Formally Dual?...?

**Showcase_22** · November 25th 2009, 09:46 AM

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

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

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

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

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

$T_U: U \rightarrow U^*$ defined by $T_U(w)(u)=<w,u> \ \forall w,u \in U$ (I know that $T_U(e_i)=e^i$ )

It's probably better to start with the RHS and get the LHS.

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

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

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

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

We get:

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

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

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

Apply $T$ to both sides to get:

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

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

(P.S This is very hard!)

**tonio** · November 25th 2009, 01:41 PM

Originally Posted by Showcase_22

It's probably better to start with the RHS and get the LHS.

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

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

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

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

We get:

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

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

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

Apply $T$ to both sides to get:

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

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 $T_U\,,\,T_V$ ...

Tonio

**Shanks** · November 25th 2009, 07:40 PM

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

**Shanks** · November 26th 2009, 08:52 AM

For further information, please refer to "Adjoint Operator".

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