How do I tell the rank of the electric susceptibility tensor (and others)? I understand that a tensor is a multilinear map from 
$V^*\times\cdots\times V^*\times V\times\cdots\times V$ to $V$'s underlying field, where $V$ is a vector space and $V^*$ its dual. This is fine, and I know that I can also see it as an element in the space $V\otimes\ldots\otimes V\otimes V^*\otimes\cdots\otimes V^*$. I also know that a tensor has rank $(p,q)$ if its domain is the cartesian product of $p$ instances of $V^*$ and $q$ instances of $V$.
However, I cannot recognize the type of tensors in examples from examples in textbooks. For instance, the electric susceptibility tensor. I know that, given a vector $E$ (from the electric field, which is a vector field), it will give me another vector $P$ (the density of electric dipole moments), which may have direction different from $E$:
$$ P = \epsilon_0\chi E $$
So $\chi$ is a tensor (and $\epsilon_0$ is the permittivity of vacuum, "included so $\chi$ is dimensionless" -- says the textbook from where the example was taken).
So, I see that it must have order 2, but is it $(2,0)$, $(1,1)$ or $(0,2)$? How would I, just from a formula like that and a description of what the tensor means, tell its rank?
My guess is that it is just a $(1,1)$ tensor, hence no more than an ordinary linear map that we study in basic Linear Algebra (because both $E$ and $P$ are vectors that should represent directions and magnitude in the same space). Is this correct? But from the information given, "it transforms a vector into another vector", how can one conclude that it is linear, and then write the equation like that?
Also, in this case, how can I tell if $\epsilon_0$ is a tensor, and what type it has if it is really a tensor?
(The example above is from "Tensor Calculus for Physics", by Dwight Neuenschwander)
 A: Sorry, long answer, but I think this should explain some questions in the comments as well. Feel free to ask more questions.
Your conclusion seems correct to me, admittedly, this is my first time learning of this definition of rank. But, since $\chi$ is finite dimensional, and since we are dealing with a physical problem, we can assume that our underlying vector space is $\mathbb{R}^n$. 
The dual of the vectors in $\mathbb{R}^n$ would simply be the transpose. That is, column vectors map to row vectors and vice versa. Now, due to some properties about $\mathbb{R}^n$ (namely completeness), we can construct the operator $\chi$ by taking the outer product of two vectors in $V=\mathbb{R}^n$. The outer product is sort of the opposite of the inner product in that it maps $V \times V \mapsto V\times V$ rather than $V \times V \mapsto \mathbb{R}$. That is, in terms of vectors and matrices, the outer product maps two vectors to a matrix, whereas the inner product maps two vectors to a scalar.
Now, to bring this all together, the inner product can be represented as the juxtaposition of an $1\times n$ vector and an $n \times 1$ vector (assuming a finite dimensional vector space). Outer products on the other hand can be represented as the products of an $n \times 1$ vector and an $1 \times m$ vector, resulting in an $n\times m$ matrix. 
Thus, for your case we have a matrix $\chi$ that can be constructed as the outer product of two vectors in $\mathbb{R}^n$, due to the definition of the outer product, we require that one of the matrices be in the dual (transpose) space, and the other be in the normal space (non-transpose) space. That leads me to assume, using your provided definition of rank, that $\chi$ must have rank $(1,1)$.
