$T: V \to V$ 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    
 A: 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 $T$.
A: New version of the answer
Let $T$ be a real $n$ by $n$ matrix whose minimal polynomial $p\in\mathbb R[X]$ splits over $\mathbb R$ as
$$p=(X-\lambda_1)^{m(1)}\cdots(X-\lambda_k)^{m(k)},$$ 
the $\lambda_i$ being distinct and real, and the $m(i)$ positive. Writing $\mathbb R[X]$-algebra canonical isomorphisms as equalities, we have, by the Chinese Remainder Theorem, 
$$\mathbb R[T]=\frac{\mathbb R[X]}{(p)}=
\prod_{i=1}^k\ \frac{\mathbb R[X]}{(X-\lambda_i)^{m(i)}}=
\prod_{i=1}^k\ A_i=A.$$ 
Denote by $x_i\in A_i$ the image of $X$, and adhere to the following somewhat symbolical notation: an element of $A$ is denoted by $f(x)$, its $i$th coordinate by 
$$f(x_i)=\sum_{j=0}^{m(i)-1}\ \frac{f^{(j)}(\lambda_i)}{j!}\ \ (x_i-\lambda_i)^j,$$ 
and its image in $\mathbb R[T]$ by $f(T)$. 
For $f(x)$ in $A$ and $v$ in the $\lambda_i$-generalized eigenspace $V_i$ of $T$ we have 
$$f(T)v=\sum_{j=0}^{m(i)-1}\ \frac{f^{(j)}(\lambda_i)}{j!}\ \ (T-\lambda_i)^jv.$$
In particular, $V_i$ is contained in the $f(\lambda_i)$-generalized eigenspace of $f(T)$. 
Assume that the $\lambda_i$ are positive, let $g(t)$ be the positive sixth root of $t > 0$, and let $f_6(x)$ be the element of $A$ whose $i$th coordinate is 
$$f_6(x_i):=\sum_{j=0}^{m(i)-1}\ \frac{g^{(j)}(\lambda_i)}{j!}\ \ (x_i-\lambda_i)^j.$$ 
Then $S:=f_6(x_i)$ fits the bill. 
Old version of the answer
Let $T$ be a real $n$ by $n$ matrix with positive eigenvalues, let $p\in\mathbb R[X]$ be the minimal polynomial of $T$. We have a natural $\mathbb R[X]$-algebra isomorphism $\mathbb R[X]/(p)\overset\sim\to\mathbb R[T].$ 
Let $A$ be the algebra of $\mathbb R$-valued analytic functions on the interval $(0,\infty)$. By the Chinese Remainder Theorem, the inclusion of $\mathbb R[X]$ into $A$ induces an $\mathbb R[X]$-algebra isomorphism $\mathbb R[X]/(p)\overset\sim\to A/(p).$ [See this answer.]
Thus, we get an $\mathbb R[X]$-algebra epimorphism $A\to\mathbb R[T]$, which we'll denote by $f(x)\mapsto f(T)$. In particular $x$ is mapped to $T$, and $\sqrt[6]{T}$ is a sixth root of $T$ in $\mathbb R[T]$. Moreover, the eigenvalues of $f(T)$ are the $f(\lambda)$, where $\lambda$ runs over the eigenvalues of $T$. So, the eigenvalues of $\sqrt[6]{T}$ are positive. [$\sqrt[6]{T}$ is the image of $\sqrt[6]{x}$, and $\sqrt[6]{x}$ is the positive sixth root of $x$.] 
[One could also use $C^\infty$ functions.] 
Thanks to Henri for having pointed out the fact that the previous version was incomplete. 
EDIT 1. The above answer shows this: 
Let $d$ be the degree of the minimal polynomial $p$. Then there is a unique polynomial $q$ of degree less than $d$ such that $q(T)$ is a sixth root of $T$, and all the eigenvalues of $q(T)$ are positive. Moreover $q$ is given by the formula 
$$q=\sum_{i=1}^k\ \heartsuit_i\left(q_i\ 
\frac{(X-\lambda_i)^{m(i)}}{p}\right)\frac{p}{(X-\lambda_i)^{m(i)}}\quad,$$ 
where the $\lambda_i$ are the eigenvalues, where $m(i)$ is the multiplicity if $\lambda_i$ as a root of $p$, where $\heartsuit_i(?)$ means "degree less than $m(i)$ Taylor approximation of ? at $X=\lambda_i$", where $q_i$ is $\heartsuit_i(\sqrt[6]{x})$ viewed as an element of $\mathbb R[X]$. 
That's concrete mathematics ;) 
Thank you to Henri and to Didier Piau. 
EDIT 2. 

On verra notamment apparaître, au cours de notre Traité, le rôle technique très important des anneaux locaux artiniens, qui intuitivement représentent des "voisinages infinitésimaux" de points sur des variétés algébriques. 

My poor translation:

We will see appear in particular, in our Treatise, the very important technical role of Artin local rings, which intuitively represent "infinitesimal neighborhoods" of points on algebraic varieties. 

Éléments de géométrie algébrique I, Volume 166 of Grundlehren der mathematischen Wissenschaften in Einzeldarstellungen mit besonderer Berücksichtigung der Anwendungsgebiete, A. Grothendieck, Jean Alexandre Dieudonné, Springer-Verlag, 1971, p. 11. 
The "mental picture" is this. Let $K$ be a field, let $p\in K[X]$ be a nonconstant polynomial which splits over $K$. By the Chinese Remainder Theorem, $K[X]/(p)$ is a product of rings of the form $A:=K[X]/(X-\lambda)^m$ with $\lambda\in K$, $m > 0$. If $x$ denotes the image of $X$ in $A$, then the image of $f\in K[X]$ in $A$ is 
$$\sum_{i=0}^{m-1}\ \frac{f^{(i)}(\lambda)}{i!}\ (x-\lambda),$$ 
and we think of this object as being "the restriction of $f$ to the order $m-1$ infinitesimal neighborhood of $\lambda$ in $K$". 
