What type of mathematical object is the "Pauli vector"? The three pauli matrices $\sigma_x$, $\sigma_y$, $\sigma_z$ are sometimes combined in the "Pauli vector", usually denoted $\boldsymbol{\sigma} = \sigma_{x} \boldsymbol{e_x} + \sigma_y \boldsymbol{e_y} + \sigma_z \boldsymbol{e_z}$, the intuitive meaning of this is somewhat clear, for example, if one multiplies, for instance, this object by $\frac{\hbar}{2}$ one gets the operator that represents the spin angular momentum "vector" of an electron. This is a "vector operator", angular momentum is a vector in classical mechanics, kinda makes sense. On the other hand, what is the formal meaning of this? Is this a vector belonging to a 3-dimensional vector space over a "pseudo-field" of hermitian matrices? Is this a rank 3 tensor, some sort of 3D grid of numbers? What type of mathematical object is the Pauli Vector?
Apologies if the question is obvious or trivial, something inside me tells me that it may be, but I don't recall working with "vectors of matrices" in any linear algebra course and my efforts to makes sense of this have been in vain so far. Perhaps I have seen this but with a different name or context and I'm just not seeing the connection when written in this notation? 
 A: The matrix element of $\vec \sigma$ sandwiched between spinor wave functions is a vector, which makes it vector operator.
A: Rather than abstractly saying what the Pauli vector $\vec{\sigma}$ is,
it is more useful to say what you can actually do with it.
$\vec{\sigma}$ is a 3D vector operator.
That means calculating the expectation value
$$\vec{n}=\langle\Psi|\vec{\sigma}|\Psi\rangle$$
for a state $|\Psi\rangle$ gives a 3D vector.
When the state $|\Psi\rangle$ is normalized (i.e. $\langle\Psi|\Psi\rangle=1$),
then the resulting vector $\vec{n}$ is a unit vector (i.e. having length $1$).
Intuitively speaking the direction of $\vec{n}$
is the spinning axis of the particle represented by $|\Psi\rangle$.
A: From the physical perspective, Pauli vector is a special case of a quantum mechanical operator corresponding to an observable of vectorial nature. Any classical vectorial quantity will have such an operator.
If you have an $N-$dimensional Hilbert space every operator $\hat{O}$ can be written down as a collection of $N\times N=N^2$ complex numbers $O_{nm}$ subject to certain conditions (e.g. hermiticity, if you write down these numbers as a matrix): $\hat{O}=\sum_{n,m} O_{nm} |n\rangle\langle m| $. If the physical observable $O$ in question is a vector $\vec{O}=\sum_{\alpha=1}^d O^\alpha {\bf e}_\alpha$ with corresponding transformation rules for the components, then in fact there are $d$ observables, $d$ being the dimensionality of your geometrical (not Hilbert) space. Each observable $O^\alpha $ has its own corresponding operator $\hat{O}^\alpha=\sum_{n,m} O^{\alpha}_{n,m} |n\rangle\langle m|$.
To summarize, in an $N-$dimensional Hilbert space, an operator corresponding to a vectorial observable can be thought of as a collection of $N^2$ complex-valued vectors $ \vec{R}_{i \in \{1,...,N^2 \}}$.
A: This is best understood using group theory. The rotation algebra $so(3)$ contains both spinor and vector representations. Multiplying two copies of the spin 1/2 representation yields the vector and scalar representations, and the vector part is the Pauli vector.
In ordinary terms: a "vector" is defined as anything with 3 indices that transforms as a vector under rotations: $v^i \rightarrow \sum_j R^{ij}(\theta^1, \theta^2, \theta^3) v^j$, where $R$ is a rotation matrix. You can verify that the quantity $A^i \equiv \psi^\dagger \sigma^i \psi$ is a vector, where $\psi$ is a 2-component spinor and $\sigma^i$, $i=1,2,3$ are the Pauli matrices.
For example, consider an infinitesimal rotation around the z-axis:
$$\begin{aligned}
A^i=\psi^\dagger \sigma^i \psi \rightarrow \psi^\dagger e^{-i\sigma^3 \theta^3/2} \sigma^i e^{i\sigma^3 \theta^3/2} \psi &= \psi^\dagger (1-i\sigma^3 \theta^3/2) \sigma^i (1+i\sigma^3 \theta^3/2) \psi \\
&=\psi^\dagger \sigma^i \psi - i\theta^3/2\psi^\dagger\left[\sigma^3,\sigma^i\right]\psi \\
&=\psi^\dagger \sigma^i \psi + \theta^3\epsilon_{3ik}\psi^\dagger\sigma^k\psi
\end{aligned}$$
so $A^3\rightarrow A^3$, $A^2 \rightarrow A^2 -\theta^3A^1$, and $A^1\rightarrow A^1+\theta^3 A^2$.
