I'm finding the steady state of this matrix \begin{bmatrix}0&1/3&0&1/4\\1/2&0&1/3&1/4\\0&1/3&0&1/2\\1/2&1/3&2/3&0\end{bmatrix} I know I first have to solve $\operatorname{det}(A-\lambda I) = 0$. So I know that I need to find the determinant of
\begin{bmatrix}-\lambda&1/3&0&1/4\\1/2&-\lambda&1/3&1/4\\0&1/3&-\lambda&1/2\\1/2&1/3&2/3&-\lambda\end{bmatrix}
When I took the determinant I got $\lambda^4-59/72\lambda^2-7/36\lambda+1/72$. When I set that equal to zero I got $\lambda_1 = 1$ and $\lambda_2 = .0575$
But when I tried to plug those into the matrix above and solve for the null space I only got the trivial solution. I don't know what that means for the steady-state, or even what to do next. I was thinking that the only answer being the trivial solution means that the steady-state maybe whatever the original distribution was.
edit: when $\lambda = 1$ the eigenvector is \begin{bmatrix}1/2\\3/4\\3/4\\1\end{bmatrix} dividing by three I get \begin{bmatrix}1/6\\1/4\\1/4\\1/3\end{bmatrix} which I believe to be my final answer.