Dual Representation, dimension mismatch? I am working out of Representation Theory, A First Course, Fulton and Harris. In particular I am in Part 1, Section 1.1 where all groups are taken to be finite and all vector spaces are finite dimensional $\mathbb{C}$- vector spaces.
They define the (..a?) representation on $\mathrm{Hom}(V,W)$ by identifying $\mathrm{Hom}(V,W) = V^{*} \otimes W$, since they have already shown how the dual of a representation and the tensor product of representations can be made into a representation. I am familiar with the identification mentioned above (as vector spaces), but I would like to verify the induced action (as a representation) on $\mathrm{Hom}(V,W)$. In order to do so I first need to understand the representation on $V^{*}$. This leads to my first question. 
Given a $G$-representation $V$, with homomorphism $\rho$, the text defines the dual representation (verbatim, page 4) as 
$$\rho^{*}(g) = \, ^{t}\rho(g^{-1})\colon V^{*} \to V^{*}.$$

First of all, is there a reason the transpose is written on the left? and further if $\mathrm{dim}(V) = n$, then if I am interpreting the above
  correctly $ \, ^{t}\rho(g^{-1})$ is (can be identified with) an $n$ x
  $n$ invertible matrix. Given $\lambda \in V^{*}$, $\lambda \colon V
 \to \mathbb{C}$, $\lambda$ can be identified with a 1 x $n$ matrix. Is
  the action of $g\lambda$ computed by identifying $\lambda$ with its
  transpose which would appear as a vector in $V$, then computing the
  product $^{t}\rho(g^{-1})$$\lambda$ which outputs a new vector in $V$,
  which can be transposed to be seen as a new map $V \to \mathbb{C}$?
  Hopefully my question here is clear. I will reformulate if it is not.

Now, assuming I do understand the representation on $V^{*}$ and the representation on a tensor product of representations, in the case of $V^{*} \otimes W$ we would have $$\begin{align*}g(\lambda \otimes w) &= g\lambda \otimes gw\\
&= \rho_{1}^{*}(g) \lambda \otimes \rho_2(g)w\\
&= \, ^{t}\rho_{1}(g^{-1})\lambda \otimes \rho_2(g)w.\\
\end{align*}$$
The final line should justify why we define the action of $g$ on $\phi \in \mathrm{Hom}(V,W)$ as $g\phi = g\phi(g^{-1}v)$?. This definition should be such that the appropriate diagram commutes. Going one direction through the diagram I get that $\lambda \otimes w$ maps to the linear map from $V$ to $W$ given by $\lambda(v)w$ then after being acted on by $g$ becomes the map that takes $v$ to $g(\lambda(g^{-1}v)w$, I do not see how this is the same as if I traveled the other direction in the diagram. 
 A: I like the following way of thinking about the representation of $V^*$:
suppose $\lambda:V \to \Bbb{C}$ is a linear functional and $g \in G$. Then $\rho^*(g)(\lambda)=\lambda \circ \rho(g^{-1})$. Also, if we are suppressing $\rho$ and $\rho^*$ and if $v \in V$, then $(g\lambda)(v)=\lambda(g^{-1}v)$. This is the same as what is written in the book and a little easier to think about. 
I do not know why they wrote the $^t$ on the left. If you are thinking in terms of matricies, then $[\rho^*(g)\lambda]=[\lambda][\rho(g^{-1})]$
In this case, we get that:
$$\begin{align*}g(\lambda \otimes w) &= g\lambda \otimes gw\\
&= \lambda \circ g^{-1}\otimes gw\\
\end{align*}$$
which would correspond to the map which takes $v$ to $\lambda(g^{-1}v)gw$. Call the map corresponding to $\lambda \otimes w$, $\phi$. then $g\phi(g^{-1}v)=g(\lambda(g^{-1}v)w)=\lambda(g^{-1}v)gw$ (since $\lambda(g^{-1}v)$ is a scaler) which is what we expected. Note that not every map in $\text{Hom}(V,W)$ corresponds to a pure tensor $\lambda \otimes w$. But they do correspond to linear combinations of such tensors (you may be familiar with this though).
