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I'm testing around this new concepts and found the field "Split-Complex numbers" (Call it F) where they include a new number $j$ such that $j^2 = 1$.

I'm working with vectors in $F^2$. Consider this matrix:

$A =\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array} \right)$

and it has a characteristic polynomial $\lambda^2 - 1$ but has 4 solutions, $1, -1, j, -j$. I'm confused because this corresponds to 4 eigenvectors. I'm trying to figure what leads to this and I'm suspecting that the field $F$ is already "2 dimensional", that $F^2$ is 4 dimensional. I'm not sure of the reason though.

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    $\begingroup$ $F$ is not a field; it has zero divisors. $\endgroup$
    – Misha Lavrov
    Commented Dec 31, 2023 at 18:49
  • $\begingroup$ I see. I assume that these problems should not appear in fields right? $\endgroup$
    – Habouz
    Commented Dec 31, 2023 at 18:51
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    $\begingroup$ Someone more knowledgeable than I should speak to what happens when trying to do linear algebra over something that's not a field, but that's definitely the root (hah!) of the problem. $\endgroup$
    – Misha Lavrov
    Commented Dec 31, 2023 at 18:52
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    $\begingroup$ Lots of linear algebra work because $F[x]$ is a PID when $F$ is a field. This is, for example, why various normal form/canonical form exists. Since $F[x]$ is a PID if and only if $F$ is a field, most things break down without that assumption. That doesn’t mean you can’t do linear algebra over arbitrary commutative rings. After all, we can consider things such as modules over a ring and matrix algebras over a ring - but lots of things won’t carry over and it’s somewhat of a different subject altogether. $\endgroup$
    – David Gao
    Commented Dec 31, 2023 at 19:35
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    $\begingroup$ (PID is important because it allows the application of structure theorem of f.g. modules over a PID.) $\endgroup$
    – David Gao
    Commented Dec 31, 2023 at 19:36

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