• CrazyIvan

## Homework Statement

Prove or give a counterexample: the product of any two self-adjoint operators on a finite-dimensional inner-product space is self-adjoint.

## Homework 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
$$\left(T \circ S \right)^{*} = S^{*} \circ T^{*}$$

## 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.

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.

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