Does the definition of tensor include basis? From my understanding, a vector is a geometric object, which can be expressed as
$$ v = v_i e^i $$
where $v_i$ and $e^i$ are components and basis, respectively. 
It seems to me that many people, e.g. in wikipedia https://en.wikipedia.org/wiki/Tensor#Definition
regard tensor as its component, e.g.
$$T_{i_1...i_n}^{j_1 ... j_n}$$
After all, is the tensor defined as the component,
$$T_{i_1...i_n}^{j_1 ... j_n}$$
or including basis?
$$T = T_{i_1...i_n}^{j_1 ... j_n} e^{i_1...i_n}_{j_1 ... j_n}  $$
 A: Given any $\mathbb{R}$-vector space $V$ and its dual space $V^\ast$, a tensor of rank $(k,l)$ is a map
$$ T :  V\times\dots V\times V^\ast\times\dots\times V^\ast \to \mathbb{R}\tag{1}$$
where $V$ occurs $k$ times and $V^\ast$ occurs $l$ times, that is linear in each argument. Equivalently, it is an element of the tensor product $V^\ast\otimes\dots\otimes V^\ast\otimes V\otimes\dots\otimes V$.
Now, if we choose a basis $e_i$ for $V$ and the corresponding dual basis $e^i$ of $V^\ast$ defined by $e^i(e_j) = \delta^i_j$, the components of the tensor are obtained as
$$ {T_{i_1\dots i_k}}^{j_1\dots j_l} := T(e_{i_1},\dots,e_{i_k},e^{j_1},\dots,e^{j_l})\tag{2}$$
If we think of the tensor as an element of the tensor product, then the tensor product has a natural basis in the tensors $e^{i_1}\otimes\dots\otimes e^{i_k}\otimes e_{j_1}\otimes\dots\otimes e_{j_l}$ and the full tensor $T$ is described by
$$ T = {T_{i_1\dots i_k}}^{j_1\dots j_l}\left(e^{i_1}\otimes\dots\otimes e^{i_k}\otimes e_{j_1}\otimes\dots\otimes e_{j_l}\right)\tag{3}$$
with a sum over all indices on the r.h.s. implied. That both of these descriptions of the tensor are equivalent follows directly from using the dual relations $e^i(e_j) = e_j(e^i) = \delta^i_j$.
It does not matter whether a tensor is defined "in components" as in $(2)$, as an abstract map as in $(1)$ or as the sum of basic tensors as in $(3)$. All these notions are equivalent.
A: This isn't really a physics question, but while it stays up I can give a quick response:
Your are correct that a tensor is often defined in terms of its coordinate transformations, but this is not the only or most modern definition. Here is a coordinate-free definition:
A tensor of type (r,s) on a vector space $V$ and its dual $V^*$ is a complex-valued function on
$V_1\times V_2 \times ...V_r \times V^*_1 \times V^*_2 \times ... V^*_s$
(where $V_i$ are just numbered copies of the same vector space and likewise for $V^{*}_i$) that is linear in each argument.
Here $r+s$ is the rank of the tensor.
When you have an inner product, this formalizes the way that Einstein summation naturally suggests you think about tensors: a tensor of rank $n$ is an object that, when 'fed' $n$ vectors, spits out a scalar. The transformation properties can also be derived from this definition quite directly.
This definition is taken from the book An Introduction to Tensors and Group Theory for Physicists, which can offer much more elaboration on what this means.
