By chance I've decided to calculate the following svd composition, which is the example shown here in figure 3. I've typed in MATLAB's command window:
$S=\begin{pmatrix} 0 & 1/2 & 1/3& 0 &1/5\\ 1& 0& 1/3& 1/3& 1/5\\ 0& 1/2& 0& 1/3& 1/5\\ 0& 0 &1/3& 0& 1/5\\ 0& 0& 0& 1/3& 1/5\\ \end{pmatrix}$
$G = 0.85*S + 0.15*(1/5)*ones(5,1)*ones(1,5)$;
If you type the command:
$[U,S,V,flag]=svds(G)$
MATLAB will return the singular values: $1.1159, 0.7068, 0.4007, 0.2864, 0.0000 $ and $flag 0$ (converged).
I've done the same for other small Google matrices and they returned JUST ONE singular value greater than one.
MY QUESTION IS:
How many singular values are greater than $1$ in a Google matrix?
Here's a partial answer: if the matrix is $n\times n$, it can have as many as $\lfloor\frac n2\rfloor$ singular values greater than $1$. I don't know whether there can be more (although I guess not), but $\lfloor\frac n2\rfloor$ is always achievable if you don't impose a positive lower bound to the entries of $G$.
By a continuity argument, it doesn't matter whether the matrix is entrywise positive or merely nonnegative. So, it's OK to allow zero entries. Let $M= \lceil\frac n2\rceil $ and $m=\lfloor\frac n2\rfloor$ (so that $M\ge m$ and $n=M+m$). Note that when $G$ is a generic stochastic matrix in the form of $$ G=\pmatrix{I_M&0\\ X&Y}, $$ the leading principal $M\times M$ submatrix of $G^TG$ is given by $I_M+X^TX$. Since $X$ is $m\times M$, we can always find $G$ such that $X$ has full row rank. Therefore $I_{\lceil n/2\rceil}+X^TX$ has $m$ eigenvalues greater than $1$ (and $M-m$ eigenvalues exactly equal to $1$, but that is irrelevant here). Consequently, by Cauchy's interlacing inequality, $G^TG$ must have at least $m$ eigenvalues greater than $1$, i.e. $G$ has $m$ singular values greater than $1$.
So, to generate an entrywise positive stochastic matrix with $m$ singular values greater than $1$, you may consider some $G$ of the form $\pmatrix{A&B\\ X&Y}$ where $A$ is close to $I_M$.