Timeline for Rank of a linear combination of quadratic forms
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20 events
| when toggle format | what | by | license | comment | |
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| Apr 23 at 23:05 | history | edited | KConrad |
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| Apr 23 at 21:47 | answer | added | David Leep | timeline score: 1 | |
| Sep 1, 2010 at 17:38 | vote | accept | sobe86 | ||
| Aug 31, 2010 at 23:27 | comment | added | darij grinberg | Uhm... If $n\geq 5k$ and all forms are of rank $< 5$, then the codimension of the kernel of each form is $< 5$ (since the codimension of the kernel of a linear map is always equal to the rank), which means the same as $\leq 4$, and thus the codimension of the kernel of their sum is $\leq 4k$ which is quite a lot smaller than $5k=n$. So the kernel at least has dimension $k$. | |
| Aug 31, 2010 at 22:41 | comment | added | sobe86 | Oh yeah, sorry I should be more clear: I'm not expecting to be able to get full rank anymore. For my purposes all I actually need is rank 5. Will edit OP. How about something like this: we only need one of our original forms to be of rank 5 and we will be done, if $n \geq 5k$ then either one of the forms must be of rank $\geq5$ or the dimension of the intersection of the nullspaces will be greater than 0 ? | |
| Aug 31, 2010 at 22:15 | history | edited | sobe86 | CC BY-SA 2.5 |
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| Aug 31, 2010 at 20:10 | comment | added | darij grinberg | Can't you get a $n=3\nu$, $k=2$ example from Dustin's $n=3$, $k=2$ example by copypasting his matrices as diagonal blocks? | |
| Aug 31, 2010 at 18:47 | answer | added | Dustin Cartwright | timeline score: 3 | |
| Aug 31, 2010 at 18:18 | comment | added | sobe86 | I don't understand the problem with Noah's comment. It would seem to me that $x_1^2, x_1 x_2 , \dots , x_1 x_n$ are a counterexample to my original question since: $x_1^2 + 2\lambda_2 x_1 x_2 + \dots 2\lambda_n x_1 x_n$ equals $(x_1+ \lambda_2 x_2 + \lambda_n x_n^2)^2 - (\lambda_2 x_2 + \lambda_n x_n^2)^2$ so this will always have rank 2 The 2nd part of my question - what if we take $k \ll n ?$ | |
| Aug 31, 2010 at 18:04 | comment | added | Will Jagy | Keivan, I see, "Darij, your formulation does not quite capture the problem." $$ $$ As always, I would like for the OP to show what happens in 2 by 2 and perhaps 3 by 3. | |
| Aug 31, 2010 at 17:41 | comment | added | Keivan Karai | Noah: I don't think so: think of the forms $x_1x_2$ and $x_3x_4$ whose sum is non-degenerate. | |
| Aug 31, 2010 at 17:31 | comment | added | Noah Stein | Will: If you symmetrize Keivan's counterexample (replace $E_{1j}$ with $E_{1j}+E_{j1}$) you get a symmetric counterexample. | |
| Aug 31, 2010 at 17:30 | comment | added | Keivan Karai | Will: I agree and I don't know how to do it for symmetric matrices. My comment was in refernce to darji's comment, basically saying that the extra assumption is needed. | |
| Aug 31, 2010 at 17:23 | comment | added | Will Jagy | Keivan, the matrices would be symmetric. | |
| Aug 31, 2010 at 17:22 | comment | added | Will Jagy | Could you please show what happens with only 2 variables? Your command boldsymbol does not work for me, I am still using jsMath, but you might switch to \bf or leave it out... | |
| Aug 31, 2010 at 17:13 | comment | added | Keivan Karai | The answer to this is no. Take for instance the elementary matrices $E_{1j}$ for $1 \le j \le n$. | |
| Aug 31, 2010 at 16:53 | comment | added | darij grinberg | In matrixspeak: Given $k$ square matrices whose kernels have trivial intersection, can we find a nonsingular matrices which can be written as a linear combination of our $k$ matrices? | |
| Aug 31, 2010 at 16:09 | history | edited | sobe86 |
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| Aug 31, 2010 at 15:49 | history | edited | sobe86 | CC BY-SA 2.5 |
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| Aug 31, 2010 at 15:31 | history | asked | sobe86 | CC BY-SA 2.5 |