$(U\circ T)^{*} = T^{*}\circ U^{*}$

Question

Let $T : V \longrightarrow W$ and $U : W \longrightarrow Z$ be linear maps. How do I prove that $(U\circ T)^{*} = T^{*}\circ U^{*}$? I'm used to seeing $V^{*}$ not $(U\circ T)^{*}$. Any help is appreciated.

$^{*}$ denotes the dual map (transpose).

Answer 1

Consider the following data:

\begin{array}{lcl} T:V\rightarrow W & \leadsto & T^*:W^*\rightarrow V^* \\ U:W\rightarrow Z & \leadsto & U^*:Z^*\rightarrow W^* \\ UT:V\rightarrow Z & \leadsto & \color{red}{(UT)^*:Z^*\rightarrow W^*} \\ \\ & \Downarrow & \\ \\ V\overset{T}{\rightarrow} W & & \color{red}{Z^*}\overset{U^*}{\rightarrow}W^* \\ {\tiny{UT}}\searrow~\downarrow {\tiny{U}}& & \color{red}{{\tiny{T^*U^*}}\searrow}~\downarrow {\tiny{T^*}} \\ ~~~~~~~~~~Z & & ~~~~~~~~~~~~\color{red}{V^*} \end{array}

Note:

\begin{array}{rcl} V^*&=&\{\varphi_V:V\rightarrow F, \text{$\varphi_V$ linear}\} \\ \\ W^*&=&\{\varphi_W:W\rightarrow F, \text{$\varphi_W$ linear}\} \\ \\ Z^*&=&\{\varphi_Z:Z\rightarrow F, \text{$\varphi_Z$ linear}\} \end{array}

Thus, we have that

\begin{eqnarray} (UT)^*(h)=h(UT),~h\in Z^* \end{eqnarray}

by the definition of dual map (tranpsose) and

\begin{eqnarray} h(UT)=(hU)T \end{eqnarray}

by the associativity of transformations. Now, $hU$ is characterized by the map

\begin{eqnarray} W\overset{U}{\rightarrow}Z~~ \\ {\tiny{hU}}\searrow~\downarrow {\tiny{h}} \\ F~~, \end{eqnarray}

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so $hU\in W^*$, and it is clear then that

\begin{eqnarray} (hU)T=T^*(hU), \end{eqnarray}

but $hU=U^*h$, or simply $U^{*}$, because $h$ itself is an element of $Z^*$ for which $U^*$ maps to $W^*$.

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Hence, $(UT)^*=T^*U^*$, as desired. $\square$