Sum of vectors and the effect of applying rotation and scale factors
$begingroup$
I have a set of vectors $mathbf{V} = [mathbf{v}_{1} cdots mathbf{v}_{M}] in mathbb{R}^{N times M}$. The sum of these vectors will form another vector $mathbf{w} in mathbb{R}^{N}$, as is already known.
Now, say that for each $mathbf{v}_{m}$, I apply a scaling and rotation factor $alpha_{m}$ and $mathbf{R}_{m} in mathbb{R}^{N times N}$, the sum will give me a vector $tilde{mathbf{w}}$ which is possibly different from $mathbf{w}$.
I don't know how to prove it (hence the question), but from some simulations, it is clear that $tilde{mathbf{w}}$ is a scaled and rotated version of $mathbf{w}$ (see Fig. 1).
We have $mathbf{w} = sum_{m=1}^{M} mathbf{v}_{m}$ and $tilde{mathbf{w}} = sum_{m=1}^{M} alpha_{m}mathbf{R}_{m}mathbf{v}_{m}$.
If we suppose that $alpha_{m} = gamma_{m}alpha_{1},~m neq 1$ and $mathbf{R}_{m} = mathbf{S}_{m}mathbf{R}_{1},~m neq 1$, we can reduce the expression to $tilde{mathbf{w}} = alpha_{1}mathbf{R}_{1}Big(mathbf{v}_{1} + sum_{m=2}^{M} gamma_{m}mathbf{S}_{m}mathbf{v}_{m}Big)$, which is still not entirely a simple scaling and rotation of the original basis $mathbf{V}$ (this is where I am at currently).
I wonder if there is a way to prove this. The objective for me is not to obtain a direct relation between the different parameters, but rather to prove simply that $tilde{mathbf{w}}$ may be characterised on the overall, as a 'total' scaling and rotation of $mathbf{w}$.
Fig.1 Example of applying scale and rotation to vectors. Black: original vectors in 3-D Euclidean space. Red: scaled and rotated versions of the original. Broken lines: individual vectors. Full line: vector formed from the sum of each individual vectors
P.S. The trivial case is when $forall~m,~alpha_{m} = gamma$ and $forall~m,~mathbf{R}_{m} = mathbf{S}$, in which case we can write $tilde{mathbf{w}} = gammamathbf{S}mathbf{w}$.
vectors rotations
asked Dec 12 '18 at 9:23
KaiserHazKaiserHaz
367
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add a comment |
$begingroup$
I have a set of vectors $mathbf{V} = [mathbf{v}_{1} cdots mathbf{v}_{M}] in mathbb{R}^{N times M}$. The sum of these vectors will form another vector $mathbf{w} in mathbb{R}^{N}$, as is already known.
Now, say that for each $mathbf{v}_{m}$, I apply a scaling and rotation factor $alpha_{m}$ and $mathbf{R}_{m} in mathbb{R}^{N times N}$, the sum will give me a vector $tilde{mathbf{w}}$ which is possibly different from $mathbf{w}$.
I don't know how to prove it (hence the question), but from some simulations, it is clear that $tilde{mathbf{w}}$ is a scaled and rotated version of $mathbf{w}$ (see Fig. 1).
We have $mathbf{w} = sum_{m=1}^{M} mathbf{v}_{m}$ and $tilde{mathbf{w}} = sum_{m=1}^{M} alpha_{m}mathbf{R}_{m}mathbf{v}_{m}$.
If we suppose that $alpha_{m} = gamma_{m}alpha_{1},~m neq 1$ and $mathbf{R}_{m} = mathbf{S}_{m}mathbf{R}_{1},~m neq 1$, we can reduce the expression to $tilde{mathbf{w}} = alpha_{1}mathbf{R}_{1}Big(mathbf{v}_{1} + sum_{m=2}^{M} gamma_{m}mathbf{S}_{m}mathbf{v}_{m}Big)$, which is still not entirely a simple scaling and rotation of the original basis $mathbf{V}$ (this is where I am at currently).
I wonder if there is a way to prove this. The objective for me is not to obtain a direct relation between the different parameters, but rather to prove simply that $tilde{mathbf{w}}$ may be characterised on the overall, as a 'total' scaling and rotation of $mathbf{w}$.
Fig.1 Example of applying scale and rotation to vectors. Black: original vectors in 3-D Euclidean space. Red: scaled and rotated versions of the original. Broken lines: individual vectors. Full line: vector formed from the sum of each individual vectors
P.S. The trivial case is when $forall~m,~alpha_{m} = gamma$ and $forall~m,~mathbf{R}_{m} = mathbf{S}$, in which case we can write $tilde{mathbf{w}} = gammamathbf{S}mathbf{w}$.
vectors rotations
asked Dec 12 '18 at 9:23
KaiserHazKaiserHaz
367
$endgroup$
add a comment |
$begingroup$
I have a set of vectors $mathbf{V} = [mathbf{v}_{1} cdots mathbf{v}_{M}] in mathbb{R}^{N times M}$. The sum of these vectors will form another vector $mathbf{w} in mathbb{R}^{N}$, as is already known.
Now, say that for each $mathbf{v}_{m}$, I apply a scaling and rotation factor $alpha_{m}$ and $mathbf{R}_{m} in mathbb{R}^{N times N}$, the sum will give me a vector $tilde{mathbf{w}}$ which is possibly different from $mathbf{w}$.
I don't know how to prove it (hence the question), but from some simulations, it is clear that $tilde{mathbf{w}}$ is a scaled and rotated version of $mathbf{w}$ (see Fig. 1).
We have $mathbf{w} = sum_{m=1}^{M} mathbf{v}_{m}$ and $tilde{mathbf{w}} = sum_{m=1}^{M} alpha_{m}mathbf{R}_{m}mathbf{v}_{m}$.
If we suppose that $alpha_{m} = gamma_{m}alpha_{1},~m neq 1$ and $mathbf{R}_{m} = mathbf{S}_{m}mathbf{R}_{1},~m neq 1$, we can reduce the expression to $tilde{mathbf{w}} = alpha_{1}mathbf{R}_{1}Big(mathbf{v}_{1} + sum_{m=2}^{M} gamma_{m}mathbf{S}_{m}mathbf{v}_{m}Big)$, which is still not entirely a simple scaling and rotation of the original basis $mathbf{V}$ (this is where I am at currently).
I wonder if there is a way to prove this. The objective for me is not to obtain a direct relation between the different parameters, but rather to prove simply that $tilde{mathbf{w}}$ may be characterised on the overall, as a 'total' scaling and rotation of $mathbf{w}$.
Fig.1 Example of applying scale and rotation to vectors. Black: original vectors in 3-D Euclidean space. Red: scaled and rotated versions of the original. Broken lines: individual vectors. Full line: vector formed from the sum of each individual vectors
P.S. The trivial case is when $forall~m,~alpha_{m} = gamma$ and $forall~m,~mathbf{R}_{m} = mathbf{S}$, in which case we can write $tilde{mathbf{w}} = gammamathbf{S}mathbf{w}$.
vectors rotations
asked Dec 12 '18 at 9:23
KaiserHazKaiserHaz
367
$endgroup$
I have a set of vectors $mathbf{V} = [mathbf{v}_{1} cdots mathbf{v}_{M}] in mathbb{R}^{N times M}$. The sum of these vectors will form another vector $mathbf{w} in mathbb{R}^{N}$, as is already known.
Now, say that for each $mathbf{v}_{m}$, I apply a scaling and rotation factor $alpha_{m}$ and $mathbf{R}_{m} in mathbb{R}^{N times N}$, the sum will give me a vector $tilde{mathbf{w}}$ which is possibly different from $mathbf{w}$.
I don't know how to prove it (hence the question), but from some simulations, it is clear that $tilde{mathbf{w}}$ is a scaled and rotated version of $mathbf{w}$ (see Fig. 1).
We have $mathbf{w} = sum_{m=1}^{M} mathbf{v}_{m}$ and $tilde{mathbf{w}} = sum_{m=1}^{M} alpha_{m}mathbf{R}_{m}mathbf{v}_{m}$.
If we suppose that $alpha_{m} = gamma_{m}alpha_{1},~m neq 1$ and $mathbf{R}_{m} = mathbf{S}_{m}mathbf{R}_{1},~m neq 1$, we can reduce the expression to $tilde{mathbf{w}} = alpha_{1}mathbf{R}_{1}Big(mathbf{v}_{1} + sum_{m=2}^{M} gamma_{m}mathbf{S}_{m}mathbf{v}_{m}Big)$, which is still not entirely a simple scaling and rotation of the original basis $mathbf{V}$ (this is where I am at currently).
I wonder if there is a way to prove this. The objective for me is not to obtain a direct relation between the different parameters, but rather to prove simply that $tilde{mathbf{w}}$ may be characterised on the overall, as a 'total' scaling and rotation of $mathbf{w}$.
Fig.1 Example of applying scale and rotation to vectors. Black: original vectors in 3-D Euclidean space. Red: scaled and rotated versions of the original. Broken lines: individual vectors. Full line: vector formed from the sum of each individual vectors
P.S. The trivial case is when $forall~m,~alpha_{m} = gamma$ and $forall~m,~mathbf{R}_{m} = mathbf{S}$, in which case we can write $tilde{mathbf{w}} = gammamathbf{S}mathbf{w}$.
vectors rotations
vectors rotations
asked Dec 12 '18 at 9:23
KaiserHazKaiserHaz
367
asked Dec 12 '18 at 9:23
KaiserHazKaiserHaz
367
asked Dec 12 '18 at 9:23
KaiserHazKaiserHaz
367
asked Dec 12 '18 at 9:23
KaiserHazKaiserHaz
367
asked Dec 12 '18 at 9:23
KaiserHazKaiserHaz
367
367
add a comment |
add a comment |
1 Answer
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$begingroup$
Two arbitrary vectors can always be related to each other by a scaling and a rotation. In particular, the scaling factor is the ratio of the norms; in 3D, the rotation axis is parallel to the cross product, and in $N$D it is not uniquely defined.
Your question is somewhat vacuous.
answered Dec 12 '18 at 9:49
Yves DaoustYves Daoust
132k676229
$endgroup$
$begingroup$
Thank you for the insight.
$endgroup$
– KaiserHaz
Dec 12 '18 at 10:42
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
Two arbitrary vectors can always be related to each other by a scaling and a rotation. In particular, the scaling factor is the ratio of the norms; in 3D, the rotation axis is parallel to the cross product, and in $N$D it is not uniquely defined.
Your question is somewhat vacuous.
answered Dec 12 '18 at 9:49
Yves DaoustYves Daoust
132k676229
$endgroup$
$begingroup$
Thank you for the insight.
$endgroup$
– KaiserHaz
Dec 12 '18 at 10:42
add a comment |
$begingroup$
Two arbitrary vectors can always be related to each other by a scaling and a rotation. In particular, the scaling factor is the ratio of the norms; in 3D, the rotation axis is parallel to the cross product, and in $N$D it is not uniquely defined.
Your question is somewhat vacuous.
answered Dec 12 '18 at 9:49
Yves DaoustYves Daoust
132k676229
$endgroup$
$begingroup$
Thank you for the insight.
$endgroup$
– KaiserHaz
Dec 12 '18 at 10:42
add a comment |
$begingroup$
Two arbitrary vectors can always be related to each other by a scaling and a rotation. In particular, the scaling factor is the ratio of the norms; in 3D, the rotation axis is parallel to the cross product, and in $N$D it is not uniquely defined.
Your question is somewhat vacuous.
answered Dec 12 '18 at 9:49
Yves DaoustYves Daoust
132k676229
$endgroup$
Two arbitrary vectors can always be related to each other by a scaling and a rotation. In particular, the scaling factor is the ratio of the norms; in 3D, the rotation axis is parallel to the cross product, and in $N$D it is not uniquely defined.
Your question is somewhat vacuous.
answered Dec 12 '18 at 9:49
Yves DaoustYves Daoust
132k676229
edited Dec 12 '18 at 9:55
answered Dec 12 '18 at 9:49
Yves DaoustYves Daoust
132k676229
answered Dec 12 '18 at 9:49
Yves DaoustYves Daoust
132k676229
answered Dec 12 '18 at 9:49
Yves DaoustYves Daoust
132k676229
132k676229
$begingroup$
Thank you for the insight.
$endgroup$
– KaiserHaz
Dec 12 '18 at 10:42
add a comment |
$begingroup$
Thank you for the insight.
$endgroup$
– KaiserHaz
Dec 12 '18 at 10:42
$begingroup$
Thank you for the insight.
$endgroup$
– KaiserHaz
Dec 12 '18 at 10:42
$begingroup$
Thank you for the insight.
$endgroup$
– KaiserHaz
Dec 12 '18 at 10:42
add a comment |
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