= Solution
The other <eigenvalues> are $1/2$ and $0$. Their magnitudes are strictly less than one. All three <eigenvalues> are distinct, so their <eigenvectors> form a <basis> and the <matrix> is <diagonalizable>. Every component outside the stationary direction therefore decays under repeated application of $P$.
For an explicit justification, choose
$$
w=(0,1,-1)^{\mathsf T},\qquad u=(-2,1,1)^{\mathsf T}.
$$
Then $Pw=w/2$, $Pu=0$, and
$$
x(0)=10v-\frac72w+2u.
$$
The zero-<eigenvalue> component disappears after one step. For every $n\geq1$,
$$
\boxed{x(n)=\begin{pmatrix}5\\5/2-(7/2)2^{-n}\\5/2+(7/2)2^{-n}\end{pmatrix}\ \longrightarrow\ \begin{pmatrix}5\\5/2\\5/2\end{pmatrix}.}
$$
Thus the shares converge to $v$ and the resource <vector> converges to $10v$, since the conserved total is ten. For any other initial resource <vector> of total $T$, the same <eigenvector> argument gives the limit $Tv$.
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