Priority vector and eigenvectors - AHP method I'm reading about the AHP method (Analytic Hierarchy Process). On page 2 of this document, it says:

Given the priorities of the alternatives and given the matrix of
  preferences for each alternative over every other alternative, what
  meaning do we attach to the vector obtained by weighting the
  preferences by the corresponding priorities of the alternatives and
  adding? It is another priority vector for the alternatives. We can use
  it again to derive another priority vector ad infinitum. Even then
  what is the limit priority and what is the real priority vector to be
  associated with the alternatives? It all comes down to this: What
  condition must a priority vector satisfy to remain invariant under the
  hierarchic composition principle? A priority vector must reproduce
  itself on a ratio scale because it is ratios that preserve the
  strength of preferences. Thus a necessary condition that the priority
  vector should satisfy is not only that it should belong to a ratio
  scale, which means that it should remain invariant under
  multiplication by a positive constant c, but also that it should be
  invariant under hierarchic composition for its own judgment matrix so
  that one does not keep getting new priority vectors from that matrix.
  In sum, a priority vector x must satisfy the relation Ax = cx, c > 0

To let you quickly grasp what AHP is all about, check this simple tutorial.
The matrix of preferences for each alternative over every other alternative is obvious to me. Ideally, such a matrix should satisfy a property $a_{ij}=a_{ik}a_{kj}$ (because if I say I prefer A to B two times, and B to C three times, then I should prefer A to C six times (it makes sense I guess, but it's a very informal rule). OK, but in the quote I gave, it says:

what meaning do we attach to the vector obtained by weighting the
  preferences by the corresponding priorities of the alternatives and
  adding? It is another priority vector for the alternatives.

I'm not quite sure what it means. Alternatives can be apple, banana, cherry. Preferences are just numbers in matrix of preferences, just like here
But what are 'corresponding priorities of the alternatives'?
I'd say to obtain a priority vector (i.e. to find out which fruit is preferred the most) one could just
1) divide every element in a given column by the sum of elements in that column (normalization)
2) calculate average of elements in each row of the matrix obtained in step 1). 
The obtained vector is the priority vector, I belive.
But in the quoted text, it gets worse - the author describes raising the matrix to consecutive powers. Why do we multiply priority matrix by itself? It says the result of this multiplication is 'another priority vector of alternatives'. Why? Haven't we just lost some information by doing this?
I mean, we always can multiply matrices, but it should be justified. In case of priority matrix I can't see the justification. Later in the document I've quoted, the author uses the Perron-Frobenius theorem and other sophisticated methods. I'd be grateful for a intuitive, clear explanation of what's going on here.
And finally: WHY the eigenvector $w$ matching the maximum eigenvalue $\lambda_{max}$ of the pairwise comparison matrix $A$ is the final expression of the preferences between the investigated elements?

More articles on AHP method that might help you with answering my questions:
http://books.google.com/books?id=wct10TlbbIUC&printsec=frontcover&hl=eng&redir_esc=y#v=onepage&q&f=false
http://www.booksites.net/download/coyle/student_files/AHP_Technique.pdf
http://www.isahp.org/2001Proceedings/Papers/065-P.pdf
For example, what's the relationship between Perron-Frobenius theorem and this method?
 A: $A \in \mathbb{R}^{n \times n}$ is called a pairwise comparison matrix, if it satisfies the following three properties:
$(1)$ $a_{i,j}>0$;
$(2)$ $a_{i,i}=1$;
$(3)$ $a_{i,j} = 1/a_{j,i}$,
for all $i,j=1,\dots,n$. Of course $(1)$ and $(3)$ together imply $(2)$.
That means a pairwise comparison matrix is a positive matrix in the following shape:
$$
A= \begin{bmatrix}
      1 & a_{1,2} & a_{1,3} & \dots & a_{1,n} \\ 
      1/a_{1,2} & 1  &  a_{2,3} & \dots & a_{2,n} \\
      1/a_{1,3} & 1/a_{2,3} & 1 & \dots & a_{3,n} \\
      \vdots  & \vdots  &  \vdots  &  \ddots  & \vdots    \\
      1/a_{1,n}   &  1/a_{2,n} & 1/a_{3,n} & \dots & 1 \\ 
    \end{bmatrix}.
$$
The motivation behind the definition is that the elements of $A$ are representing pairwise comparisons, since if $a$ alternative is $2$ times better than $b$, then $b$ is $1/2$ times better than $a$. Because we can use only positive quantities, that means the measure of $a$ to $a$ is always identical. The $i$th alternative is $a_{i,j}$ times better than the $j$th alternative.
If $A$ also satisfies that $a_{i,k}a_{k,j}=a_{i,j}$ for all $i,j,k=1,\dots,n$ then $A$ is called consistent, otherwise $A$ is inconsistent. That means a cardinal transitivity.
Note that the properties $(1)\!-\!(3)$ are very natural, so it is easy to compare alternatives in that way, but it is hard to hold consistenty for all triplets.
It is easy to see (prove it!) that $A$ is consistent if and only if there exist a $w\in\mathbb{R}^n$ positive vector, for that $a_{i,j}=w_i/w_j$ for all $i,j=1,\dots,n$.
Because of the Perron–Frobenius theorem we know that $A$ has a $\lambda_{\max}$ eigenvalue which is the spectral radius of $A$, and the components of the corresponding $v$ eigenvector are nonzero and have the same sign, so we can suppose that $v$ is positive.
Another easy remark (prove it!) that if $A$ is consistent, then the $v$ eigenvector corresponding to $\lambda_{\max}$ has the property that $a_{i,j}=v_i/v_j$ for all $i,j=1,\dots,n$. This eigenvector is called Perron eigenvector of principal eigenvector.
In general we call a positive $w$ vector a weight vector if it is the Perron eigenvector if the matrix is consistent, and it is representing "somehow" the preferences of the decision maker.
In AHP the eigenvector method (EM) means that we calculate the Perron eigenvector of the matrix, and this is the weight vector. But in general there are other methods, with we can find weight vectors (for example by distance minimization).
Finally, I give an example for the eigenvector method with $4$ alternatives.
Let $(\text{apple},\text{banana},\text{pear},\text{orange})$ be the list of alternatives, and after the decision maker made the pairwise comparisons we have the following matrix:
$$ A= \left[ \begin {array}{cccc} 1&4&2&5\\ 1/4&1&1/4&3
\\ 1/2&4&1&4\\ 1/5&1/3&1/4&1
\end {array} \right]. 
$$
For example apple is $4$ times better than banana and $2$ times better than pear. $\lambda_{\max}$ of $A$ is the following:
$$\lambda_{\max} \approx 4.170149768.$$
The Perron eigenvector is:
$$ w= \left[ \begin {array}{cccc}  6.884563466,& 1.859400323,& 4.693747683,&
 1.0\end {array} \right]^T. 
$$
Which gives a preferences order: $ \text{orange} \precsim \text{banana} \precsim \text{pear} \precsim \text{apple}$.
In the example $A$ is inconsistent. AHP measures the inconsistency with $CR$ (consistancy ratio):
$$
CR := \frac{\lambda{_\max}-n}{n-1}.
$$
A matrix is acceptable, if $CR<0.1$. In the example above $CR=0.05671658933$.
A: I try to reply here to the "WHY?" question.
Let $\bf x$ be the vector containing the underlying unknown priorities
\begin{equation}
\bf x=
\begin{bmatrix}
x_{1} \\
x_{2} \\
\dots \\
x_{n}
\end{bmatrix}
\end{equation}
Ideally (with a perfect judgement), $Z$ is given by
\begin{equation} 
  Z = {\bf x} {\bf x}^{-T} 
\end{equation}
where, with a little abuse of notation, we define
${\bf x}^{-1}$ as the vector containing the inverse of each element of
$\bf x$.
Now, let us multiply both sides by $\bf x $, obtaining:
\begin{equation}
Z {\bf x} = n {\bf x}
\end{equation}
and this proves that the underlying priority vector $\bf x$ is the eigenvector of $Z$ corresponding to the eigenvalue $n$ (after normalisation this becomes 1).
