Let $V$ be a vector space of finite dimension $n$, and let $\operatorname{Bil}(V)$ be the vector space of all bilinear forms on $V$.
In some notes by Keith Conrad, he says in an exercise that $\operatorname{Bil}(V)$ has dimension $n$ as well. I am confused by this, because it seems to me that the dimension ought to be $n^2$. Here is my reasoning: we have isomorphisms $\operatorname{Bil}(V)\cong (V\otimes V)^* \cong V \otimes V$ since $V$ is finite dimensional. The dimension of $V\otimes V$ is $n^2$. Where have I gone wrong here?
You have not gone wrong at all. More concretely, a bilinear form $B$ is associated with an $n\times n$ matrix [given a basis $\{v_i\}$, its $ij$-entry is, of course $B(v_i,v_j)$].