Let V be a finite-dimensional vector space over R. Define V* to be the set of linear transformations T : V → R. Show that V is isomorphic to V*
I know that two vector spaces are isomorphic if they have the same dimension (or that there exists and invertible linear transformation from one space to the other), only I don't know how to show that these two spaces have the same dimension (or that such linear transformation exists).
I will flesh out the argument given by Matt Biesecker. You say you know that two (finite) vector spaces are isomorphic if they have the same dimension. We will show that the dimensions of $V$ and $V^*$ are the same.
Assume that the dimension of $V$ is $n$ (Note that if $n=0$ we have a trivially true case and so we will henceforth assume $n>0$). Now let us write a basis of $V$ as $\{v_1,v_2,\dots,v_n\}$. Now let us consider $n$ linear maps defined as $$w_i(v_j)=\begin{cases}1&\text{if }i=j\\0&\text{if }i\ne j\end{cases}$$
The goal is to prove that these maps serve as a basis of $V^*$. To this end, we first show that they span $V^*$. Let an arbitrary vector in $V$ be written as $$v=c_1v_1+c_2v_2+\dots+c_nv_n$$ and note that $$w_i(v)=w_i(c_1v_1+c_2v_2+\dots+c_nv_n)=c_1w_i(v_1)+c_2w_i(v_2)+\dots+c_nw_i(v_n)=c_i$$ Now, applying an arbitrary linear map $L$ to $v$ must give
$$L(v)=L(c_1v_1+c_2v_2+\dots+c_nv_n)=c_1L(v_1)+c_2L(v_2)+\dots+c_nL(v_n)$$
But $L(v_i)$ is just a real number, and we have $$L(v)=w_1(v)L(v_1)+w_2(v)L(v_2)+\dots+w_n(v)L(v_n)$$ Hence $L$ is a linear combination of the $w_i$s, so the $w_i$s span $V^*$.
It remains to show that the $w_i$s are linearly independent. To this end, suppose that $$d_1w_1+d_2w_2+\dots+d_nw_n=0$$ where $d_i\in\mathbb{R}$. Hence $$d_1w_1(v)+d_2w_2(v)+\dots+d_nw_n(v)=0$$ for every $v\in V$.
However, choosing $v=v_i$ gives us $$0=d_1w_1(v_i)+d_2w_2(v_i)+\dots+d_nw_n(v_i)=d_i$$ and so $$0=d_1=d_2=\dots=d_n$$ and so the $w_i$s are linearly independent and therefore form a basis (of $n$ elements), so $V$ and $V^*$ are isomorphic and we are done.