Sum of vectors and the effect of applying rotation and scale factors












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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



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







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    0












    $begingroup$


    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



    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







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      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



      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







      share|cite|improve this question









      $endgroup$




      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



      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






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      asked Dec 12 '18 at 9:23









      KaiserHazKaiserHaz

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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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for the insight.
            $endgroup$
            – KaiserHaz
            Dec 12 '18 at 10:42












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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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for the insight.
            $endgroup$
            – KaiserHaz
            Dec 12 '18 at 10:42
















          0












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for the insight.
            $endgroup$
            – KaiserHaz
            Dec 12 '18 at 10:42














          0












          0








          0





          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$



          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 12 '18 at 9:55

























          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


















          • $begingroup$
            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


















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