$T: V \to V$ non-negative linear transformation: There's $S: V \to V$ , so that $S^6=T$ Let $V$ be an Euclidean finite dimension, and $T: V \to V$ non-negative linear transformation.
I need to prove that there's another non-negative transformation, $S: V \to V$ , so that $S^6=T$.
If I could know that $T$ is diagonal, so It will be easy to prove, because $T$ is non-negative.
Thanks    

First of all, as the OP made precise in the comments, the assumption on $T$ means that its complex spectrum is included in $]0,+\infty[$.
By Jordan's theorem, we know that there exits non-negative real numbers $\lambda_1, \ldots, \lambda_r$, $\mu_1, \ldots, \mu_s$ such that the matrix of $T$ in some basis is given by $\mathrm{diag}(\lambda_1, \ldots, \lambda_r,J(\mu_1), \ldots, J(\mu_s))$, where
$J(\mu) = \begin{pmatrix} \mu & 1 & 0 & \ldots & 0 \\
0 & \mu & 1 & \ldots &0 \\
\vdots & & & &\vdots \\
0 &  & \ldots && \mu \end{pmatrix}$
(the size of the matrix may vary with the index, of course).
We want to find a sixth-root of $T$, so it is enough to do it block by block. For the first part, $\mathrm{diag}(\lambda_1, \ldots, \lambda_r)$, it is clear. So we have to do it for $J(\mu)$ now. 
The key observation is that $J(\mu)^2$ is similar to $J(\mu^2)$. Indeed, $J(\mu)^2=(\mu I+J)^2=\mu^2 I + (\mu J + J^2)$, where $J=J(0)$ with the previous notations. To see that the nilpotent matrices $J$ and $\mu J+J^2$ lie in the same conjugacy class, it suffices to check that $\dim(\mathrm{Ker}(J)^k)=\dim(\mathrm{Ker}(\mu J+J^2)^k)$ for all $k$ integer. But this is clear because $\mu J + J^2= J(\mu I + J))$ and $\mu I + J$ is invertible.
So by iteration $J(\mu^6)$ is similar to $J(\mu)^6$, and therefore, $J(\mu)$ admits a sixth root, which is gonna be conjugated to $J(\sqrt[6]{\mu})$.
Remark. Note that it is important to assume that no $\mu$ is zero. Else, the result fails to be true. The optimal assumption would thus be something like $0$ has maximal multiplicity as eigenvalue of $T$.
