$(1,1)$ tensor vs a linear transformation (matrix) Take $d$-dimensional Vector space $V$ with Field $R$. 
A typical linear algebra linear transformation $V \to V$ can be represented by a $d \times d$ matrix $A$ such that for some $v,w \in V$, $Av=w$.
I'm learning about tensors, and I understand that a $(1,1)$ tensor $T$ is a linear transformation $V^* \times V \to R$. I've read that such a $(1,1)$ tensor is equivalent to such a matrix. 
However, I find it very difficult to imagine what $V^*$ (the dual space, i.e. set of all maps $V\to R$) has to do with a simple linear transformation from $R^d$ to $R^d$. 
Moreover, the tensor components apparently are defined as $T^i_{\space \space j}=T(\epsilon_i, e^j)$, where $e^j, \epsilon _i$ are the $d$ bases of $V$ and $V^*$ respectively. This means that if we would write $T$ as a 2-dimensional array, it would have nothing to do with a matrix as in linear algebra.
So how are these two concepts connected?

This post is related to my question, but it doesn't really go into the difference between the matrix and tensor form.
 A: Given a linear map $\alpha:V\to V$ we can construct a bilinear form $\tau:V^*\times V\to R$, by taking $\tau(f,v)=f(\alpha v)$. (Note that $f(\alpha v)$ makes sense because $v\in V$ and $\alpha:V\to V$ so $\alpha v\in V$, and then $f\in V^*$ means $f:V\to R$, so $f(\alpha v)\in R$.)
Similarly given a bilinear form $\tau':V^*\times V\to R$ we can construct a map $\alpha': V\to V$ by noting that if $v\in V$ then $\tau'(-,v):V^*\to R$, and hence $\tau'(-,v)\in V^{**}$. Since $V$ is finite dimensional we have $V^{**}\cong V$ and hence we can define $\alpha'(v)$ to be the element of $V$ corresponding to $\tau'(-,v)$ in $V^{**}$. This means that $f(\alpha' v)=\tau'(f,v)$.
Hence given a map $V\to V$ we get a map $V^*\times V\to R$ and given a map $V^*\times V\to R$ we get a map $V\to V$ (and furthermore if we translate back and forth we end up where we started). So we can view linear maps $V\to V$ as "the same as" bilinear maps $V^*\times V\to R$.
Finally lets check the matrices are the same. Given a map $\alpha:V\to V$ its matrix is defined by $A^i_{\;j}=\epsilon_j(\alpha e^i)$, and given a map $\tau:V^*\times V\to R$ its matrix is defined by $T^i_{\;j}=\tau(\epsilon_j,e^i)$. So if we have $\tau(f,v)=f(\alpha v)$ then $T^i_{\;j}=\tau(\epsilon_j,e^i)=\epsilon_j(\alpha e^i)=A^i_{\;j}$.
A: Take $v\in V$ and $w_*\in V^*$. You have a natural bijection $i$ between $V$ and $V^*$ (because the dimension is finite), hence you can obtain the corresponding $i(w_*)=w\in V$.
Consider a map $Q:V\times V\to \Bbb  R$ given by
$$Q(x,y) = T(x,i(y)).$$
Obviously, $Q$ is a bilinear map, hence it can be given by a matrix $M$:
$$Q(x,y) = x^TMy.$$ We will call this matrix - by extension - the matrix of $T$.
