1. Apr 18, 2008

### CrazyIvan

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

2. Relevant equations
The only two equations I've used so far are:
$$\left\langle T v, w\right\rangle = \left\langle v, T^{*}w\right\rangle$$
and
$$\left(T \circ S \right)^{*} = S^{*} \circ T^{*}$$

3. The attempt at a solution
I started with the definition of adjoint, using $$T \circ S$$ :
$$\left\langle \left(T \circ S \right) v, w \right\rangle = \left\langle v, \left(T \circ S \right)^{*} w \right\rangle$$
Using the second relevant equation, I got
$$\left\langle \left(T \circ S \right) v, w \right\rangle = \left\langle v, \left(S \circ T \right) w \right\rangle$$
So the composition is self-adjoint only if $$T$$ and $$S$$ commute. But I don't know where to go from there.

2. Apr 18, 2008

### Dick

Take R^2 with the usual basis and inner product. Then self adjoint operators are the symmetric 2x2 matrices. Can you find two of those that don't commute? It's not that hard. When you find them also notice that their product isn't symmetric.

3. Apr 18, 2008

### CrazyIvan

It was, in fact, quite easy to find two such matrices. It only took a few minutes .