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Let $T:V \rightarrow V$ a linear map such that $Tv = <v,u>w$ then the adjoint linear map $T^*$ is $T^*v = <v,w>u. \forall u,v,w \in V $.

My professor defined the linear map $T^*$ as follow:

For any linear operator $T:V \rightarrow V$ on a finite-dimensional inner product space $V$, there is a unique operator $T^*$ on $V$ such that:

$<Tv,u> = <v,T^*u>$ for all $u,v \in V.$ $\hspace{4cm}(1)$

So my objective is arrives in $T^*v = <v,w>u. \forall u,v,w \in V $ using the equation $(1).$

Can you give me a hint about this? (I have tried substitute $Tv = <v,u>w$ in $(1)$ but isn't enough.)

Joãonani
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  • Try substituting both $\langle v, u \rangle w$ in for $Tv$ and $\langle v', w \rangle u$ in for $T^v'$ into the definition: $\langle Tv, v' \rangle = \langle v, T^v' \rangle$. I think maybe you could be getting confused, having $u$ appear as a dummy variable in $(1)$, yet it is an important fixed vector when defining $T$. – Theo Bendit May 03 '19 at 04:14
  • If I substitute both then they satisfies (1) so can I conclude the result or no? – Joãonani May 03 '19 at 04:25
  • If you can show that $(1)$ (or, the equivalent expression $\langle Tv, v' \rangle = \langle v, T^v' \rangle$ for all $v, v' \in V$) is satisfied for the given definition of $T$ and the proposed $T^$, then you've shown that the proposed operator is the adjoint of $T$, by definition of the adjoint of $T$. – Theo Bendit May 03 '19 at 04:27
  • I think there is a lot confusion in the notations and that is why you are facing difficulty. When $T$ is given $u$ and $w$ are fixed vectors. So 'for all $u$' in (1) causes confusion. If notations are set properly you will be able to write down a proof yourselves. – Kavi Rama Murthy May 03 '19 at 06:12

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