1. ## Formally Dual?...?

Consider two finite dimensional vector spaces $\displaystyle U$ and $\displaystyle V$ over complex numbers $\displaystyle \mathbb{C}$, their bases $\displaystyle e_i \in U, \ i=1,2,....,n$ and $\displaystyle f_i \in V, \ i=1,2,....,m$, their dual spaces $\displaystyle U^*$ and $\displaystyle V^*$, and their dual bases $\displaystyle e^i$ and $\displaystyle f^i$.

Two linear maps $\displaystyle T:U \rightarrow V$ and $\displaystyle S: V \rightarrow U$ are formally dual if:

$\displaystyle <T(u),v>=<u,S(v)> \ \forall v \in V, \ u \in U$

Prove that the linear maps $\displaystyle T$ and $\displaystyle T_U^{-1} T^*T_V$ are formally dual.

$\displaystyle T: U \rightarrow V$
$\displaystyle T^*: V^* \rightarrow U^*$ so that for each $\displaystyle \alpha \in V^*, \ T^*(\alpha)$ is an element of $\displaystyle U^*$ defined by $\displaystyle T^*(\alpha)(u)=\alpha(T(u)) \ \forall u \in U$.

$\displaystyle T_U: U \rightarrow U^*$ defined by $\displaystyle T_U(w)(u)=<w,u> \ \forall w,u \in U$ (I know that $\displaystyle T_U(e_i)=e^i$)
It's probably better to start with the RHS and get the LHS.

ie. $\displaystyle <T(u),v>=<u,T_U^{-1}T^*T_V(v)>$

It's also clear that $\displaystyle S=T_U^{-1}T^*T_V:V \rightarrow V^* \rightarrow U^* \rightarrow U$ or $\displaystyle S(v)=T_U^{-1}T^*T_V(v)$

Take a basis vector of $\displaystyle V$, call it $\displaystyle f_i$. Do the same for $\displaystyle U$, call the basis vector $\displaystyle e_j$.

I have to get $\displaystyle <T(e_j), f_i>=<e_i,T_U^{-1}T^*T_V(f_i)>$

We get:

$\displaystyle <e_j, S(f_i)>=<e_j, T_U^{-1}T^*f^i(v)>$ since $\displaystyle T_U(e_i)=e^i$ and i'm now working on a vector in $\displaystyle V$. The linear map $\displaystyle f^i$ also has to act on a vector, so it's acting on some fixed $\displaystyle v \in V$.

$\displaystyle <e_j, S(f_i)>=<e_j, T_U^{-1}T^*f^i(v)>=<e_j, T_U^{-1}T^* f^i T(u)>$

Here $\displaystyle f^iT \in U^*$ but linear maps are defined on how they work on a vector so I need to put a $\displaystyle u$ in there.

Apply $\displaystyle T$ to both sides to get:

$\displaystyle <e_j, S(f_i)>=<e_j, T_U^{-1}T^*f^i(v)>=<e_j, T_U^{-1}T^*f^i T(u)>=<T(e_j), T T_U^{-1}T^*f^i T(u)>$

From here i'm absolutely lost! Does anyone have any ideas??

(P.S This is very hard!)

2. Originally Posted by Showcase_22
It's probably better to start with the RHS and get the LHS.

ie. $\displaystyle <T(u),v>=<u,T_U^{-1}T^*T_V(v)>$

It's also clear that $\displaystyle S=T_U^{-1}T^*T_V:V \rightarrow V^* \rightarrow U^* \rightarrow U$ or $\displaystyle S(v)=T_U^{-1}T^*T_V(v)$

Take a basis vector of $\displaystyle V$, call it $\displaystyle f_i$. Do the same for $\displaystyle U$, call the basis vector $\displaystyle e_j$.

I have to get $\displaystyle <T(e_j), f_i>=<e_i,T_U^{-1}T^*T_V(f_i)>$

We get:

$\displaystyle <e_j, S(f_i)>=<e_j, T_U^{-1}T^*f^i(v)>$ since $\displaystyle T_U(e_i)=e^i$ and i'm now working on a vector in $\displaystyle V$. The linear map $\displaystyle f^i$ also has to act on a vector, so it's acting on some fixed $\displaystyle v \in V$.

$\displaystyle <e_j, S(f_i)>=<e_j, T_U^{-1}T^*f^i(v)>=<e_j, T_U^{-1}T^* f^i T(u)>$

Here $\displaystyle f^iT \in U^*$ but linear maps are defined on how they work on a vector so I need to put a $\displaystyle u$ in there.

Apply $\displaystyle T$ to both sides to get:

$\displaystyle <e_j, S(f_i)>=<e_j, T_U^{-1}T^*f^i(v)>=<e_j, T_U^{-1}T^*f^i T(u)>=<T(e_j), T T_U^{-1}T^*f^i T(u)>$

From here i'm absolutely lost! Does anyone have any ideas??

(P.S This is very hard!)

To begin with, a good idea would be, perhaps, to tell us what are $\displaystyle T_U\,,\,T_V$ ...

Tonio

3. Let U be finite dimention Hilbert space, for any $\displaystyle t \in U$, associates $\displaystyle t^{*} \in U^*$ defined by $\displaystyle t^*(x) = < t, x >$. (Define $\displaystyle v^*$ Similarly in V)
Denote the mapping $\displaystyle {\pi} : U^* --> U$ :
$\displaystyle v^{*}T=t^*-->t$ ,Since Hilbert space is self-dual,the mapping is injective,
then RHS
$\displaystyle =<u,T_{U}^{-1}T^{*}T_{V}(v)>$
$\displaystyle =<u,T_{U}^{-1}T^{*}(v^{*})>=<u,T_{U}^{-1}(v^{*}T)>$
$\displaystyle =< T_{U}^{-1}(v^{*}T),u>=<t,u>=t^*(u)$
$\displaystyle = v^*T(u) = < v, T(u)> = < T(u), v>$
= LHS