= Solution
The factorization of the <characteristic polynomial> shows that $\lambda=1$ is an <eigenvalue>. To find its normalized <eigenvector>, solve $(P-I)v=0$ together with $v_C+v_H+v_B=1$. The first two independent <linear equations> are
$$
-4v_C+v_H+v_B=0,\qquad v_C-2v_H+2v_B=0.
$$
The first equation and normalization give $5v_C=1$. The second gives $v_H-v_B=v_C/2=1/10$, while $v_H+v_B=4/5$. Consequently,
$$
\boxed{v=\begin{pmatrix}1/5\\9/20\\7/20\end{pmatrix},\qquad Pv=v.}
$$
All entries are positive and sum to one. This <probability vector> is the <stationary distribution>.
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