# Formally Dual?...?

• Nov 25th 2009, 10:46 AM
Showcase_22
Formally Dual?...?
Quote:

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:

$= \ \forall v \in V, \ u \in U$

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

ie. $=$

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 $=$

We get:

$=$ 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 $v \in V$.

$==$

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:

$===$

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

(P.S This is very hard!)
• Nov 25th 2009, 02:41 PM
tonio
Quote:

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

ie. $=$

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 $=$

We get:

$=$ 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 $v \in V$.

$==$

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:

$===$

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
• Nov 25th 2009, 08:40 PM
Shanks
Let U be finite dimention Hilbert space, for any $t \in U$, associates $t^{*} \in U^*$ defined by $
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
$=$
$==$
$=< T_{U}^{-1}(v^{*}T),u>==t^*(u)$
$= v^*T(u) = < v, T(u)> = < T(u), v>$
= LHS
• Nov 26th 2009, 09:52 AM
Shanks