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I am trying to solve this problem but I fail in the converse part...

Show that the product of two self-adjoint operators is self-adjoint if and only if the two operators commute.

($\Rightarrow$) If we suppose that $T,U,V$ are self-adjoint operators such that $T=UV$ and $U^*=U,V^*=V,T^*=T$ we have by Theorem that:

$$(UV)^*=V^*U^*=VU$$

but we know, by hypothesis, that: $$(UV)^*=T^*=T=UV$$.We already know that the adjoint operator is unique, so we hav that $UV=VU$ and we are done with this part.

So my troubles are this the converse, can anyone help please.

Björn Friedrich
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ViKaN-HND504
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3 Answers3

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Assume $UV = VU$, and $U,V$ are self-adjoint. $$(UV)^*=V^* U^* = V U = UV \implies (UV)^* = UV$$ i.e, $UV$ is self adjoint. Hopefully that helps.

Jeb
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  • But why are you assuming thath $U$ and $V$ are self-adjoint? for the if and only if you just have to suppose that $UV=VU$ right? – ViKaN-HND504 Nov 03 '14 at 03:48
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    No no, let U and V be self adjoint. Then their product is self adjoint iff U and V commute. – Jeb Nov 03 '14 at 04:12
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The statement of the theorem could be drastically improved. It should read:

Let $U$ and $V$ be self-adjoint operators.

Theorem. The product $UV$ is self-adjoint if and only if $UV = VU$.

Proof. ($\impliedby$) Suppose that $U$ and $V$ commute. We aim to show that $(UV)^* = UV$. The defining characteristic of the adjoint is the following equation: $$ \langle UV x,y\rangle = \langle x,(UV)^*y\rangle,\qquad\text{for all $x,y$.} $$ By our assumption that $U$ and $V$ commute, we have \begin{align*} \langle UV x,y\rangle &= \langle VU x,y\rangle \\ &= \langle U x, V^*y\rangle \\ &= \langle x,U^*V^*y\rangle, \end{align*} where we have applied the defining characteristic of the adjoint twice. Using the fact that $U$ and $V$ are self-adjoint, so that $U^* = U$ and $V^* = V$, we deduce $$ \langle UVx,y\rangle = \langle x,UVy\rangle,\qquad \text{for all $x,y$.} $$ By uniqueness of the adjoint, $UV = (UV)^*$.$\qquad\square$

Alex Ortiz
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Suppose $UV=VU$. Then, for any $x,y$ in the given space $$ 0=\langle UVx,y\rangle - \langle VUx,y\rangle=\langle x,V^*U^*y\rangle - \langle x,U^*V^*y\rangle = \langle x, (V^*U^* - U^*V^*)y\rangle .$$ By the Zero-Operator Lemma (i.e., $T=0$ iff $\langle Tx, y\rangle =0$ for all $x,y$), $V^*U^* - U^*V^*=0$. Since $U^*=U$ and $V=V^*$, we have $0=V U - U^*V^*= VU - (VU)^*.$

Teodorism
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