Timeline for Commuting Matrices and the Weak Nullstellensatz
Current License: CC BY-SA 2.5
4 events
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| Oct 24, 2010 at 10:37 | vote | accept | Holger Partsch | ||
| Oct 23, 2010 at 17:44 | comment | added | user6976 | Just remember that the standard proof that commuting matrices have a common eigenvector is trivial. If $A, B$ commute and $V_\alpha$ is the $\alpha$-eigenspace of $A$ (non-zero for some $\alpha$), then $\alpha Bv=BAv=A(Bv)$ for every $v\in V$, so $V_\alpha$ is invariant with respect to $B$. Now if $V_\alpha$ is the whole space, we can forget about $A$ and reduce the number of matrices. If $V_\alpha$ is not the whole space, we can reduce the dimension. No Nakayama or Hilbert are needed. | |
| Oct 23, 2010 at 17:04 | comment | added | David E Speyer |
To elaborate: By Nakayma, $V_p \supsetneq \mathfrak{m} V_p \supsetneq \mathfrak{m}^2 V_p \supsetneq \cdots$, with strict inequality until we reach the zero module. So, there is some $n$ such that $\mathfrak{m}^n V_p \neq (0)$ but $\mathfrak{m}^{n+1} V_p=0$. For $v \in \mathfrak{m}^n V_p \setminus \{ 0 \}$, the vector $v$ is annihilated by $\mathfrak{m}$.
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| Oct 23, 2010 at 16:43 | history | answered | Piotr Achinger | CC BY-SA 2.5 |