I have some confusions regarding index notation in tensor algebra.
Let's assume $\vec{v}$ is a vector belonging to vector space $V$. Choosing a basis set $\{\vec{e}_\nu\}$, $\vec{v}=x^\nu\vec{e}_\nu$ where $x^\nu$ is the corresponding contravariant coordinate. Now, let's say I want to express the same vector in another basis set $\{\vec{e'}_\mu\}$. Then, $\vec{v}=x'^\mu\vec{e'}_\mu$ where $x'^\mu$ is the coordinate in the new basis. We know how to transform from one set of coordinates to another one. It is given by:
$$x'^\mu=\frac{\partial x'^\mu}{\partial x^\nu} x^\nu \tag I $$
The transformation $T$ is known to be a (1,1) tensor with components $\frac{\partial x'^\mu}{\partial x^\nu}$. Just by looking at the components $\frac{\partial x'^\mu}{\partial x^\nu}$ how can one tell what the order of the (1,1) tensor is? I mean, are the components ${T^{\mu}}_{\nu}$ or ${T_{\mu}}^{\nu}$? Equivalently, does it belong to tensor product space $e_{\mu}\otimes e^{\nu}$ or $e^{\mu}\otimes e_{\nu}$? I know the answer is ${T^{\mu}}_{\nu}$, is this because $\nu$ is summed over in equation (I)?
The way I justify that ${T^{\mu}}_{\nu}$ is correct is as following. The tensor $T$ can be written as $T={T^{\mu}}_{\nu}\ e_{\mu}\otimes e^{\nu}$. Applying $T$ on a vector $\vec{v}=x^{\nu}e_{\nu}$ results in a geometrical object which I call $f$. ${f=T^{\mu}}_{\nu}\ e_{\mu}\otimes e^{\nu} \ x^{\nu}e_{\nu} $, now because the one-form $e^{\nu}$ acting on $e_{\nu}$ quals $1$ (or by defenition $e^{\nu}(e_{\nu})=1$), the final product is a vector in the vector space V which can be written as $x'^{\mu}e_{\mu}$. Is this way of thinking correct? Is this basically the reason behind contraction of indices?
My second question is regarding the inverse transformation or transforming the basis vectors. we know that:
$$e'_\mu=\frac{\partial x^\nu}{\partial x'^\mu} e_\nu \tag{II}$$
The inverse transformation has components $\frac{\partial x^\nu}{\partial x'^\mu}$ . Let's call it $Q$ for now, this is also a (1,1) tensor. Should the component be written as ${Q_\mu}^\nu$? This does not make sens to me, because the tensor $Q={Q_\mu}^\nu \ e^\mu \otimes e_\nu$ can not be applied on the vector $e_\nu$! Hence contraction operation can not be performed.
Now, let's recall that $Q$ is the inverse of $T$. Then ${(T^{-1})^\mu}_\nu$ makes sense to me because $T^{-1}$ would be in the space $e_\mu \otimes e^\nu$ and can be applied to the vector $e_{\nu}$ (contraction would be possible). However, there is a contradiction here. If $T^{-1}$ is the inverse, shouldn't it live in the reversed order tensor product space? I mean if $T$ belongs to $e_\mu \otimes e^\nu$ space, shouldn't $T^{-1}$ belong to $e^\mu \otimes e_\nu$?
The reason that I am very confused is because Physicists usually only care about components of tensors. Contraction is an operation defined blindly as disappearing of an index if it occurs both as an upper and lower index. However, I would like to see the reasoning behind this contractions. I would like to think of it as, dual vectors (linear one forms) acting on vectors from the original vector space therefore the index vanishes. Or as elements of double dual vectors $V^{**}$ acting on elements of the dual vector $V^*$, in case we have expressions like $e_\mu(e^\nu)$ (this is only the case when $V$ is finite dimensional, so $V$ and $V^{**}$ are isomorphic and can be thought of equal spaces). I am even not sure if I have to go that far and think of a contraction as elements of double dual acting on elements of the dual space!! Is my way of construction correct or wrong? Thinking this way has made me confused. I am not sure if this is even a correct way of thinking.
Thanks.