Correct way of viewing a (1,1)-tensor returning a vector I am currently watching this excellent video series building up to general relativity. We have finally started looking at tensors and a question came up from the audience which (to my understanding) was asking why tensors are defined as multi-linear maps from sets of vectors and covectors to the real numbers yet (1,1)-tensors are often viewed as linear maps from vectors to vectors.
He says that the two ways of viewing the (1,1)-tensor contain the same information, but I don't understand how. If anyone could provide another explanation for why this is true I would really appreciate it.
 A: This kind of trick is common when thinking about tensors. 
Let $v$ be a vector, $\alpha$ a one-form, and $c$ a number. A vector gives a map from one-forms to numbers,
$$\alpha \mapsto v(\alpha)$$
but it also gives a map from numbers to vectors,
$$c \mapsto c v.$$
Similarly, a $(1, 1)$ tensor $T$ gives a map from a vector and covector to numbers,
$$(\alpha, v) \mapsto T(\alpha, v).$$
But it also gives a map from vectors to vectors,
$$v \mapsto T(\cdot, v).$$
To see that the output here is a vector, note that it gives a map from one-forms to numbers,
$$\alpha \mapsto T(\alpha, v).$$
By continuing with this, you can also use a $(1, 1)$ tensor to define a map from covectors to covectors, or numbers to a vector and covector, and so on. 
To avoid having to associate the exact same object with a ton of different maps, physicists use index notation. In this notation, the five preceding maps take the form
$$\alpha_i \mapsto v^i \alpha_i, \quad c \mapsto c v^i, \quad (\alpha_i, v^j) \mapsto T^i_j \alpha_i v^j, \quad v^i \mapsto T^i_j v^j, \quad \alpha_i \mapsto (T^i_j v^j) \alpha_i.$$
From this you can get the general pattern, which is that a $(n, m)$ tensor can act on a $(n', m')$ tensor to yield a $(n+n'-k, m+m'-k)$ tensor where $k$ is the number of upstairs/downstairs index pairs that are contracted.
A: A $(n,m)$ tensor eats $n$ vectors and $m$ co-vectors and spits out a real number. A $(1,1)$ tensor eats one vector and one co-vector and spits out a real number. Let's call our $(1,1)$ tensor $f(.,.)\colon V \times V^{*} \mapsto \mathbb{R} $. The first argument is a vector and the second argument a co-vector. So for generic vectors $a \in V, b \in V^{*}$, $f(a,b) \in \mathbb{R}$. 
What about $f(a,.)$? Clearly, it eats a co-vector (in the empty argument) and spits out a real number. But this is precisely the definition of a vector. So, $f(a,.)$ is a vector. Therefore $f(.,.)$ ate a vector(i.e $a$) and produced a vector $f(a,.)$.
