v,B^{*}A^{*}w>=<v,(BA)w>$$

Therefore $(AB)^*=BA\neq AB$. When does $BA=AB$ hold?

Last edited: Jul 22, 2009
8. Jul 22, 2009

### cipher42

Right. That's exactly what the question is asking you to find out. You cannot assume you know the answer, you either have to find a counterexample to show that it is not true in some specific case, or make a logical argument (aka find a proof) to show that it must be true in all situations.