Timeline for Eigenvectors of sum of SO(3) matrices
Current License: CC BY-SA 4.0
16 events
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| Sep 5, 2018 at 16:53 | comment | added | myorbs | uh yes, as the question states, under what circumstances is the real eigenvector independent of the rotation angles. That is what i meant in the previous comment, not simply if it exists (because you are right there is always a real eigenvector). | |
| Sep 5, 2018 at 16:16 | comment | added | user35593 | Every 3x3 matrix has a real eigenvector since the characteristic polynomial has odd degree and therefore a real root... | |
| Sep 5, 2018 at 14:58 | comment | added | myorbs | Yes $\alpha=0$ makes $A=I$. It is a bit trivial. But if one can show that the real eigenvector exists only when either of $\beta,\alpha$ is zero then that is interesting to me | |
| Sep 5, 2018 at 14:35 | review | Close votes | |||
| Sep 6, 2018 at 7:37 | |||||
| Sep 5, 2018 at 14:34 | comment | added | user35593 | If you set $\alpha=0, \beta\neq 0$ the only real eigenvector is $\bar{b}$ and vice versa. So x | |
| Sep 5, 2018 at 14:33 | history | edited | myorbs | CC BY-SA 4.0 |
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| Sep 5, 2018 at 14:18 | history | edited | myorbs | CC BY-SA 4.0 |
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| Sep 5, 2018 at 14:16 | comment | added | myorbs | If $B=B'+P$, where $B'$ commutes with $A$ and $K$ is a small pertubation, that too might be useful. I do not know how to approach it though :) | |
| Sep 5, 2018 at 14:12 | comment | added | Steven Landsburg | Duplicate: math.stackexchange.com/questions/2883868/… | |
| Sep 5, 2018 at 13:40 | history | edited | myorbs | CC BY-SA 4.0 |
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| Sep 5, 2018 at 11:31 | comment | added | myorbs | You are right, I was sloppy asking about the SO(n) case. This, in the SO(3) case is then the requirement that the matrices commute. However, is this the only possibility? I.e. A+B has a real eigenvector if and only if $W$ is non-empty? | |
| Sep 5, 2018 at 11:14 | comment | added | Stefano Gogioso | You can make this argument arbitrarily general, by the way: if $A$ rotates about $k_A$ orthogonal planes and $B$ rotates about $k_B$ orthogonal planes, then taking $n \geq 2k_A+2k_B+1$ always guarantees the existence of at least on eigenvector for $A+B$ independent of all rotation angles. | |
| Sep 5, 2018 at 11:11 | comment | added | Stefano Gogioso | For an example of conditions under which such a real eigenvector exists independently of the angles, let $n\geq 5$ and let $A,B$ be rotations on a single plane $P_A,P_B$ each (by which I mean that they are the identity on the $(n-2)$-dimensional subspaces $P_A^\bot,P_B^\bot$). Then such eigenvectors always exist: any vector in the subspace $W:= P_A^\bot \cap P_B^\bot$, which is at least $(n-4)$-dimensional, is a real eigenvector of $A+B$ with eigenvalue 2. To see this, choose diagonalisations of both $A$ and $B$ which agree on $W$. | |
| Sep 5, 2018 at 11:10 | comment | added | Stefano Gogioso | If $n>3$, it makes no sense to say that matrices describe rotations about axes: it only makes sense to say that they describe rotations inside planes (2-dimensional subspaces). Also, you should be aware that this is no longer the general form of $SO(n)$ matrices: they can simultaneously rotate inside two or more orthogonal planes, with independent angles for each plane. See e.g. this wikipedia article. | |
| Sep 5, 2018 at 9:55 | review | First posts | |||
| Sep 5, 2018 at 10:25 | |||||
| Sep 5, 2018 at 9:51 | history | asked | myorbs | CC BY-SA 4.0 |