Have question about finding invariant factors of integer matrix: $\begin{pmatrix} 6 & 2 \\ -2 & 6 \end{pmatrix}$
Was sick during lectures, and not completely that what I do is right.
What I do:
Well, first lets make
$x\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}-\begin{pmatrix} 6 & 2 \\ -2 & 6 \end{pmatrix}=\begin{pmatrix} x-6 & 2 \\ -2 & x-6 \end{pmatrix}$ such matrix,and discover its determinant, like this: $det\begin{pmatrix} x-6 & 2 \\ -2 & x-6 \end{pmatrix}=(x-6)^2+4=x^2-12x+40$
So far so good, but after this.. I am not sure if it is what is asked for:
$x_1=6+2i$ and $x_2=6-2i$, thus $x^2-12x+40=(x-6+2i)(x-6-2i)$
Are $(x-6+2i)$ and $(x-6-2i)$ invariant factors of above mentioned integer matrix? (Basing my thoughts on http://www.math.uh.edu/~klaus/08spt03_solu.pdf , and very well might be way-way off, please dont get too upset)