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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

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    Is T symmetric?2011-08-22
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    No. if it was symmetric it was diagonalable as well.2011-08-22
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    What do you mean by non-negative if it is not symmetric (or Hermitian in the complex case)?2011-08-22
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    I guess he means $\langle Tx, x \rangle \geqslant 0$ for every $x\in V$.2011-08-22
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    Non negative is symmetric? I'm not sure now, come to think about that what does it exactly mean.2011-08-22
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    Just saying you have to define your terms2011-08-22
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    It has only eigenvalues 0 or positive..2011-08-22
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    So it has no complex non-real eigenvalues? In that case the result is easy using the Jordan normal form. Indeed, for $\lambda \geq 0$, the Jordan matrix $J(\lambda)$ satisfies $\exists P \in \mathrm{GL}_n(\mathbb R)$; $J(\sqrt[6]{\lambda})^6=P^{-1}J(\lambda) P$, which concludes then. (take $S$ to be defined on each Jordan block by $PJ(\sqrt[6]{\lambda})P^{-1})$.2011-08-22
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    Can you please extend it to an answer? I can't understand what exactly you wrote and mean.2011-08-22
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    So wait, your saying is $T$ non-negative means that all eigenvalues of $T$ lie in $\mathbb{R}_{\geq 0}?$ But the matrix $$ T = \left[ {\begin{array}{cc} 0 & 1 \\ 0 & 0 \\ \end{array} } \right] $$ has this property and has no square, let alone sixth, root in $GL_2(\mathbb{R})$2011-08-22
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    what does $G L_2$ stand for?2011-08-22
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    Oops. I meant to write $\mathfrak{M}_2(\mathbb{R})$ the set of 2 by 2 matrices over $\mathbb{R}$, good catch.2011-08-22
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    also meant to write you're. embarrassing2011-08-22

2 Answers 2

3

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 $T$.

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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 $K$".

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    I don't understand the isomorphism between $\mathbb R[X]/(p)$ and $A/(p)$. And how does the surjective map $A \to \mathbb R[T]$ enable you to conclude?2011-08-22
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    @Henri: Let's take the 2nd point first. The morphism $A \to \mathbb R[T]$ maps the function $x$ to $T$. Since $x$ has a 6th root in $A$, $T$ has a 6th root in $\mathbb R[T]$. Do you agree with that?2011-08-22
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    Ok, but why would have this root only positive eigenvalues?2011-08-22
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    We want a 6th root in $M_n(\mathbb R)$. We get one in $\mathbb R[T]$. [It **has** positive eigenvalues, but I think this plays no role.]2011-08-22
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    That's not the OP question. And I still don't see why your isomorphism holds. Where precisely appears the fact that $p$ has only positive roots in your argument?2011-08-22
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    OK! I see. I'll correct that. Thank you very much!2011-08-22
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    @Henri: Sorry, I forgot to put the "@Henri" in my previous comments. I've made the correction you suggested.2011-08-22
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    @Pierre-Henri: Swell. Seems a low-tech version of this is to consider each Jordan block of T, say the block xI+J of dimension n with x real and positive, and to use the fact that, since J^n=0, a sixth root of xI+J is a sixth root of x times the polynomial in J/x which is the series expansion of the function sixth-root-of 1+t at t=0, truncated at t^n and evaluated at t=J/x.2011-08-22
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    Hi @Didier! I agree. Anyway, it’ll always be the same argument: $\mathbb R[T]$ is the algebra of functions on the spectrum (taking the multiplicities into account). That’s what I tried to emphasize. It only uses the Chinese Remainder Theorem and Taylor’s formula, which are (IMHO) more elementary than Jordan normal forms. But again, only the package can change, not the product.2011-08-22
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    @Pierre-Yves: Thanks for your answer. (And sorry for the mispelling of your name.)2011-08-22
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    @Pierre-Yves: Ok, now I completely agree with your answer! However, this method does not treat the case where $0$ is eigenvalue with maximal multiplicity.2011-08-22
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    @Henri: I don’t understand: The eigenvalues of $T$ being $ > 0$, so are those of its (positive) 6th root. No?2011-08-22
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    @Didier: No problem for the misspelling (I wonder if you didn’t misspell the word misspelling). One way of saying things is this: Compute $f(T)$ one generalized eigenspace at a time. On the $\lambda$-generalized eigenspace, use the degree less than $m$ Taylor approximation of $f$ at $\lambda$, where $m$ is the multiplicity of $\lambda$ as a root of the minimal polynomial. Does this look reasonable to you?2011-08-22
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    @Pierre-Yves: Yep.2011-08-22
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    Dear @Henri: You're right! I changed "if $\lambda$ is an eigenvalue of $T$, then $f(\lambda)$ is an eigenvalue of $f(T)$" to "the eigenvalues of $f(T)$ are the $f(\lambda)$, where $\lambda$ runs over the eigenvalues of $T$".2011-08-22
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    I removed some confusing spacing characters because on my machine [it looked like this](http://i.stack.imgur.com/eqavz.png). By the way: the Publ. Math. IHÉS editions of EGA are freely available on http://www.numdam.org, the [first volume is here](http://www.numdam.org/item?id=PMIHES_1960__4__5_0). Moreover, you may have noticed that your answer was converted into Community Wiki. I've flagged it for moderator attention to undo that. See [this meta thread](http://meta.math.stackexchange.com/questions/1770/) for further elaboration on this point.2011-08-24
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    Dear @Theo: Thank you very much!!! Spacing: now I know. EGA: I know Numdam, but the Springer version of EGA 1 is substantially different from the IHES one. In particular the excerpt I quote is *not* in the IHES version [it's in an added Introduction]. CW: I thank you of course, but you've probably noticed that there hasn't been a single upvote so far, and if it's simpler to keep it as a CW, I don't mind. [BTW in EGA 1 Springer 1971, the Yoneda Lemma (not called that way) is contained in the *very first lines* of the whole treatise.]2011-08-24
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    Yes, I noticed that the quote is not in volume one of the IHES edition and I'm aware of the fact that they are substantially different from the Springer ed. CW: the point is that after 10 or 12 edits (I don't remember the exact number, my edit was the 12th) an answer is *automatically* converted into CW (I can't turn answers by other people into CW except by editing sufficiently often). [Yoneda: duly noted; neither is that name mentioned in SGA4/1, Exp I, 1.3,1.4, especially p.8, but you certainly noticed already...]2011-08-24