Minimum number of random vectors needed to span a space












0












$begingroup$


I am working with $mathbb{R}^{nm}$ for some $n,mgeq3$.
What is the minimum number of random vectors I need for them to span $mathbb{R}^{nm}$ (in terms of $n$ and $m$)?



I'm happy for this to involve any sampling distribution (whichever is more convenient), but Uniform would be preferred.






linear-algebra random geometric-probability







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$endgroup$








  • 2




    $begingroup$
    Isn't the probability that you pick a hyper coplanar vector $0$, since until the end the hyperplanes in which you pick vectors have measure $0$. Maybe a 3d example will illustrate it better. When you pick your first vector everything's fine, when you pick your second the probability that it is collinear is $0$ since the line has measure $0$ wrt to your volume pdf, then when you pick your third vector the probability is once more $0$ for it to be coplanar with the others, yielding that you'll always get vectors that span the whole space.
    $endgroup$
    – lightxbulb
    Jan 24 at 2:01












  • $begingroup$
    Thanks @lightxbulb, I dont know why I didn't think of that.
    $endgroup$
    – Abby
    Jan 24 at 2:31


















0












$begingroup$


I am working with $mathbb{R}^{nm}$ for some $n,mgeq3$.
What is the minimum number of random vectors I need for them to span $mathbb{R}^{nm}$ (in terms of $n$ and $m$)?



I'm happy for this to involve any sampling distribution (whichever is more convenient), but Uniform would be preferred.






linear-algebra random geometric-probability







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Isn't the probability that you pick a hyper coplanar vector $0$, since until the end the hyperplanes in which you pick vectors have measure $0$. Maybe a 3d example will illustrate it better. When you pick your first vector everything's fine, when you pick your second the probability that it is collinear is $0$ since the line has measure $0$ wrt to your volume pdf, then when you pick your third vector the probability is once more $0$ for it to be coplanar with the others, yielding that you'll always get vectors that span the whole space.
    $endgroup$
    – lightxbulb
    Jan 24 at 2:01












  • $begingroup$
    Thanks @lightxbulb, I dont know why I didn't think of that.
    $endgroup$
    – Abby
    Jan 24 at 2:31
















0












0








0





$begingroup$


I am working with $mathbb{R}^{nm}$ for some $n,mgeq3$.
What is the minimum number of random vectors I need for them to span $mathbb{R}^{nm}$ (in terms of $n$ and $m$)?



I'm happy for this to involve any sampling distribution (whichever is more convenient), but Uniform would be preferred.






linear-algebra random geometric-probability







share|cite|improve this question











$endgroup$




I am working with $mathbb{R}^{nm}$ for some $n,mgeq3$.
What is the minimum number of random vectors I need for them to span $mathbb{R}^{nm}$ (in terms of $n$ and $m$)?



I'm happy for this to involve any sampling distribution (whichever is more convenient), but Uniform would be preferred.






linear-algebra random geometric-probability




linear-algebra random geometric-probability






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 24 at 2:31







Abby

















asked Jan 24 at 1:55









AbbyAbby

928




928








  • 2




    $begingroup$
    Isn't the probability that you pick a hyper coplanar vector $0$, since until the end the hyperplanes in which you pick vectors have measure $0$. Maybe a 3d example will illustrate it better. When you pick your first vector everything's fine, when you pick your second the probability that it is collinear is $0$ since the line has measure $0$ wrt to your volume pdf, then when you pick your third vector the probability is once more $0$ for it to be coplanar with the others, yielding that you'll always get vectors that span the whole space.
    $endgroup$
    – lightxbulb
    Jan 24 at 2:01












  • $begingroup$
    Thanks @lightxbulb, I dont know why I didn't think of that.
    $endgroup$
    – Abby
    Jan 24 at 2:31
















  • 2




    $begingroup$
    Isn't the probability that you pick a hyper coplanar vector $0$, since until the end the hyperplanes in which you pick vectors have measure $0$. Maybe a 3d example will illustrate it better. When you pick your first vector everything's fine, when you pick your second the probability that it is collinear is $0$ since the line has measure $0$ wrt to your volume pdf, then when you pick your third vector the probability is once more $0$ for it to be coplanar with the others, yielding that you'll always get vectors that span the whole space.
    $endgroup$
    – lightxbulb
    Jan 24 at 2:01












  • $begingroup$
    Thanks @lightxbulb, I dont know why I didn't think of that.
    $endgroup$
    – Abby
    Jan 24 at 2:31










2




2




$begingroup$
Isn't the probability that you pick a hyper coplanar vector $0$, since until the end the hyperplanes in which you pick vectors have measure $0$. Maybe a 3d example will illustrate it better. When you pick your first vector everything's fine, when you pick your second the probability that it is collinear is $0$ since the line has measure $0$ wrt to your volume pdf, then when you pick your third vector the probability is once more $0$ for it to be coplanar with the others, yielding that you'll always get vectors that span the whole space.
$endgroup$
– lightxbulb
Jan 24 at 2:01






$begingroup$
Isn't the probability that you pick a hyper coplanar vector $0$, since until the end the hyperplanes in which you pick vectors have measure $0$. Maybe a 3d example will illustrate it better. When you pick your first vector everything's fine, when you pick your second the probability that it is collinear is $0$ since the line has measure $0$ wrt to your volume pdf, then when you pick your third vector the probability is once more $0$ for it to be coplanar with the others, yielding that you'll always get vectors that span the whole space.
$endgroup$
– lightxbulb
Jan 24 at 2:01














$begingroup$
Thanks @lightxbulb, I dont know why I didn't think of that.
$endgroup$
– Abby
Jan 24 at 2:31






$begingroup$
Thanks @lightxbulb, I dont know why I didn't think of that.
$endgroup$
– Abby
Jan 24 at 2:31












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$begingroup$

Just so that you can mark it as answered:



Isn't the probability that you pick a hyper coplanar vector 0, since until the end the hyperplanes in which you pick vectors have measure 0. Maybe a 3d example will illustrate it better. When you pick your first vector everything's fine, when you pick your second the probability that it is collinear is 0 since the line has measure 0 wrt to your volume pdf, then when you pick your third vector the probability is once more 0 for it to be coplanar with the others, yielding that you'll always get vectors that span the whole space.



Note that due to machine precision (since you're effectively sampling a discrete probability in a computer) you may get a degenerate basis with a very small probability if you implement this on a computer.






share|cite|improve this answer









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    $begingroup$

    Just so that you can mark it as answered:



    Isn't the probability that you pick a hyper coplanar vector 0, since until the end the hyperplanes in which you pick vectors have measure 0. Maybe a 3d example will illustrate it better. When you pick your first vector everything's fine, when you pick your second the probability that it is collinear is 0 since the line has measure 0 wrt to your volume pdf, then when you pick your third vector the probability is once more 0 for it to be coplanar with the others, yielding that you'll always get vectors that span the whole space.



    Note that due to machine precision (since you're effectively sampling a discrete probability in a computer) you may get a degenerate basis with a very small probability if you implement this on a computer.






    share|cite|improve this answer









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      2












      $begingroup$

      Just so that you can mark it as answered:



      Isn't the probability that you pick a hyper coplanar vector 0, since until the end the hyperplanes in which you pick vectors have measure 0. Maybe a 3d example will illustrate it better. When you pick your first vector everything's fine, when you pick your second the probability that it is collinear is 0 since the line has measure 0 wrt to your volume pdf, then when you pick your third vector the probability is once more 0 for it to be coplanar with the others, yielding that you'll always get vectors that span the whole space.



      Note that due to machine precision (since you're effectively sampling a discrete probability in a computer) you may get a degenerate basis with a very small probability if you implement this on a computer.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Just so that you can mark it as answered:



        Isn't the probability that you pick a hyper coplanar vector 0, since until the end the hyperplanes in which you pick vectors have measure 0. Maybe a 3d example will illustrate it better. When you pick your first vector everything's fine, when you pick your second the probability that it is collinear is 0 since the line has measure 0 wrt to your volume pdf, then when you pick your third vector the probability is once more 0 for it to be coplanar with the others, yielding that you'll always get vectors that span the whole space.



        Note that due to machine precision (since you're effectively sampling a discrete probability in a computer) you may get a degenerate basis with a very small probability if you implement this on a computer.






        share|cite|improve this answer









        $endgroup$



        Just so that you can mark it as answered:



        Isn't the probability that you pick a hyper coplanar vector 0, since until the end the hyperplanes in which you pick vectors have measure 0. Maybe a 3d example will illustrate it better. When you pick your first vector everything's fine, when you pick your second the probability that it is collinear is 0 since the line has measure 0 wrt to your volume pdf, then when you pick your third vector the probability is once more 0 for it to be coplanar with the others, yielding that you'll always get vectors that span the whole space.



        Note that due to machine precision (since you're effectively sampling a discrete probability in a computer) you may get a degenerate basis with a very small probability if you implement this on a computer.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 24 at 2:34









        lightxbulblightxbulb

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