s*******y
发帖数: 558 | 1 Let Y = R X, where X is an n-dimensional vector, and R is an nxn orthogonal
matrix.
My question is:
Given X and Y, can we compute the original R?
My thoughts:
If both X and Y are two-dimensional vectors, and R is a rotation matrix (
which is also an orthogonal matrix), then given Y and X, it seems we can
compute the original rotation matrix R by finding the angle between X and Y.
However, for other cases, I don't know whether we are able to find out the
original R. | D*******a
发帖数: 3688 | 2 |X| may not equal to |Y|
【在 s*******y 的大作中提到】
: Let Y = R X, where X is an n-dimensional vector, and R is an nxn orthogonal
: matrix.
: My question is:
: Given X and Y, can we compute the original R?
: My thoughts:
: If both X and Y are two-dimensional vectors, and R is a rotation matrix (
: which is also an orthogonal matrix), then given Y and X, it seems we can
: compute the original rotation matrix R by finding the angle between X and Y.
: However, for other cases, I don't know whether we are able to find out the
: original R.
| s*******y
发帖数: 558 | 3 Thank you very much.
What if I have many pairs X1, Y1; X2, Y2; ..., Xk, Yk
such that Y1 = R X1, Y2 = R X2, ..., Yk = R Xk, where
R is an ( n x n ) dimensional orthogonal matrix.
Then how many such pairs are needed in order to re-identify R?
Obviously, if I have n such pairs, I can compute the inverse
of matrix X and get R. But do I really need n pairs?
Thanks a lot | x*****d
发帖数: 427 | 4 In general you cannot "recover" R because there are many matrices
mapping X to Y. The reason you can recover this matrix in two dimensions
is that the set S^1 of 2-dim unit vectors is equivalent to the set SO(2) of
rotation matrices. But in higher dimensions, S^{n-1} is "smaller" than
SO(n). The extra degrees of freedom is (n-1)(n-2)/2 if you look at
n-dim vectors.
【在 s*******y 的大作中提到】
: Let Y = R X, where X is an n-dimensional vector, and R is an nxn orthogonal
: matrix.
: My question is:
: Given X and Y, can we compute the original R?
: My thoughts:
: If both X and Y are two-dimensional vectors, and R is a rotation matrix (
: which is also an orthogonal matrix), then given Y and X, it seems we can
: compute the original rotation matrix R by finding the angle between X and Y.
: However, for other cases, I don't know whether we are able to find out the
: original R.
| D*******a
发帖数: 3688 | 5 |X| may not equal to |Y|
【在 s*******y 的大作中提到】
: Let Y = R X, where X is an n-dimensional vector, and R is an nxn orthogonal
: matrix.
: My question is:
: Given X and Y, can we compute the original R?
: My thoughts:
: If both X and Y are two-dimensional vectors, and R is a rotation matrix (
: which is also an orthogonal matrix), then given Y and X, it seems we can
: compute the original rotation matrix R by finding the angle between X and Y.
: However, for other cases, I don't know whether we are able to find out the
: original R.
| x*****d
发帖数: 427 | 6 This (n-1)(n-2)/2 is not the number of matrices which
carry X to Y but the dimension of the space of such matrices.
There are uncountably infinitely many matrices taking X to Y.
【在 x*****d 的大作中提到】
: In general you cannot "recover" R because there are many matrices
: mapping X to Y. The reason you can recover this matrix in two dimensions
: is that the set S^1 of 2-dim unit vectors is equivalent to the set SO(2) of
: rotation matrices. But in higher dimensions, S^{n-1} is "smaller" than
: SO(n). The extra degrees of freedom is (n-1)(n-2)/2 if you look at
: n-dim vectors.
| l*****i
发帖数: 3929 | 7 this is exactly the point!
【在 x*****d 的大作中提到】
: This (n-1)(n-2)/2 is not the number of matrices which
: carry X to Y but the dimension of the space of such matrices.
: There are uncountably infinitely many matrices taking X to Y.
| s*******y
发帖数: 558 | 8 Thank you very much.
What if I have many pairs X1, Y1; X2, Y2; ..., Xk, Yk
such that Y1 = R X1, Y2 = R X2, ..., Yk = R Xk, where
R is an ( n x n ) dimensional orthogonal matrix.
Then how many such pairs are needed in order to re-identify R?
Obviously, if I have n such pairs, I can compute the inverse
of matrix X and get R. But do I really need n pairs?
Thanks a lot
【在 x*****d 的大作中提到】
: This (n-1)(n-2)/2 is not the number of matrices which
: carry X to Y but the dimension of the space of such matrices.
: There are uncountably infinitely many matrices taking X to Y.
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