Proving that $f ^{*} $ is a linear map Here's another that I want to see if my reasoning was correct. 

Let $f:E \rightarrow F$ be a linear map and let $$f^* : \text{Hom}_K(F,K) \rightarrow \text{Hom}_K(E,K) $$ if we define  $f^*(u) = u \circ f$, prove that $f^*$ is a linear map. 

Originally my reasoning was that since $f^*(u) = u \circ f$ then by definition the two functions were inverse of each other (f and u I mean). f was onto and one-to-one and that's the only way that equality works But I think now that could be wrong.  
 A: Duality is very different from inversion. The problem does not mention anything about injectivity or surjectivity of $f$ either.
The problem asks you to show that $f^*$ is linear. You need to go back to what it means by $f^*$ being linear. That will involve the vector space structures on $\text{Hom}_K(F, K)$ and $\text{Hom}_K(E, K)$. I can expand a bit here:
$f^*$ is linear if for any given $u, v \in \text{Hom}_K(F, K)$ and any $c \in K$,
$$
f^*(u + cv) = f^*(u) + cf^*(v).
$$
The operations (addition and multiplication by a scalar) on the left-hand side are in $\text{Hom}_K(F, K)$, while the operations on the right-hand side are in $\text{Hom}_K(E, K)$. The equality sign is in $\text{Hom}_K(E, K)$. (You might want to think a little about what it means when two elements of $\text{Hom}_K(E, K)$ are equal.)
I hope this is enough hint.
A: No that isn't quite right.  
First note: $\text{Hom}_K(F, K)$ are linear maps from $F$ to $K$ (these maps are called linear functionals).  Similarly, for $\text{Hom}_K(E,K)$.  Moreover, these each form vector spaces over $K$ of their own!  So the "vectors" in these spaces are the linear functionals.  $\text{Hom}_K(F, K)$ is called the dual space to $F$ and sometimes denoted $F^*$ (that is where the $"*"$ is coming from in the problem) see here.
So now $f^*$ in the question is a linear map that is mapping a linear functional in $\text{Hom}_K(F, K)$ to a linear functional in $\text{Hom}_K(E, K)$.  So the $u$ in the problem is a linear functional in $\text{Hom}_K(F, K)$ and $f^*(u)$ is a linear functional in $\text{Hom}_K(E, K)$. This means that $f^*(u) : E \to K$ i.e. its domain is $E$ and notice that also $f : E \to F$ so that its domain is also $E$  
Now, $f^*(u) = u \circ f$ means that they agree on all points of their domain (notice that we do have the same domain).  In other words, for all $v \in E$ we have that
$$f^*(u)(v) = u \circ f (v) = u(f(v))$$
So now that we understand the maps a bit more, we want to check the conditions of being a linear map.  That is, we want for all $u_1, u_2 \in \text{Hom}_K(F, K)$ $$f^*(u_1 + u_2) = f^*(u_1) +f^*(u_2)$$
and for all $\alpha \in K$ $$f^*(\alpha u) = \alpha f^*(u)$$ 
I will do the first one and maybe you can do the second.
Let $u_1, u_2 \in \text{Hom}_K(F, K)$.  By the above to show $f^*(u_1 + u_2) = f^*(u_1) +f^*(u_2)$ we need to show they agree on all points of their domain, so let $v \in E$.  Then we have
$$f^*(u_1 + u_2)(v) = (u_1 + u_2) \circ f(v) = (u_1 + u_2)(f(v)) = u_1(f(v)) + u_2(f(v))$$ $$= u_1 \circ f(v) + u_2 \circ f(v) = f^*(u_1)(v) + f^*(u_2)(v)$$
which is what we wanted.  Do you see why $(u_1 + u_2)(f(v)) = u_1(f(v)) + u_2(f(v))$?  Remember, $u_1, u_2$ are functions . . . how do we define addition of functions?
