# Properties of quasiregular elements in a matrix ring

I've been puzzling over one of the properties of quasiregular elements listed in the wikipedia article on the topic.

An element of a ring $x$ is quasiregular (left, right) when $1-x$ has a multiplicative inverse (left, right). The article claims that, in a matrix ring, this is equivalent to the element (matrix) not possessing $-1$ as an eigenvalue.

There is no reference given for this and there seem to be some pretty obvious counterexamples. For instance, if $x = \left( \begin{smallmatrix} -1 & 0 \\ 0 & -1 \end{smallmatrix} \right)$ then $1 - x$ (where 1 is the $2 \times 2$ identity matrix) has a multiplicative inverse in the ring of 2 by 2 matrices but $-1$ is clearly an eigenvalue of $x$. In fact, for matrices, having or not having $-1$ as an eigenvalue doesn't seem to say anything definite about quasiregularity.

Just wondering if I'm missing something obvious or if it's simply an error in the article. Thanks!

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See the talk page where this was discussed and not resolved back in 2009. – Jack Schmidt Jul 25 '13 at 19:42

An element $x$ of a ring with 1 is left quasiregular if $1+x$ has a left multiplicative inverse. Looks like Wikipedia has it wrong. I've edited the Wikipedia page to have the correct definition.
With this definition, it is immediate that $-1$ can not occur as an eigenvalue in the case you mention. Indeed, if $-1$ were an eigenvalue, then $\det(x - (-1)I) = 0$ making $x + I$ not invertible.
@Jack: Good point! I didn't even think about the references. Grove (the Dover series) uses $1+x$. – RghtHndSd Jul 25 '13 at 19:51