Timeline for When is $\ker AB = \ker A + \ker B$?
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14 events
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| Nov 6, 2011 at 9:08 | comment | added | Ariyan Javanpeykar | Even though your question has been answered already, I thought you might find Michael Stoll´s notes on Linear Algebra 2 helpful. See Lemma 4.6 for example: faculty.iu-bremen.de/stoll/schrift.html#JacobsNotes | |
| Nov 6, 2011 at 7:04 | answer | added | Manoj | timeline score: 7 | |
| Nov 6, 2011 at 3:13 | vote | accept | Manoj | ||
| Nov 6, 2011 at 3:10 | comment | added | Manoj | Thanks, Andrew. Actually that's the book I'm teaching from. He does avoid determinants, but he essentially goes through the classification theorem for finite abelian groups (putting matrices into upper triangular form: Thm 8.10). I wanted to avoid that. | |
| Nov 5, 2011 at 22:34 | comment | added | Andrew Stacey | I believe you might be interested in the book Linear Algebra Done Right by Sheldon Axler. He has a similar dislike of determinants. | |
| Nov 5, 2011 at 12:48 | comment | added | Boris Novikov | @Matthew: Sorry, I'm not right. I delete my answer. – Boris Novikov 34 mins ago | |
| Nov 5, 2011 at 12:37 | answer | added | user6976 | timeline score: 8 | |
| Nov 5, 2011 at 12:10 | answer | added | user91132 | timeline score: 29 | |
| Nov 5, 2011 at 11:51 | comment | added | Manoj | 6. Therefore, $dim(\ker h(\gamma)) + dim(\ker g(\gamma)) >= dim W$. But hold on, these kernels intersect trivially, as we already showed. So we're done! | |
| Nov 5, 2011 at 11:51 | comment | added | Manoj | 3. $\alpha$ commutes with $W=\ker (gh)(\alpha)$, so $W$ is invariant under $\alpha$. Let $\gamma$ be the restriction of $\alpha$ to $W$. We have already shown that the two other kernels of interest live inside $W$. 4. Restricted to $W$, we know that $(gh)(\gamma)$ is identically zero. Therefore the image of $h(\gamma)$ must be contained in the kernel of $g(\gamma)$. [The proof seems to get this backwards. Perhaps they are using left multiplication?] 5. But NOW! $dim(\ker h(\gamma)) + dim(im h(\gamma)) = dim W$ by the rank-nullity theorem! 6. Therefore, $dim(\ker h(\gamma)) + dim(\ker g(\ga | |
| Nov 5, 2011 at 11:50 | comment | added | Manoj | Thanks, Martin. Satz 1 would certainly give me the kind of proof I am looking for. If I'm not mistaken, it says that: Claim: If g,h are polynomials in one variable whose gcd is 1, then for every endomorphism $\alpha$, the kernel $\ker (gh)(\alpha)$ is a direct sum of $\ker g(\alpha)$ and $\ker h(\alpha)$. Proof: 1. If there is a vector $v$ in the intersection of the kernels of $g(\alpha)$ and $h(\alpha)$, then its annihilator (in the polynomial ring) is trivial, so $v=0$. 2. Proves the easy direction of the inclusion: $\ker g(\alpha) \subseteq \ker (gh) (\alpha)$. 3. $\alpha$ commutes wi | |
| Nov 5, 2011 at 10:07 | comment | added | Martin Brandenburg | A neat proof of the Jordan canonical form can be found here: matheplanet.com/matheplanet/nuke/html/article.php?sid=1028 (in german). In particular, Satz 1 might interest you. | |
| Nov 5, 2011 at 8:58 | history | edited | Manoj | CC BY-SA 3.0 |
Added note to show why conditions are necessary
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| Nov 5, 2011 at 8:48 | history | asked | Manoj | CC BY-SA 3.0 |