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I have to show the following:

Let $K$ be a field such that $\mbox{char } K \neq 2$ and each element of $K$ is a square (i.e. $K^2=K$) and let $V$ be a finite-dimensional vector spaces over $K$. Then, for every automorphism $\tau \in \mbox{Aut}_K V$ there exists an endomorphism $\rho \in \mbox{End}_K V$ such that $\tau = \rho^2$.

I have proved (according to the hint given in the problem) that if $\sigma$ is a nilpotent endomorphism, then there exists an endomorphism $\rho$ such that $\rho^2=1_V+\sigma$.

So, I guess (although I am not sure) that under our assumptions one could show the automorphism $\tau$ can be represented as $\tau=1_V+\sigma$, where $\sigma$ is nilpotent. I'll be grateful for your help.

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    Hint: "upper triangular matrices" are one of the most basic examples of matrices that have the form $I + \sigma$.2012-04-22
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    So, do you suggest that $\tau$ should be triangularizable under these assumptions?2012-04-22
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    Yep. IMO it is a pleasant exercise to actually produce an algorithm for triangularizing a matrix, but I would be surprised if you couldn't just look it up with the right search term. (Or maybe there is some canonical form you could use here)2012-04-22
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    It was the first thing I thought about but I didn't believe in it:) Thanks for the hint2012-04-22
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    Of course, not every triangular matrix has 1's down the diagonal, so it doesn't completely solve your problem. Try triangularizing a 2x2 matrix; if I've thought it through correctly, you can triangularize an nxn matrix by iterating the 2x2 version.2012-04-22
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    But now that I think about it again, I'm no longer convinced. :( I searched for "triangularization", and it suggested you need the eigenvalues in the field.2012-04-22
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    I know I should to try to decompose the minimal polynoimial into linear factors.. but now I dont know how to do it...2012-04-22
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    I dont see how to make use of the assumption $K^2=K$ to do it2012-04-22
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    You seem to have left out a hypothesis somewhere. Over the field with two elements, I don't think that $\left( \begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix} \right)$ has a square root. This is of the form identity plus nilpotent, so the omitted assumption must be in the part you've already done.2012-04-22
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    Yes, you are right! It should be $char K \neq 2$, thank you very much, I have used this assumpion in the first part.2012-04-22
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    Your guess cannot be right, as a decomposition $\tau=1_V+\sigma$ would automatically be the Dunford-Jordan decomposition of $\tau$ and would imply that it is trigonalizable and has spectrum $\lbrace 1\rbrace$.2012-04-22
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    @Olivier Bégassat Ok, but how do you know that $\tau$ is not triangularizable or it has not spectrum $\{1\}$ under our assumptions?2012-04-22
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    @Olivier Bégassat If $\tau$ has spectrum $\{1\}$ the problem would be solved since in this case the minimal polynomial of $\tau$ would be $(X-1)^m$ and then we set $\sigma=\tau-1_V$.2012-04-22
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    $\tau$ might very well be triangularizable, but unless $K$ is assumed algebraically closed it will not be true for arbitrary automorphisms which is what you set about to prove. and if it is both trigonalisable and has only $1$ in its spectrum its determinant is equal to $1$.2012-04-22
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    @dawid if its spectrum in some algebraic closure of $K$ is one, you are indeed done, but if you are only talking about its spectrum in $K$, you wouldn't be done I think.2012-04-22
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    I think I have a proof if I can show that for $P(X)$ irreducible polynomial over $K$, $P(X^2)$ cannot be irreducible.2012-04-22
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    Yes, when $K$ is algebraically closed there is no problem since the minimal polynomial of $\tau$ can be decomposed into linear factors, so $\tau$ is triangularized and main diagonal of its matrix consists only non-zero elements. But the assumption $K^2=K$ has to somehow work..2012-04-22
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    If I understand correctly what you need for your proof it is not true in generality, for example $P(X)=X+1$ over $\mathbb{R}$ $P$ is irreducible and $P(X^2)=X^2+1$ is also irreducible over $\mathbb{R}$ but I guess you claim that it should follows from the assumption $K^2=K$.2012-04-23
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    i think I have a proof, but it uses the fundamental theorem of Galois theory and group theory... is this even reasonable, in case this question is homework, have you covered these topics?2012-04-23
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    I am not a specialist in the Galois theory, but I know the basics, so I hope I will understand:) I would be grateful if I could see your proof.2012-04-23
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    However, it would be suprising for me if there was no elementary proof since it is one of exercises that that I got from my professor. I have only two hits: the first one is that I have written above and the socond one is to use the desomposition theorem (I mean decomposition into $\tau$-ivariant subspaces)2012-04-23
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    I'll write down what I dound tomorow in case noone gives an answer in the meamtime, right now I'm off to bed. I just found out my Galois thing was false -__-2012-04-23
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    This is tricky. If $K$ were algebraically closed, we all would have solved this by considering one Jordan block at the time. So if we temporarily extend scalars to $\overline{K}$, we can do it. But I'm not sure about coming back down to $K$? Need to group conjugate blocks, take their square roots, and glue the results together in such a way that the square root, too, is a linear mapping gotten from extending scalars...2012-04-23
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    @dawid do you have some email adress i can send you a scan of my partial proof? I am writing this from my phone and it would take me too long. You can create a dummy email address and send me this.2012-04-23
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    I sent you an email, unfortunately the pictures are low resolution. There's a total of 6 of them, numbered 1 through 6.2012-04-23

1 Answers 1

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If my answer to the question Reducibility of $P(X^2)$ appears to be right, then I think the statement is false.

1 — Counter example

Consider $A$ the companion matrix of a polynomial $P(x)$, and let $B$ be a square root of $A$. It is clear — take a triangulation in the algebraic closure — that $\chi_B(x) \chi_B(-x) = \chi_A(x^2)$ and that $\chi_A(x) = P(x)$.

This implies that $P(x^2)$ is reducible over the coefficient field of $B$. If one take $P = x^5 + 20x - 16$, I think I have proved that $P(x^2)$ is irreducible over the quadratic closure of $\Bbb Q$, which implies that $A$ has no square root over this field.

2 — Complement

Let $K$ be a field, with characteristic not two, and $A$ a square matrix with coefficients in $K$. It is not easy to see whether of not $A$ admit a square root.

We can assume that $\chi_A$ is the power of a irreducible polynomial. Indeed, $A$ stabilize its eigen spaces associated to each irreducible factor, and so does every matrix $B$ which commutes with $A$, which is the case if $B^2 = A$.

There nilpotent case — corresponding to $\chi_A = x^n$ — is particular, there is some combinatorial condition on the size of the nilpotents Jordan blocks.

Let consider the non-singular case. We can assume that $A$ is diagonalizable. Indeed, we can always write $A$ as $D+N$, with $D$ diagonalizable and $N$ nilpotent, both with coefficients in $K$, and with $DN = ND$. The matrix $A$ has a square root if and only if $D$ has a square root.

So we are reduced to the case of a matrix with blocks along the diagonal all equal to the companion matrix $C_P$, where $\chi_A = P^d$. This matrix has a square root if and only if one of the following holds :

  • $n$ is even ;
  • The decomposition field of $\chi_A$ contains the roots of $\chi_A(x^2)$.