# Subset of elements of finite order of a group

Consider a group $$G$$ and have a look at the question: is the subset $$S$$ of elements of finite order a subgroup of $$G$$?

The answer is positive when any two elements of $$S$$ commute. For the proof, consider $$x,y \in S$$ of order $$m,n$$ respectively. Then $\left(xy\right)^{mn} = x^{mn} y^{mn} = (x^m)^n (y^n)^m = e$ where $$e$$ is the identity element. Hence $$xy$$ is of finite order (less or equal to $$mn$$) and belong to $$S$$.

### Example of a non abelian group

In that cas, $$S$$ might not be subgroup of $$G$$. Let’s take for $$G$$ the general linear group over $$\mathbb Q$$ (the set of rational numbers) of $$2 \times 2$$ invertible matrices named $$\text{GL}_2(\mathbb Q)$$. The matrices $A = \begin{pmatrix}0&1\\1&0\end{pmatrix},\ B=\begin{pmatrix}0 & 2\\\frac{1}{2}& 0\end{pmatrix}$ are of order $$2$$. They don’t commute as $AB = \begin{pmatrix}\frac{1}{2}&0\\0&2\end{pmatrix} \neq \begin{pmatrix}2&0\\0&\frac{1}{2}\end{pmatrix}=BA.$ Finally, $$AB$$ is of infinite order and therefore doesn’t belong to $$S$$ proving that $$S$$ is not a subgroup of $$G$$.

# Complex matrix without a square root

Consider for $$n \ge 2$$ the linear space $$\mathcal M_n(\mathbb C)$$ of complex matrices of dimension $$n \times n$$. Is a matrix $$T \in \mathcal M_n(\mathbb C)$$ always having a square root $$S \in \mathcal M_n(\mathbb C)$$, i.e. a matrix such that $$S^2=T$$? is the question we deal with.

First, one can note that if $$T$$ is similar to $$V$$ with $$T = P^{-1} V P$$ and $$V$$ has a square root $$U$$ then $$T$$ also has a square root as $$V=U^2$$ implies $$T=\left(P^{-1} U P\right)^2$$.

### Diagonalizable matrices

Suppose that $$T$$ is similar to a diagonal matrix $D=\begin{bmatrix} d_1 & 0 & \dots & 0 \\ 0 & d_2 & \dots & 0 \\ \vdots & \vdots & \ddots & 0 \\ 0 & 0 & \dots & d_n \end{bmatrix}$ Any complex number has two square roots, except $$0$$ which has only one. Therefore, each $$d_i$$ has at least one square root $$d_i^\prime$$ and the matrix $D^\prime=\begin{bmatrix} d_1^\prime & 0 & \dots & 0 \\ 0 & d_2^\prime & \dots & 0 \\ \vdots & \vdots & \ddots & 0 \\ 0 & 0 & \dots & d_n^\prime \end{bmatrix}$ is a square root of $$D$$. Continue reading Complex matrix without a square root

# Two matrices A and B for which AB and BA have different minimal polynomials

We consider here the algebra of matrices $$\mathcal{M}_n(\mathbb F)$$ of dimension $$n \ge 1$$ over a field $$\mathbb F$$.

It is well known that for $$A,B \in \mathcal{M}_n(\mathbb F)$$, the characteristic polynomial $$p_{AB}$$ of the product $$AB$$ is equal to the one (namely $$p_{BA}$$) of the product of $$BA$$. What about the minimal polynomial?

Unlikely for the characteristic polynomials, the minimal polynomial $$\mu_{AB}$$ of $$AB$$ maybe different to the one of $$BA$$.

Consider the two matrices $A=\begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix} \text{, } B=\begin{pmatrix} 0 & 0\\ 0 & 1\end{pmatrix}$ which can be defined whatever the field we consider: $$\mathbb R, \mathbb C$$ or even a field of finite characteristic.

One can verify that $AB=A=\begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix} \text{, } BA=\begin{pmatrix} 0 & 0\\ 0 & 0\end{pmatrix}$

As $$BA$$ is the zero matrix, its minimal polynomial is $$\mu_{BA}=X$$. Regarding the one of $$AB$$, we have $$(AB)^2=A^2=0$$ hence $$\mu_{AB}$$ divides $$X^2$$. Moreover $$\mu_{AB}$$ cannot be equal to $$X$$ as $$AB \neq 0$$. Finally $$\mu_{AB}=X^2$$ and we verify that $X^2=\mu_{AB} \neq \mu_{BA}=X.$