Timeline for On similar matrices and polynomial matrices
Current License: CC BY-SA 4.0
11 events
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| Nov 16, 2019 at 21:26 | comment | added | user20948 | Did you think of replacing matrix rings by arbitrary noncommutative rings? Seemingly Denis' proof still works, I wonder whether this also works. So let us fix notations. Let $A$ be a ring and $a,b\in A$. If we have $(t-a)p(t)=q(t)(t-b)$, then we have a commutative diagram of right $A[t]$-modules ($t$ is a central element in $A[t]$), where morphisms are given by left multiplications by $t-a,p(t),q(t),t-b$ respectively. Passing to the cokernel of $t-a$ and $t-b$, we get a map $\phi$ of right $A[t]$-modules $A\to A$. It follows that $\phi(a)=b\phi(1)$. | |
| Oct 17, 2018 at 11:06 | history | edited | Amritanshu Prasad | CC BY-SA 4.0 |
added 1 character in body
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| Oct 17, 2018 at 11:05 | comment | added | Amritanshu Prasad | @FrankScience Thanks for your comment, I modified my answer. | |
| Oct 17, 2018 at 10:59 | history | edited | Amritanshu Prasad | CC BY-SA 4.0 |
deleted 97 characters in body
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| Oct 17, 2018 at 10:54 | comment | added | Amritanshu Prasad | @FrankScience Your'e right. Matrices of relations give isomorphic modules if and only if they are equivalent. There is no need to use Smith form. However, from the Smith normal form of $A-It$ you will be able to read off the rational canonical form of A, and if you first separate the primary parts, and then do Smith form, you can read off the Jordan canonical form. | |
| Oct 17, 2018 at 9:45 | comment | added | user20948 | It should have been the cokernel of $T\otimes1_R-1_M\otimes T$, not the kernel. | |
| Oct 16, 2018 at 15:30 | comment | added | user20948 | Sorry for a late comment, but I don't understand where you use Smith normal form. Let $R=k[T]$, and I will write $M,N$ instead of $M_A,M_B$ in your post. What you have shown is that, $A,B$ are similar iff $M\cong N$ as $R$-modules. The presentation for $M$ you gave is essentially saying that $M$ is $R$-isomorphic to the kernel of the $R$-endomorphism $T\otimes1_R-1_M\otimes T$ on $M\otimes_kR$, and a similar description for $N$ holds. If $A-T,B-T$ are equivalent, then these kernels are isomorphic (as $R$-modules). These arguments are quite formal, not dependent on the fact that $R$ is a PID. | |
| May 11, 2012 at 16:39 | comment | added | Amritanshu Prasad | Did you find an elementary AND simple answer, please post ir here; I too would like to know. | |
| May 11, 2012 at 12:45 | comment | added | Wei Wang | thanks. but it's for freshmen, and they certainly don't know what are modules. | |
| May 5, 2012 at 6:05 | history | edited | Amritanshu Prasad | CC BY-SA 3.0 |
added link to Smith normal form
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| May 5, 2012 at 5:27 | history | answered | Amritanshu Prasad | CC BY-SA 3.0 |