# If $K=K^2$ then every automorphism of $\mbox{Aut}_K V$, where $\dim V< \infty$, is the square of some endomorphism.

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 $I + \sigma$. – Hurkyl Apr 22 '12 at 17:17
So, do you suggest that $\tau$ should be triangularizable under these assumptions? – dawid Apr 22 '12 at 17:25
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) – Hurkyl Apr 22 '12 at 17:27
It was the first thing I thought about but I didn't believe in it:) Thanks for the hint – dawid Apr 22 '12 at 17:30
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. – David Speyer Apr 22 '12 at 17:50

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 $\chi_A(x^2)$.
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