Tag Archives: matrices

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 \[
d_1 & 0 & \dots & 0 \\
0 & d_2 & \dots & 0 \\
\vdots & \vdots & \ddots & 0 \\
0 & 0 & \dots & d_n
\end{bmatrix}\] 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 \[
d_1^\prime & 0 & \dots & 0 \\
0 & d_2^\prime & \dots & 0 \\
\vdots & \vdots & \ddots & 0 \\
0 & 0 & \dots & d_n^\prime
\end{bmatrix}\] 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 \[
0 & 1\\
0 & 0\end{pmatrix} \text{, }
0 & 0\\
0 & 1\end{pmatrix}\] which can be defined whatever the field we consider: \(\mathbb R, \mathbb C\) or even a field of finite characteristic.

One can verify that \[
0 & 1\\
0 & 0\end{pmatrix} \text{, }
0 & 0\\
0 & 0\end{pmatrix}\]

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 \[X^2=\mu_{AB} \neq \mu_{BA}=X.\]