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Let $R$ be a commutative ring (not necessarily integral), fix a positive integer $n$, and let $A\in M_n(R)$. We say $\alpha\in R$ is a eigenvalue of $A$ if there exists a nonzero $x\in R^n$ such that $Ax=\alpha x.$ We say $\alpha$ is a eigenroot (this is not a word maybe, but I cannot find a more appropriate) of $A,$ if $\alpha$ is a root of the eigenvalue polynomials of $A.$ Now I have two questions:

(i) Does there exist a eigenvalue of $A,$ which is not a eigenroot of $A?$

(ii) If $\alpha,\beta$ are eigenroots of $A,B\in M_n(R),$ then can we conclude $\alpha\beta$ is a eigenroots of the Kronecker product $A\otimes B?$

qinxs
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1 Answers1

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Using the adjugate matrix and $\det(AB)=\det(A)\det(B)$, we can show that a matrix $A$ over a commutative ring $R$ is invertible iff $\det(A)\in R^{\times}$.

  • Yes. First we analyze a bit. If $Ax=\alpha x$, then $(A-\alpha I)\alpha=0$, hence $A-\alpha I$ is not invertible, equivalently $\det(A-\alpha I)$ is not invertible. When $R$ is not a field, this is not the same as to say $\det(A-\alpha I)=0$. In fact, we can find the almost trivial example $R=\mathbb Z/6\mathbb Z, n=1, A=2, x=3, \alpha=0$.

  • Yes. $\det(\alpha I - A)=0$ and $\det(\alpha I - B)=0$ imply $\det(\alpha\beta I - A\otimes B)=0$ is true for any field, and hence is true in the fraction field of $\mathbb Z[\{x_i\}_{i\in I}]$, hence hold in the ring itself and further in any of its quotient ring, which would be all rings.

To elaborate the last point a bit more. Consider the elements $p_1=\det(\alpha I - A), p_2=\det(\beta I - B), q=\det(\alpha\beta I - A\otimes B)$ as polynomials (over $\mathbb Z$) where all of $\alpha, \beta$ and entries of $A, B$ are just variables. Then we know that as polynomials over $\mathbb C$, $p_1=0$ and $p_2=0$ imply $q=0$ when all entries are replace by complex numbers. Therefore, by Hilbert's Nullstellensatz, $q^n=p_1'p_1+p_2'p_2$ over $\mathbb C$. And we may assume $p_1', p_2'\in \mathbb Z[\alpha, \beta, A_{ij}, B_{ij}]$. If we know $p_1, p_2$ generate a prime ideal, then we may assume $n=1$. That is $p_1=0$ and $p_2=0$ imply $q=0$ in the universal ring $\mathbb Z[\alpha, \beta, A_{ij}, B_{ij}]$, hence it must hold in any commutative ring $R$. Adapt this argument, we can show that $p_1, p_2$ are irreducible, and since they don't share any common variables, we know the ideal genrated by them is prime.

Just a user
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  • Thanks. But from $\det(\alpha I-A)=0,$ can we conclude that $\det(\alpha I-A)=0$ in $Z[{x_i}_{i\in I}]?$ – qinxs Oct 21 '22 at 03:34
  • I mean $R=\mathbb Z[x_I]$, so yes. – Just a user Oct 21 '22 at 03:38
  • If your $x_i\in R,$ then $\mathbb{Z}[x_I]$ is not an integral domain. If your $x_i$ is an intermediate, we don't know $\det(\alpha I-A)=0$ in it. – qinxs Oct 21 '22 at 03:46
  • $x_i$ are just formal variables, and $\mathbb Z[x_I]$ is just a polynomial ring, and I used $I$ instead of $x_1, x_2, \cdots$, because there might be infinitely (even uncountably) many of them. What I had in mind is similar to the proof of Cayley-Hamilton using Zariski topology... I may rewrite this part later. – Just a user Oct 21 '22 at 03:56
  • If we define a surjective map $\phi\colon \mathbb{Z}[x_I]\rightarrow R,$ the map is not injective in general. So we don't know $\phi^{-1}(\det(\alpha I-A))=0.$ – qinxs Oct 21 '22 at 04:00
  • $0$ is sent to $0$ in homomorphism. I agree if you haven't seen the argument, it's a bit difficult to swallow... the idea is to think in terms of "universal" polynomial identities, I don't have time now to write up for this particular case, but the idea can be found in https://kconrad.math.uconn.edu/blurbs/linmultialg/univid.pdf – Just a user Oct 21 '22 at 04:09
  • Does we need first consider the case $A,B$ are diagonalizable? – qinxs Oct 21 '22 at 05:07
  • I updated the answer a bit to reflect my thought. It's not as easy as I thought. – Just a user Oct 24 '22 at 02:54
  • Thanks a lot for your careful reply. – qinxs Oct 24 '22 at 11:10