= Solution
The factor $(\lambda-1)$ proves that one is an <eigenvalue>. To find its normalized <eigenvector>, solve $Pv=v$. The second and third rows give
$$
\frac14v_1+\frac12v_2=v_2,\qquad\frac14v_1+\frac12v_3=v_3,
$$
so $v_2=v_3=v_1/2$. The normalization $v_1+v_2+v_3=1$ now gives
$$
\boxed{v=(1/2,1/4,1/4)^{\mathsf T}.}
$$
Its entries are nonnegative and sum to one, and direct <matrix multiplication> verifies $Pv=v$. As a <probability vector> it is the <stationary distribution>; as resource shares it assigns half to the USA and a quarter to each other region.
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