# Show that $f \circ g$ is self-adjoint iff $f \circ g = g \circ f$ in a euclidean vector space.

This question has been asked before HERE, but I could not understand the following result from the linked post:

$$\langle F(G(v)), w \rangle = \langle G(v), F(w) \rangle = \langle v, G(F(w)) \rangle$$

The rest of the proof of the "$$\Rightarrow$$" direction is clear to me, but I am also not so sure how to approach the "$$\Leftarrow$$" part of the proof, that is: $$f \circ g = g \circ f \Rightarrow f \circ g = (f \circ g)^*$$

## 1 Answer

For the first part, the adjoint of a linear operator $$T$$ is defined to be the unique operator $$T^*$$ such that: $$\langle Tv,w\rangle=\langle v,T^*w\rangle,\ \forall v,w\in V$$ Now, since $$F,G$$ are assumed to be selfadjoint, that is $$F=F^*$$ and $$G=G^*$$, we get: $$\langle F(G(v)),w\rangle=\langle G(v),F^*(w)\rangle=\langle v,G^*(F^*(w))\rangle=\langle v,G(F(w))\rangle$$

For the other direction, we have in general that $$(T\circ S)^*=S^*\circ T^*$$, so: $$F\circ G=F^*\circ G^*=(G\circ F)^*=(F\circ G)^*$$

• Hello! Thank you for your answer, but the statement only assumes $f \circ g$ to be self-adjoint, can we from this already conclude that $f$ and $g$ are self-adjoint? EDIT: I have to take this back, since we are actually assuming both to be self adjoint and not only the composite... I should have read more carefully! – The Tralfamadorian May 8 at 16:52
• Case closed then! Glad I could help! – SPetrakos May 8 at 17:05
• Yes, I think I'd still be an interesting question that I am not able to answer, but as far as this exercise is concerned, I am done. – The Tralfamadorian May 8 at 17:07
• Well, it has an easy answer: consider any $T\neq 0$ such that $T^2=0$ (e.g. an upper triangular $2\times 2$ matrix with zero diagonal and a non-zero element in the upper right corner). Then of course $T\circ T=T^2=0$ is selfadjoint, but $T$ itself is not. – SPetrakos May 8 at 17:28