Problem. Let $$A=\begin{pmatrix}i&0\\0&1\end{pmatrix}$$ be a matrix over the real quaternion ring $\mathbb{H}.$ Prove that it is similar to its inverse, but is not a product of two involutions. Can it be a product of commutators of involutions?
A square matrix $A$ over a ring is called similar to $B$ if there exists an invertible matrix $P$ such that $A=PBP^{-1}.$ A square matrix $T$ over a ring is called an involution if $T^2=I$
is the identity matrix. A square matrix $A$ over a ring is called a commutator of involution if there exist involutions $T, S$ such that $A=TST^{-1}S^{-1}.$
I have shown $$A^{-1}=\begin{pmatrix}-i&0\\0&1\end{pmatrix},$$ and $$A=\begin{pmatrix}ij&0\\0&1\end{pmatrix}A^{-1}\begin{pmatrix}ij&0\\0&1\end{pmatrix}^{-1}.$$
But I don't answer this remaining. I don't know why it is not a product of two involutions. If this is true, then it can be a product of commutators of involutions, can't it? Thanks for all your support.
Best Answer
If we had $A = UV\ldots UV$, with $U$, $V$ involutions, then $V(A^{-1})V^{-1} = A$. Therefore, there would exists an involution $V$ that conjugates $A^{-1}$ to $A$. Now, we know that $$A=\begin{pmatrix}ij&0\\0&1\end{pmatrix}A^{-1}\begin{pmatrix}ij&0\\0&1\end{pmatrix}^{-1}$$
Moreover, it is easy to see that the centralizer of $A$ equals $$\begin{pmatrix}a+ b i&0\\0&\star\end{pmatrix}$$, with $a$, $b$ not both $0$. So now the question is: can a matrix of the form $$\begin{pmatrix}a+ b i&0\\0&\star\end{pmatrix}\cdot \begin{pmatrix}ij&0\\0&1\end{pmatrix}$$ be an involution? Equivalently, can we have $$((a+bi) ij)^2 = 1$$ Let's calculate: $$(-b + a i) j (- b + a i) j= \cdots = - a^2 - b^2 < 0$$ so it cannot equal $1$.