# Diagonalizing a matrix where the pairwise difference of eigenvalues are units

Let $$R$$ be a commutative ring (with $$1$$), and $$M$$ be an $$n\times n$$ matrix over $$R$$ whose characteristic polynomial is $$\det(tI - M) = (t-\alpha_1)\cdots(t-\alpha_n)$$ with $$\alpha_i \in R$$ for all $$i$$. Assume that $$\alpha_i - \alpha_j \in R^\times$$ whenever $$i\ne j$$. Is it true that there is some invertible $$P \in GL_n(R)$$ with $$M = P\begin{pmatrix}\alpha_1 & & & \\ & \alpha_2 & & \\ & & \ddots & \\ & & & \alpha_n\end{pmatrix}P^{-1}?$$

This is true if $$R$$ is a field. The proof here shows that we must have $$R^n = \oplus_{i=1}^n \ker(M - \alpha_i I_n)$$, at least in the case when $$R$$ is a local ring (but also it seems to work for general $$R$$). But would it follow that we can write $$M$$ in the form above? It is not clear to me that each $$\ker(M - \alpha_i I_n)$$ is free (if $$R$$ is not local), so I am not sure if we can necessarily find a basis of $$R^n$$ consisting of eigenvectors of $$M$$.

If the above is false, might there be some conditions on $$R$$ that would make it true, e.g. if we assume $$R$$ is an algebra over a field?

Let $$A_i$$ be the product of all $$M-\alpha_j$$ $$j\neq i$$. Let $$K_i\subset R^n$$, $$K_i=A_i(R^n)$$. Then $$M$$ acts as $$\alpha_i$$ on $$K_i$$. Next show that $$\sum K_i=R^n$$ using your condition $$\alpha_i-\alpha_j$$ are units and $$K_i\cap \sum_{j\neq i} K_j=0$$, so $$R^n=\oplus K_i$$. This gives you more or less what you want, except, the $$K_i$$ are projective modules of rank one, but may not be free, so not sure you will get your diagonalization globally.
• Do you have an example where those $K_i$ may not be free? Also, how do we know that the $K_i$ are of rank $1$? – Ligo Feb 3 at 1:06
Take $$R = \mathbb{Z}[\sqrt{-5}]$$ and $$M = \begin{bmatrix}3 & 1+\sqrt{-5} \\ -1+\sqrt{-5} & -2\end{bmatrix}$$. This has characteristic polynomial $$t(t-1)$$.
I claim that we cannot diagonalize $$M$$ over $$R$$. If we could, then the $$0$$-eigenspace and $$1$$-eigenspace must both be free of rank $$1$$ over $$R$$. Say $$\begin{bmatrix} x \\ y\end{bmatrix} \in R^2$$ generates the $$0$$-eigenspace. The $$0$$-eigenspace contains both $$\begin{bmatrix} 1+\sqrt{-5} \\ -3\end{bmatrix}$$ and $$\begin{bmatrix} 2 \\ -1+\sqrt{-5}\end{bmatrix}$$, so there must be some $$r,s \in R$$ such that $$\begin{bmatrix} 1+\sqrt{-5} \\ -3\end{bmatrix} = r\begin{bmatrix} x \\ y\end{bmatrix}, \quad \begin{bmatrix} 2 \\ -1+\sqrt{-5}\end{bmatrix} = s\begin{bmatrix} x \\ y\end{bmatrix}.$$ In particular, $$ry = -3$$ and $$sx = 2$$. Since $$2$$ and $$3$$ are irreducible in $$R$$ (and $$\pm 1$$ are the units of $$R$$), then $$x,y$$ must lie in $$\mathbb{Z}$$. But $$\begin{bmatrix} 0 \\ 0\end{bmatrix} = \begin{bmatrix}3 & 1+\sqrt{-5} \\ -1+\sqrt{-5} & -2\end{bmatrix} \begin{bmatrix} x \\ y\end{bmatrix} = \begin{bmatrix} 3x + (1+\sqrt{-5})y \\ (-1+\sqrt{-5})x - 2y\end{bmatrix}$$ has no integer solutions for $$(x,y)$$ other than $$(0,0)$$ (because $$\sqrt{-5}$$ is irrational). Contradiction.