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!
 A: Fix $u_{1},u_{2}\in V$. $T(v)=\langle v,u_{1}\rangle u_{2}$.
Then
$$\langle x,T^*(y)\rangle =\langle T(x),y\rangle=\langle \langle x,u_{1}\rangle u_{2},y\rangle= \langle cu_{2},y\rangle $$. Where $c=\langle x,u_{1}\rangle$
Then you can bring $c$ outside:-
$$\langle x,u_{1}\rangle\langle u_{2},y\rangle=\langle x,u_{1}\rangle\cdot c'=c'\langle x,u_{1}\rangle $$
Where $c'=\langle u_{2},y\rangle$.
Now again you can bring $c'$ inside but do it for the second component. i.e
$$\langle x,\overline{c'}u_{1}\rangle $$.
Then  you have $\langle x,z_{1}\rangle =\langle x,z_{2}\rangle$ for all vectors $x\in V$ . So $z_{1}=z_{2}$ by positive definiteness .
So $T^*(y)=\bar{c'}u_{1}=\overline{\langle u_{2},y\rangle} u_{1} = \langle y,u_{2}\rangle u_{1},\,\,\forall y\in V$
So the unique formula for $T^*$ is given by
$$T^*(y)=\langle y,u_{2}\rangle u_{1}$$
I am going to just give you the standard construction for the adjoint:-
First fix an orthonormal basis $\{v_{1},v_{2},...v_{n}\}$ of $V$.
Then you can observe that for any linear functional $g:V\to\mathbb{F}$. you can uniquely define it to be $\displaystyle g(x)=\langle x,\sum_{i=1}^{n}\overline{g(v_{i})}v_{i}\rangle$ .
You can easily verify this. as if I define $\displaystyle h(x)=\langle x,\sum_{i=1}^{n}\overline{g(v_{i})}v_{i}\rangle$.
Then $\displaystyle h(v_{j})=\langle v_{j},\sum_{i=1}^{n}\overline{g(v_{i})}v_{i}\rangle=g(v_{j}),\forall 1\leq j\leq n$
Now fix a vector $y\in V$ and Define $g:V\to\mathbb{F}$ by $g(x)=\langle T(x),y\rangle$ . From the previous observation we have the unique vector $y'=\sum_{i=1}^{n}\overline{g(v_{i})}v_{i}$ .
Then define the operator $T^*:V\to V$ such that  $T^*(y)=y'=\sum_{i=1}^{n}\overline{g(v_{i})}v_{i}$.
Then $$\displaystyle T^*(y)=\sum_{i=1}^{n}\overline{g(v_{i})}v_{i}=\sum_{i=1}^{n}\langle y,T(v_{i})\rangle v_{i}$$.
Then $$\displaystyle \langle x,T^*(y)\rangle = \left\langle x,\sum_{i=1}^{n}\langle y,T(v_{i})\rangle v_{i}\right\rangle$$
$$=\sum_{i=1}^{n}\overline{\left\langle y,T(v_{i})\right\rangle}\langle x, v_{i}\rangle=\sum_{i=1}^{n}\left\langle T(v_{i}),y\right\rangle\langle x,v_{i}\rangle$$
Now if $\displaystyle x=\sum_{i=1}^{n}c_{i}v_{i}$. We would have $\displaystyle \langle x,v_{i}\rangle = c_{i}$
Then we get
$$\sum_{i=1}^{n}\left\langle T(v_{i}),y\right\rangle\langle x,v_{i}\rangle=\sum_{i=1}^{n}\langle c_{i}T(v_{i}),y\rangle\\=\langle T(\sum_{i=1}^{n}c_{i}v_{i}),y\rangle=\langle T(x),y\rangle$$ as needed.
I did this extra step so that you can understand what was going on. Otherwise it just follows from how we defined $g$ and from the above observation that $T^*$ satisfies the condition $\langle T(x),y\rangle=\langle x,T^*(y)\rangle$
You can verify that this $T^*$ is linear and all that.
With all this said . If you want to calculate $T^*$ when you are given $T$. The easiest and standard way of doing it would be using matrices. $[T^*]_{\mathcal{B}}=[T]_{\mathcal{B}}^*$. That is the matrix of $[T^*]_{\mathcal{B}}$ equals the adjoint(conjugate transpose of the matrix of $[T]_\mathcal{B}$ with respect to an orthonormal basis $\mathcal{B}$. You can directly prove this as the $ij$-th entry of this matrix is $\langle T^*(v_{j}),v_{i}\rangle=  \langle v_{j},T(v_{i})\rangle = \overline \langle T(v_{i}),v_{j}\rangle= \overline{a_{ji}}$ which is the ij-th entry of the conjugate transpose of $[T]_{\mathcal{B}}$.
So given a linear operator $T$. You should first find the matrix of $T$ wrt some orthogonal basis(The standard ordered basis would work fine). And then calculate the conjugate transpose of that matrix and you will get the matrix of $T^*$ and using it you can figure out the actual explicit description of $T^*$. Try out some excercises to clear your concepts.
A: $$\langle T^*x,y\rangle = \langle x,Ty\rangle=\langle  u_1,y\rangle \langle x,u_2\rangle =
\langle\langle x,u_2 \rangle u_1,y\rangle
$$ We know that if $\langle u,y\rangle = \langle v,y\rangle,$ for any $y,$ then $u=v.$ Using that we may conclude
$$T^*x=\langle  x, u_2\rangle u_1$$
