Define a linear map $T:V \to V$ by $T(v)=\left(v,u_{1}\right)u_{2}$. Here $v,u_{1},u_{2}\in V$. Write down a formula for the adjoint map $T^{*}$. Problem: Let $V$ denote an inner product vector space. Define a linear map $T:V \rightarrow V$ by $T(v)=\left(v,u_{1}\right)u_{2}$. Here $v,u_{1},u_{2}\in V$. Write down a formula for the adjoint map $T^{*}$.
Attempt: Let $ x,y \in V $ be arbitrary. $ ( Tx,y ) =  ( (x,u_1)u_2,y ) = (x,u_1)(u_2,y) = \overline{ (u_1,x) }(u_2,y) =(x,T^*y)$
Now, $ \overline{ (u_1,x) }(u_2,y) =(x,T^*y) \iff \overline{ (u_1,x) }(u_2,y) - (x,T^*y) = 0 \iff \overline{ (u_1,x) }(u_2,y) - \overline{(T^*y,x)} = 0 \iff \overline{ (u_1,x)}\cdot \overline{\overline{(u_2,y)}} - \overline{(T^*y,x)} = 0 \iff  \overline{ (u_1,x) \cdot \overline{(u_2,y)} - (T^*y,x) } = 0  \iff \overline{ (\overline{(u_2,y)} \cdot u_1,x)  - (T^*y,x)} = 0 \iff \overline{ (\overline{(u_2,y)} \cdot u_1-T^*y,x) } =0 \iff  (x,\overline{(u_2,y)} \cdot u_1-T^*y)  =0   $
[ From here I've stopped ]
Official solution:
Soultion: For all $v, w \in V$ we have
$$
\left(v, T^{*} w\right)=(T v, w)=\left(\left(v, u_{1}\right) u_{2}, w\right)=\left(v, u_{1}\right)\left(u_{2}, w\right)=\left(v, \overline{\left(u_{2}, w\right)} u_{1}\right)=\left(v,\left(w, u_{2}\right) u_{1}\right)
$$
Since this is true for all $v \in V$, we conclude that
$$
T^{*} w=\left(w, u_{2}\right) u_{1}
$$
Questions:

*

*How do I continue from my own attempt to find the formula?

*In the official solution, what $v$ did the professor plug into $ ( v,T^* ) = ( v, (w,u_2) u_1 ) $ in order to conclude that $ T^{*} w=\left(w, u_{2}\right) u_{1} $ ?

Thanks in advance for help!

$$\langle T^*x,y\rangle = \langle x,Ty\rangle=\langle  u_1,y\rangle \langle x,u_2\rangle =
\langle\langle x,u_2 \rangle u_1,y\rangle
$$ We know that if $\langle u,y\rangle = \langle v,y\rangle,$ for any $y,$ then $u=v.$ Using that we may conclude
$$T^*x=\langle  x, u_2\rangle u_1$$
