# How to compute the Perron vector

I'm new to spectral graph theory and I just read Laplacians and the Cheeger inequality for directed graphs. In this paper, the author proposed transition probability matrix using Perron-Frobenius Theorem.

To understand what they say, I tried to compute the Perron vector with simple example. But there's one thing I don't understand.

Suppose we have irreducible matrix $$A$$, $$A=\begin{pmatrix} 0 &1 &0 &0 \\ 0 &0 &1 &0 \\ 0&0&0&1 \\ 1&0&0&0 \end{pmatrix}$$ then, the probability transition matrix $$P$$ of $$A$$ is $$P=\begin{pmatrix} 0 &1 &0 &0 \\ 0 &0 &1 &0 \\ 0&0&0&1 \\ 1&0&0&0 \end{pmatrix}$$ (same as $$A$$ in this case).

Then the eigenvalues of $$P$$ is $$\lambda=1, -1, i, -i$$. The Perron-Frobenius Theorem states that an irreducible matrix with non-negative entries has a unique left eigenvector with all entries positive. In this case, let $$\rho$$denote the eigenvalue of all positive eigenvector of $$P$$.

Then $$P$$ has a unique left eigenvector $$\phi$$ and $$\phi P = \rho P$$. In this case, $$\rho =1$$ and $$\phi = \begin{pmatrix} 1&1& 1&1\end{pmatrix}$$.

I don't understand from this step. Then we have to normalize and choose $$\phi_{norm}$$ so that $$\sum_v \phi_{norm}(v)=1$$.
What does "choose" actually mean and how should I compute this? I thought I should simply normalize $$\phi$$ and which will be $$\phi_{norm}= \begin{pmatrix} \frac{1}{2}&\frac{1}{2}&\frac{ 1}{2}&\frac{1}{2}\end{pmatrix}$$. Is this incorrect? Can someone explain this?

Then we have to normalize and choose $$\phi_{norm}$$ so that $$\sum_v \phi_{norm}(v)=1$$.
That means you have to scale $$\phi$$ so that its entries sum to $$1$$. That is, you are normalising $$\phi$$ with respect to the $$1$$-norm (rather than the $$2$$-norm). So, $$\phi_{\text{norm}}=\frac14(1,1,1,1)$$.