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!

  • 1
    $\begingroup$ 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 $\endgroup$
    – kade
    Jan 3, 2021 at 20:13
  • 1
    $\begingroup$ Yes your last step is correct due to the definition of adjoint operator! $\endgroup$ Jan 3, 2021 at 20:18

1 Answer 1


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 $(ST)^*=T^*S^*$.


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