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Let $D=\mathbb{C}[x]$ the polynomial ring in one variable over the complex numbers.

Let $$A=\begin{bmatrix} 1 & 1 \\ 0 & x \\ \end{bmatrix}$$

We can view $A$ as a matrix over $\mathbb{C}(x)$ and have the usual linear algebra theory, or we can vies $A$ as a matrix over $\mathbb{C}[x]$:

$1$ and $x$ are roots of $p(t)=\det(A-tI)=t^2-(1+x)t+x=(t-1)(t-x)$, namely, they are eigenvalues, with corresponding eigenvectors $$u=\begin{bmatrix} 1 \\ 0 \\ \end{bmatrix}$$ and $$v=\begin{bmatrix} 1 \\ x-1 \\ \end{bmatrix}$$ $u$ and $v$ are 'linearly independent' over $\mathbb{C}[x]$.

However, $u$ and $v$ do not span $\mathbb{C}[x]^2$; indeed, if there exist $\lambda,\mu \in \mathbb{C}[x]$ such that $$\begin{bmatrix} 0 \\ 1 \\ \end{bmatrix}= \lambda \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix}+ \mu\begin{bmatrix} 1 \\ x-1 \\ \end{bmatrix} $$ then $\lambda+\mu=0$ and $\mu(x-1)=1$, but the second equation never holds in $\mathbb{C}[x]$.

Question: What can be said in such situations where both eigenvalues are in an integral domain (which is not necessarily a UFD like $\mathbb{C}[x]$)? Is it possible to apply some module theory and how? See, this question or this one.

Thank you very much!

user237522
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It seems like "situations where both eigenvalues are in an integral domain" is more simply stated as "eigenvalues of matrices in $M_n(R)$ where $R$ is a domain."

The only thing that occurs to me is that, given the eigenvalues of a matrix over an integral domain, the matrix is an injective transformation iff the eigenvalues are all nonzero.

More generally the matrix would be injective iff the eigenvalues are all regular (meaning they are not zero divisors.)

You must be aware of the theory useful for matrices over principal ideal domains that gives us things like the Jordan canonical form. I think also Kaplansky discusses the theory of matrices over local rings in Linear algebra and geometry, a second course somewhere near the end (or perhaps it was just local domains.)

rschwieb
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