# Fields where the number of eigenvalues is greater than the dimension

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.

• $F$ is not a field; it has zero divisors. Commented Dec 31, 2023 at 18:49
• I see. I assume that these problems should not appear in fields right? Commented Dec 31, 2023 at 18:51
• 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. Commented Dec 31, 2023 at 18:52
• 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. Commented Dec 31, 2023 at 19:35
• (PID is important because it allows the application of structure theorem of f.g. modules over a PID.) Commented Dec 31, 2023 at 19:36