Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 1 down vote accepted

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 $\chi_A(x^2)$.
share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.