Anti-correlations in the EPR experiment I'm trying to understand the section "EPR and the Bell inequality" from the book Quantum Computation and Quantum Information by Nielson and Chuang. While going through the box titled "Anti-correlations in the EPR experiment," I have trouble understanding the following line:

It turns out that no matter what choice of $v$ we make, the results of the two measurements are always opposite to one another. That is, if the measurement on the first qubit yields $+1$, then the measurement on the second qubit will yield $−1$, and vice versa.

While the authors go on explaining why this is true, I doesn't quite fit with my understanding by performing computations.
Specifically, as the authors show, first we show that $$\frac{|01\rangle-|10\rangle}{\sqrt{2}} = \frac{|ab\rangle-|ba\rangle}{\sqrt{2}},$$ up to an unobservable global phase factor, where $|a\rangle$ and $|b\rangle$ are the eigenstates of $v.\sigma$. Now, the text says 

...if a measurement of $v.\sigma$ is performed on both qubits, then we can see that a result of $+1$ ($−1$) on the first qubit implies a result of $−1$ ($+1$) on the second qubit.

That means, if we apply $v.\sigma\otimes v.\sigma$ on the state we get $$(v.\sigma\otimes v.\sigma)\frac{|ab\rangle-|ba\rangle}{\sqrt{2}} = -\frac{|ab\rangle-|ba\rangle}{\sqrt{2}}.$$ How does this justify the above statement?
Also, on the same page of the book, it's mentioned that 

Alice performs a measurement of spin along the $v$ axis, that is, she measures the observable $v.\sigma$.

Shouldn't then we be using the operator $v.\sigma\otimes I$ instead of $v.\sigma \otimes v.\sigma$?
May be I'm misunderstanding something. Please help me sort this out. Thanks!
 A: To measure the a single-qubit observable, you rotate the qubit to align that observable's axis with the computational basis (e.g. align it along the Z axis) then do a measurement. In other words, for any single-qubit observable $O$ you measure $M_O$ by instead performing a single-qubit operation $u$ then a Z-basis measurement $M_Z$. For all $M_O$, there exists a $u$ such that $M_Z \cdot u \equiv M_O$.
Now consider the case where two parties are measuring $M_O$ on their respective parts of a singlet state. They perform $M_O$ by applying $u$ and then measuring $M_Z$. However, the singlet state is not changed by both parties applying $u$, i.e. $(u \otimes u) \cdot (|01\rangle - |10\rangle)$ gives $|01\rangle - |10\rangle$ again (up to global phase). Therefore you can just drop those operations without changing the expected outcome:
$$\begin{align}
(M_O \otimes M_O) \cdot (|01\rangle - |10\rangle)
&=
(M_Z \cdot u) \otimes (M_Z \cdot u) \cdot (|01\rangle - |10\rangle)
\\
&=
(M_Z \otimes M_Z) \cdot (u \otimes u) \cdot (|01\rangle - |10\rangle)
\\
&\propto
(M_Z \otimes M_Z) \cdot (|01\rangle - |10\rangle)
\\
&\rightarrow
\text{measurements give opposite answers}
\end{align}$$
Because the singlet state gives opposite answers in the computation basis, and the singlet state is not affected by doing a basis change to both qubits, it must give opposite answers in every basis.
A: I found that section very difficult to follow myself and their claims are not at all obvious. The way I understand it is as follows. We start with
\begin{equation}
\psi = \frac{|01> - |10>}{\sqrt 2}
\end{equation}
which actually is a shorthand for following probability amplitude vector:
\begin{equation}
\frac{1}{\sqrt 2} \begin{pmatrix}
0 \\ 1 \\ -1 \\ 0
\end{pmatrix}
\end{equation}
Now as an exercise it can be verified that there is no way above vector can be written as a tensor product of two unit vectors $\begin{pmatrix} a \\ b \end{pmatrix}$ and $\begin{pmatrix} c \\ d \end{pmatrix}$. This shows the two qubits are entangled.
The purpose and objective of the subsequent discussion seems to be to show that no matter what unitary transformation one applies to $\psi$ it is impossible to un-entangle the two qubits. So ideally this claim should be proved in context of a general unitary matrix $U$.
But the section asks us to consider measuring the observable $\vec{v} \cdot \vec{\sigma}$ on each qubit. What does this statement even mean?
First, we have to construct the unitary matrix corresponding to the transformation. We start with:
\begin{equation}
\vec{v} \cdot \vec{\sigma} = v_1 X + v_2 Y + v_3 Z = \begin{pmatrix} v_3 && v_1 - iv_2 \\
v_1 + iv_2 && -v_3 
\end{pmatrix}
\end{equation}
where $X, Y, Z$ are the Pauli matrices.
Next "to measure the observable $\vec{v} \cdot \vec{\sigma}$ on each qubit" we need to apply following matrix:
\begin{equation}
M = \vec{v} \cdot \vec{\sigma} \otimes \vec{v} \cdot \vec{\sigma} =
\begin{pmatrix}
v_3^2 && v_3(v_1-iv_2) && v_3(v_1-iv_2) && (v_1-iv_2)^2 \\
v_3(v_1+iv_2) && -v_3^2 && v_1^2 - i^2v_2^2 && -v_3(v_1-iv_2) \\
v_3(v_1+iv_2) && v_1^2 - i^2v_2^2 && -v_3^2 && -v_3(v_1-iv_2) \\
(v_1+iv_2)^2 && -v_3(v_1+iv_2) && -v_3(v_1+iv_2) && v_3^2
\end{pmatrix}
\end{equation}
No guarantee above is correct but if it is, it turns out that:
\begin{equation}
M \psi = \frac{1}{\sqrt 2}\begin{pmatrix} 0 \\ -1 \\ 1 \\ 0 \end{pmatrix}
\end{equation}
irrespective of what $v_1, v_2, v_3$ are (of course $v_1^2 + v_2^2 + v_3^2 = 1$).
Again this (the RHS) cannot be factored out into a tensor product of two 2x1 vectors and this tells me that the qubits cannot be un-entangled. If the qubit register is measured it will be 01 or 10. A result of +1(-1) on the first qubit implies a result of -1(+1) on the second qubit.
