Finding the corresponding Perron eigenvalue Find the Perron root and the corresponding Perron eigenvector of A. $\begin{bmatrix} 0 &1 &1 \\\ 1&0&1 \\\ 1&1&0 \end{bmatrix}$ I figured out the Perron root which happens to be $ \lambda = 2 $. And I tried to figure out the eigenvector and got $\begin{bmatrix} 1 \\\ 1 \\\ 1 \end{bmatrix}$, but that turned out to be incorrect. Can someone show me how to find the Perron eigenvector for the Perron root 2? !The wrong eigenvector

The problem is that you forgot to normalize your vector. The answer should indeed be $$ \pmatrix{1/3\\\1/3\\\1/3} $$