Thinking of a linear operator as a (1,1) tensor I am reading that a linear operator $A$ can be thought of as a (1,1) tensor [where $(r,s)$ corresponds to $r$ vectors and $s$ dual vectors]. This can be done by saying
$$A(v,f) \equiv f(Av)$$
where $v$ is a vector and $f$ is a dual vector.
I assume the $Av$ in the argument on the rhs corresponds to the linear operator $A$ acting on the vector $v$. But this is where I get mixed up. What is the dual vector for $Av$ if $A$ is a (1,1) tensor? And isn't a tensor supposed to be a multilinear map that results in a number? If that is the case, why does $Av$ presumably result in a vector that goes in the argument of $f$ on the rhs? What is $Av$ here?
 A: Let $V$ be a finite-dimensional vector space over a field $\Bbb{F}$, and $\alpha$ be an endomorphism on $V$, meaning $\alpha:V\to V$ is a linear map. This gives rise to a $(1,1)$ tensor $A:V\times V^*\to\Bbb{F}$ defined for each $(v,f)\in V\times V^*$ as
\begin{align}
A(v,f):=f(\alpha(v)).
\end{align}
On the RHS, $v\in V$, so $\alpha(v)\in V$ so plugging into $f$ gives $f(\alpha(v))\in\Bbb{F}$. You can now easily check that $A$ is bilinear.
Conversely, given a $(1,1)$ tensor $A$ on $V$, we can obtain an endomorphism $\alpha$ by a two step process

*

*first, consider the linear map $\tilde{A}:V\to V^{**}$ defined as $\tilde{A}(v)= A(v,\cdot)$. Note that the slot which is left empty is where we input an element $f$ of $V^*$ and get a scalar output, so this object really does belong to the double dual.

*Next, it is a wonderful fact about finite-dimensionality that the map $\iota:V\to V^{**}$, $\iota(v) = [f\mapsto f(v)]$ (i.e $\iota(v)$ is the 'evaluation on $v$' map... which btw is actually a finite-dimensional version of formally defining the Dirac delta) is an isomorphism. Actually, this map is always injective, and in finite-dimensions, the two spaces have the same finite dimension, so the map is an isomorphism (a simple consequence of the rank-nullity theorem).

*We combine the previous statements to get the endomorphism $\alpha=\iota^{-1}\circ \tilde{A}:V\to V$. So, $\alpha(v)=\iota^{-1}(A(v,\cdot))$, i.e $\alpha(v)$ is the vector which corresponds naturally to the double dual vector $A(v,\cdot)$.


So, I've told you how given an endomorphism $\alpha$, we can get a corresponding $(1,1)$ tensor $A$, or more formally, I've defined a mapping $\Phi:\text{End}(V)\to T^1_1(V)$. Next, I've told you how to go the other way, i.e I told you how given a $(1,1)$ tensor, we get an endomorphism, so more formally I gave you a mapping $\Psi:T^1_1(V)\to \text{End}(V)$.
To really complete the argument, you should prove that $\Phi,\Psi$ are both linear maps, and they are inverses to each other $\Phi\circ\Psi=\text{id}_{T^1_1(V)}$ and $\Psi\circ \Phi=\text{id}_{\text{End}(V)}$. I leave it to you to verify these details. It is only once we establish this fact that we can say that $(1,1)$ tensors can be naturally identified with endomorphisms (note that the two spaces have the same dimension, namely $(\dim V)^2$, so are isomorphic, but the emphasis here is on the naturality of the isomorphism).
A: I believe the confusion lies in a common abuse of notation of denoting by the same symbol the $(1,1)$ tensor and the associated linear operator. Let $A:V\to V$ be a linear operator on the vector space $V$. Define the following $(1,1)$ tensor $\widetilde{A}:V\times V^\ast \to \mathbb{K}$, where $\mathbb{K}$ is the underlying field of scalars,
$$\widetilde{A}(v,\phi)=\phi(Av)\tag{1}.$$
In this scenario we started with some linear operator $A$. Given that, we know how to evaluate $Av$ for any $v\in V$. Therefore we know how to evaluate $\phi(Av)$ for any $\phi\in V^\ast$. So we have everything we need to evaluate (1) for any pair $(v,\phi)\in V\times V^\ast$. That this is a multilinear map can be easily checked.
This establishes a linear map ${\cal I}:{\rm End}(V)\to T^1_1(V)$ where ${\rm End}(V)$ is the space of linear operators in $V$ and $T^1_1(V)$ the space of $(1,1)$ tensors in $V$. The map is just ${\cal I}(A)=\tilde{A}$.
Now we would like to find its inverse. To do so we rewrite (1) in a convenient way. Recall that covectors are defined to be linear functionals $\phi:V\to \mathbb{K}$, in that case they naturally act on vectors $v\mapsto \phi(v)$. The thing is that if you fix a vector $v$ and define $\Phi(v):V^\ast\to \mathbb{K}$ as $\Phi(v)(\phi)=\phi(v)$ then the map $\Phi(v)$ is a linear functional on $V^\ast$. This is just to say that we can either view $\phi(v)$ as $\phi$ acting on $v$ or as $v$ acting on $\phi$. In fact, in some texts authors even write this as $\langle \phi,v\rangle$ to notationally emphasize this point.
So look to (1) and, instead of viewing it as the action of $\phi$ on $Av$, look at it as the action of $Av$ on $\phi$. In other words, using the map $\Phi$ we defined
$$\widetilde{A}(v,\phi)=\Phi(Av)(\phi)\tag{2}.$$
Since this is valid for any $\phi$ we are left with an equality of maps $V^\ast \to \mathbb{K}$:
$$\widetilde{A}(v,\cdot )=\Phi(Av)\tag{3}.$$
Now observe that this $\Phi$ is a map $\Phi:V\to (V^{\ast})^\ast$. In finite-dimensional vector spaces one may show that this map is a linear isomorphism. In that case, applying its inverse $\Phi^{-1}$ to the above equation one obtains $Av$ starting with $\widetilde{A}$:
$$Av = \Phi^{-1}(\widetilde{A}(v,\cdot))\tag{4}.$$
This defines ${\cal I}^{-1}:T^1_1(V)\to {\rm End}(V)$. All this abstract story, however, can be written in components, so let us do so. Let $\{e_a\}$ be a basis of $V$ and $\{\omega^a\}$ be the dual basis of $V^\ast$. Start with $A:V\to V$. This operator is defined by its components in the basis, $A e_a = A_a^{\phantom a b}e_b$. Likewise, a $(1,1)$ tensor $\widetilde{A}$ is also defined by its components:
$$\widetilde{A}(v,\phi)=\widetilde{A}(v^ae_a,\phi_b\omega^b)=v^a\phi_b \widetilde{A}(e_a,\omega^b)=v^a \phi_b \widetilde{A}_{a}^{\phantom{a}b},\quad \widetilde{A}_{a}^{\phantom{a}b}=\widetilde{A}(e_a,\omega^b)\tag{5}.$$
Given $A$ we construct $\widetilde{A}$ following (1). Its components are:
$$\widetilde{A}_a^{\phantom{a}b}=\widetilde{A}(e_a,\omega^b)=\omega^b(Ae_a)=\omega^b(A_a^{\phantom{a}c}e_c)=A_a^{\phantom{a}c}\omega^b(e_c)=A_a^{\phantom{a}c}\delta^b_{\phantom{b}c}=A_a^{\phantom{a}b}\tag{6}$$
so we see the components are exactly the same. Obviously had we started with $\widetilde{A}$ and constructed $A$ following (4) the result would be the same since we have already argued that (4) construction gives exactly the inverse of the map defined in (1) we studied above.
So in the end of the day, droping notational distinction between $A$ and $\widetilde{A}$ is an abuse of notation, very common in Physics, motivated by the fact that the two objects are related by a linear isomorphism and, in particular, that they have exactly the same components when one works in terms of components (which is also very common in Physics). From the Linear Algebra perspecive, though, it is important to understand that what is really going on is what is described above. So my advice is this: know when something is an abuse of notation, learn to get used to it, but make sure you understand what is actually happening behind it.
