# Proove that $ST$ has adjoint operator $(ST)^{*}$ and $(ST)^{*}=T^{*}S^{*}$

I have to prove this

Let be $$V$$ a inner product space over $$\mathbb{C}$$, with $$T$$ and $$S$$ lineal operators in $$V$$ with adjoints operators $$T^{*}$$ and $$S^{*}$$ respectively. Prove that $$ST$$ has adjoint operator $$(ST)^{*}$$ and $$(ST)^{*}=T^{*}S^{*}$$

I've done it by this way:

Let be $$\alpha, \beta \in V$$. First we need to observe that:

\begin{align} \left \langle ST \alpha,\beta \right \rangle&=\left \langle T\alpha,S^{*}\beta \right \rangle\\&=\left \langle \alpha,T^{*}S^{*}\beta \right \rangle \end{align} So, $$ST$$ has an adjoint lineal operator $$T^{*}S^{*}$$.

Now, to prove that $$(ST)^{*}=T^{*}S^{*}$$, I did this:

\begin{align} \left \langle \alpha,T^{*}S^{*}\beta \right \rangle&=\left \langle T\alpha,S^{*}\beta \right \rangle\\&=\left \langle ST\alpha,\beta \right \rangle\\&=\left \langle \alpha,(ST)^{*}\beta \right \rangle \ \ \ \ \ \ \text{I have doubt in this last step, is it correct?} \end{align}

\begin{align} \therefore (ST)^{*}=T^{*}S^{*} \end{align}

Is it correct my proof? I would really appreciate your help!

• It seems fine to me. The step you express doubt about is justified because you showed that $ST$ has an adjoint operator and that is just the definition of the adjoint operator
So what you have shown is that $$\langle \alpha, T^*S^*\beta\rangle = \langle \alpha, (ST)^*\beta\rangle$$ for all $$\alpha, \beta\in V$$. We can rewrite this as saying that for all $$\alpha, \beta\in V$$ we have that $$\langle \alpha, [T^*S^*-(ST)^*]\beta\rangle =0$$ Fix $$\beta\in V$$ and choose $$\alpha = [T^*S^*-(ST)^*]\beta$$ then we can rewrite the inner product as $$||[T^*S^*-(ST)^*]\beta||^2 = \langle [T^*S^*-(ST)^*]\beta, [T^*S^*-(ST)^*]\beta\rangle = 0$$ Now a norm is nonnegative and is only equal to $$0$$ for the $$0$$ vector. Thus $$||[T^*S^*-(ST)^*]\beta||^2=0$$ implies that $$[T^*S^*-(ST)^*]\beta=0$$ which means that $$T^*S^*\beta=(ST)^*\beta$$ Well $$\beta\in V$$ was arbitrary so the two operators are equal on the entire vector space, which means that $$(ST)^*=T^*S^*$$.