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In Linear Algebra working with Dual space and dual transformations I've come along this very basic definition of the dual transformations:

Suppose:

$T^*$ is a dual transformations from $W^*\to V^*$
$T$ is a linear transformation from $V\to W$
$u$ is a linear functional which belongs to $W^*$
$v$ is a vector which belongs to $V$

Then the following applies: $$ (T^*u)(v) = u(Tv) $$

Why is the dual transformation defined this way? (I know this is a very problematic question, but please any intuition will be very helpful)

vondip
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  • $T^$ goes from $W^$ to $V^*$ if $T:V\to W$. – wj32 Jul 03 '13 at 08:30
  • Usually, if $T: V\rightarrow W$, then $T:W^{}\rightarrow V^{}$, with $T^{}(\psi)(v)=\psi(T(v))$. You need to invert the arrow in the definition of $T^{*}$. – Avitus Jul 03 '13 at 08:31

2 Answers2

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In the special case of matrix algebra, this turns out to be fairly obvious.

In this setting, one usually writes vectors as $n \times 1$ matrices ("column vectors"), and linear functionals as $1 \times n$ matrices ("row vectors").

If we have a matrix $A$ with suitable dimensions, then "multiplication on the left" results in a linear transformation (call it $T$): i.e. $T(v) = Av$. The dual transformation is "multiplication on the right". That is, $T^*(u) = uA$. So your identity is merely

$$ (uA)v = u(Av) $$

  • Thanks for the response, while this does give a good intuition, I don't completely understand why T* is denoted by A and so is T. Why is it so that both of them have the same vectors in their matrix in the first place? – vondip Jul 04 '13 at 06:34
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    @vondip: Try turning the question around: we know for matrices it's useful to be able to multiply on the left and on the right. So how do you translate that to the abstract setting? The value of $(uA)$ at $v$ should be the value of $u$ at $Av$. The notation $T^(u)$ is, in some sense, an artifact of using the convention that linear transformations are always written on the left. If we didn't adopt that convention, we could write $(u)T$ or even just $uT$ rather than writing $T^(u)$. Of course, $u \circ T$ makes sense either way, and $T^*(u)$ does equal $u \circ T$... –  Jul 04 '13 at 08:03
  • And sometimes, we need to think of the space of linear functionals as a vector space in its own right, rather than as a space of functionals -- so that's another setting where we need notation to reflect the fact we've "transposed" how $T$ acts on things. One particular case of this sort of thing is when we have an inner product on $V$, and we use it to convert elements of $V$ into elements of $V^*$ instead. –  Jul 04 '13 at 08:06
  • Thank you very much Hurkyl, this is an excellent answer that helped me understand this vague topic. – vondip Jul 04 '13 at 08:50
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Given a map $T: V \to W$ where $V, W$ are $K$ vector spaces. We define the transpose or (dual transformation) $T^* : W^* \to V^*$ of $T$ by $T^*(f) = f \circ T$. I think it's defined this way, because it's the most natural way to associate to each linear map a new map on the corresponding dual spaces.

There's a reason this assignment is called the transpose. If $V$ is finite dimensional, any basis $\beta = \{v_1, \dots, v_n\}$ of $V$, gives rise to $n$ linear functionals $v^*_i : V \to K$ defined by $v^*_i(w) = a_i$ where $a_i$ is the $i$'th component of $[w]_B$. The collection $B^* = \{v^*_i : 1 \le i \le n\}$ is linearly independent and spans $V^*$, and therefore forms a basis which we call the dual basis of $\beta$. Fix ordered bases $\beta$ and $\gamma$ of $V$ and W, respectively. Assuming $W$ is finite dimensional, if $T: V \to W$ is linear map, we have a matrix representation of $T$ denoted $[T]_\beta^ \gamma$. It's fairly straightforward to prove that

$$[T^*]_{\gamma^*} ^{\beta^*} = ([T]_\beta^ \gamma)^t$$

where by $( \cdots )^t$ I mean the transpose of a matrix.

Some properties of taking the dual:

The assignment $V \mapsto V^*, T \mapsto T^*$ is what's called a contravariant functor (i.e. a functor that reverses composition and morphisms (linear maps)) on $\mathbf{Vect}_K$ the category of $K$ vector spaces.

If $V$ and $W$ are finite dimensional, $T \mapsto T^*$ is an injective linear map from $\operatorname{Hom}_K(V, W)$ to $\operatorname{Hom}_K(W^*, V^*)$ and hence an isomorphism since both spaces have dimesion $\dim V \dim W$.

Ink
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