Mixing
Minimum number of random vectors needed to span a space
$begingroup$
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
asked Jan 24 at 1:55
AbbyAbby
928
$endgroup$
add a comment |
$begingroup$
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
asked Jan 24 at 1:55
AbbyAbby
928
$endgroup$
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.
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– 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
add a comment |
$begingroup$
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
asked Jan 24 at 1:55
AbbyAbby
928
$endgroup$
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
asked Jan 24 at 1:55
AbbyAbby
928
asked Jan 24 at 1:55
AbbyAbby
928
edited Jan 24 at 2:31
Abby
asked Jan 24 at 1:55
AbbyAbby
928
asked Jan 24 at 1:55
AbbyAbby
928
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
add a comment |
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
add a comment |
1 Answer
1
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oldest
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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.
answered Jan 24 at 2:34
lightxbulblightxbulb
1,125311
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add a comment |
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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.
answered Jan 24 at 2:34
lightxbulblightxbulb
1,125311
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 24 at 2:34
lightxbulblightxbulb
1,125311
$endgroup$
add a comment |
$begingroup$
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.
answered Jan 24 at 2:34
lightxbulblightxbulb
1,125311
$endgroup$
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.
answered Jan 24 at 2:34
lightxbulblightxbulb
1,125311
answered Jan 24 at 2:34
lightxbulblightxbulb
1,125311
answered Jan 24 at 2:34
lightxbulblightxbulb
1,125311
answered Jan 24 at 2:34
lightxbulblightxbulb
1,125311
1,125311
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$begingroup$
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