How to prove Hom$(U,V)\cong U^∗ \otimes V$ I know this result is well-known, but could some one give me some help with this proof - or some reference? Thank you.
Also, I have taken second year linear algebra (Axler's Linear Algebra Done Right), but I still find much trouble understanding Hom(U,V), like what is its basis; and no idea about tensor product.
So should I consider "a third year linear algebra"? Though there is not really a course named so - but is there a course or book I should read after Axler? Or I can learn more about these in some other course, like functional analysis? Is Representation theory related/necessary at this level?
Thanks!
 A: As Tyler writes, the map which is defined on decomposable tensors by
$$\phi \otimes v \mapsto (u \mapsto \phi(u) v)$$ is an isomorphism---at least, when $U$ and $V$ are finite dimensional. To see this, note that if $u_1^*,\dots,u_m^*$ is a basis for $U^*$ and $v_1,\dots,v_n$ is a basis for $V$, then the tensors $u_i^* \otimes v_j$ are a basis of the tensor space $U^* \otimes V$, and these map to the matrix units, which are a basis of the Hom space $\mathrm{Hom}(U,V)$. 
[In fact, the same argument works assuming only that $U$ is finite dimensional, and the result is false, without imposing more structure, if $U$ is infinite dimensional.]
A: As noted already in the comments, define
$$\phi: U^*\otimes V\to \text{Hom}(U,V)\;\;\text{by}\;\;\phi(f\otimes v)(u):=f(u)v\,,\,\,\text{and extend def. by linearity}$$
I'll leave it to you to check this is actually a homomorphism between vector spaces, but for example
$$\phi\left(t\sum_{k=1}^rf_k\otimes v_k\right)(u)=\phi\left(\sum_{k=1}^rf_k\otimes (tv_k)\right)(u):=\sum_{k=1}^rf_k(u)(tv_k)=t\sum_{k=1}^rf_k(u)v_k=\ldots$$
with, of course, $\,t\in\Bbb F=$ the definition field.
As an idea for the isomorphism thingy,  check that 
$$\dim_{\Bbb F}\text{Hom}(U^*,V)=\dim_{\Bbb F}U^*\cdot\dim_{\Bbb F}V=\dim_{\Bbb F}U^*\otimes V$$
so that you only need to prove the above map is either injective or surjective (because the finite dimensions). Choose basis $\,A=\{u_1,...,u_n\}\;,\;\{v_1,...,v_m\}\;$ of $\,U,V\;$ resp., and let $\,\{f_1,...,f_n\}\,$ be the dual basis of $\,A\,$ above in $\,U^*\,$ , meaning: $\,f_i(u_j)=\delta_{ij}:=$ the delta of Kronecker, then
$$\phi\left(f_i\otimes v_k\right)(u_j):=f_i(u_j)v_k=\delta_{ij}v_k$$
Check the above makes the map $\,\phi\,$ surjective and thus automatically an isomorphism.
