Timeline for Are there any known criteria for quadratic mapping from R^n to R^n being surjective?
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| Jan 13, 2014 at 15:17 | comment | added | Name | I was thinking about an explanation why the example provided by Glasby works, based on your necessary and sufficient condition for $n=1$. Thank you anyway, also for your thoughtful answer. | |
| Jan 13, 2014 at 12:11 | comment | added | Alex Degtyarev | Let me illustrate the case $n=1$. All binary forms form a plane $\mathbb{R}\mathrm{p}^2$, and the discriminant is a conic in this plane. A pencil is a line, and what matters is whether this line intersects the conic. Obviously, any line is spanned by two points outside the conic, and separately any such form is equivalent to $x_1^2-x_2^2$. Changing the coordinates in the target, you can always say that the other form is arbitrary close to the first one; it's the direction that matters! | |
| Jan 13, 2014 at 12:02 | comment | added | Alex Degtyarev | In a sense, it is in terms of the original forms, as they span the linear system. But otherwise the answer is "no", as you can always choose those forms very close to each other, so that they'll be equivalent algebraically. It's the system that matters! Concerning the example, do not mix inertia index (number of negative squares in the diagonal form) and signature: in the example, both indices are equal to 1. A binary form of index 0 is $x_1^2+x_2^2$. | |
| Jan 13, 2014 at 11:42 | comment | added | Name | In your case $n=1, 2$, which is the case $n=2,3$ in the original question, (1) would it be possible to reformulate your necessary and sufficient condition purely in terms of the original quadratic forms? (2) The example provided by @Glasby the quadratic forms $x_1^2-x_2^2$ ad $2x_1x_2$ are both of trivial inertia index, in the last paragraph before stating your theorem you are talking about the negative inertia index of which quadratic form? Thank you. | |
| Jan 9, 2014 at 11:29 | history | answered | Alex Degtyarev | CC BY-SA 3.0 |