Vector decomposition in linear space












0












$begingroup$


There is a finite set of features $F = {f_1, ..., f_n}$. System registers a signal $s$ that is a vector from a linear span of F $(1)$.



There is a set of "unit signals" that are vectors from $(1): u_1, ..., u_m, m ll n.$ Signal $s$ could be decomposed into a kind of linear combination:



$s = alpha_1 u_1 + ... + alpha_m u_m + theta (2)$



where $theta$ is a compensation vector (background noise).



The challenge is to find an optimal set of parameters $alpha_1, ..., alpha_m$ so that $theta$ has a minimal $L_2$ norm.










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$endgroup$

















    0












    $begingroup$


    There is a finite set of features $F = {f_1, ..., f_n}$. System registers a signal $s$ that is a vector from a linear span of F $(1)$.



    There is a set of "unit signals" that are vectors from $(1): u_1, ..., u_m, m ll n.$ Signal $s$ could be decomposed into a kind of linear combination:



    $s = alpha_1 u_1 + ... + alpha_m u_m + theta (2)$



    where $theta$ is a compensation vector (background noise).



    The challenge is to find an optimal set of parameters $alpha_1, ..., alpha_m$ so that $theta$ has a minimal $L_2$ norm.










    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      There is a finite set of features $F = {f_1, ..., f_n}$. System registers a signal $s$ that is a vector from a linear span of F $(1)$.



      There is a set of "unit signals" that are vectors from $(1): u_1, ..., u_m, m ll n.$ Signal $s$ could be decomposed into a kind of linear combination:



      $s = alpha_1 u_1 + ... + alpha_m u_m + theta (2)$



      where $theta$ is a compensation vector (background noise).



      The challenge is to find an optimal set of parameters $alpha_1, ..., alpha_m$ so that $theta$ has a minimal $L_2$ norm.










      share|cite|improve this question











      $endgroup$




      There is a finite set of features $F = {f_1, ..., f_n}$. System registers a signal $s$ that is a vector from a linear span of F $(1)$.



      There is a set of "unit signals" that are vectors from $(1): u_1, ..., u_m, m ll n.$ Signal $s$ could be decomposed into a kind of linear combination:



      $s = alpha_1 u_1 + ... + alpha_m u_m + theta (2)$



      where $theta$ is a compensation vector (background noise).



      The challenge is to find an optimal set of parameters $alpha_1, ..., alpha_m$ so that $theta$ has a minimal $L_2$ norm.







      linear-algebra optimization vector-spaces vectors






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 28 '18 at 22:11







      Denis Kulagin

















      asked Dec 28 '18 at 22:00









      Denis KulaginDenis Kulagin

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      1454






















          1 Answer
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          $begingroup$

          First, if the $u_i$ are not linearly independent, replace them with a maximal linearly independent subset (apply the sifting algorithm).



          Since the $L_2$ norm comes from an inner product $langle -,-rangle$, we can orthonormalise $(u_1,ldots,u_m)$ by the Gram-Schmidt process into $(v_1,ldots,v_m)$. Extend this to an orthonormal basis of the whole space (this is easy: extend it by the standard basis vectors, sift, and Gram-Schmidt whatever's left: if you're doing this calculation lots of times and need it very efficient, the extended basis doesn't actually need to be orthonormal for an actual implementation) $(v_1,ldots,v_n)$. Then there are unique coefficients $beta_1,ldots,beta_n$ such that $s = sumlimits_{i=1}^n beta_i v_i$. Choose $theta = sumlimits_{i=m+1}^nbeta_iv_i$. Then $s - theta = sumlimits_{i=1}^mbeta_iv_i$ lies in the linear span of the $v_i$, which is exactly the linear span of the $u_i$, so there are unique $alpha_1,ldots,alpha_m$ such that $s-theta = sumlimits_{i=1}^malpha_iu_i$ (and obtaining these from the $beta_i$ is easy, since we get the transition matrix from the $u_i$ to the first $m$ of the $v_i$ out of the Gram-Schmidt process for free).



          Now, $|theta|_2^2 = sumlimits_{i=m+1}^n|beta_i|^2$, since the $v_i$ are orthonormal. Further, if there is some $varphineqtheta$ and some other choice of $alpha_i$ such that $s = sumlimits_{i=1}^malpha_iu_i + varphi$, then, since the $v_i$ form a basis, $varphi$ must be of the form $theta + sumlimits_{i=1}^mgamma_iv_i$ for some scalars $gamma_i$, so $left|varphiright|^2_2 = left|sumlimits_{i=1}^mgamma_iv_i+sumlimits_{i=m+1}^nbeta_iv_iright|^2_2 = sumlimits_{i=1}^m|gamma_i|^2+sumlimits_{i=m+1}^n|beta_i|^2$ by orthonormality of the $v_i$, which is strictly greater than $sumlimits_{i=m+1}^n|b_i|^2 = |theta|_2^2$ (with the strictness since we have equality only when $|gamma_i| = 0$ for all $i$, but in that case, $theta = varphi$, a contradiction). Thus, our chosen $alpha_i$ are those that minimise the associated $theta$, as required.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks! I was thinking differentiation and your solution is obviously smarter and more native to the field of LA. Now there is a little challenge to implement it in code, but it shouldn't be too hard.
            $endgroup$
            – Denis Kulagin
            Dec 29 '18 at 8:07








          • 1




            $begingroup$
            I've just noticed two minor errors in there (firstly: I'm assuming that the $alpha_i$ are linearly independent: if not, just take a maximal linearly independent subset first; and secondly, there are a whole bunch of $^2$s missing in the proof. Neither changes the result at all, but I've fixed them.
            $endgroup$
            – user3482749
            Dec 29 '18 at 9:58












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          1












          $begingroup$

          First, if the $u_i$ are not linearly independent, replace them with a maximal linearly independent subset (apply the sifting algorithm).



          Since the $L_2$ norm comes from an inner product $langle -,-rangle$, we can orthonormalise $(u_1,ldots,u_m)$ by the Gram-Schmidt process into $(v_1,ldots,v_m)$. Extend this to an orthonormal basis of the whole space (this is easy: extend it by the standard basis vectors, sift, and Gram-Schmidt whatever's left: if you're doing this calculation lots of times and need it very efficient, the extended basis doesn't actually need to be orthonormal for an actual implementation) $(v_1,ldots,v_n)$. Then there are unique coefficients $beta_1,ldots,beta_n$ such that $s = sumlimits_{i=1}^n beta_i v_i$. Choose $theta = sumlimits_{i=m+1}^nbeta_iv_i$. Then $s - theta = sumlimits_{i=1}^mbeta_iv_i$ lies in the linear span of the $v_i$, which is exactly the linear span of the $u_i$, so there are unique $alpha_1,ldots,alpha_m$ such that $s-theta = sumlimits_{i=1}^malpha_iu_i$ (and obtaining these from the $beta_i$ is easy, since we get the transition matrix from the $u_i$ to the first $m$ of the $v_i$ out of the Gram-Schmidt process for free).



          Now, $|theta|_2^2 = sumlimits_{i=m+1}^n|beta_i|^2$, since the $v_i$ are orthonormal. Further, if there is some $varphineqtheta$ and some other choice of $alpha_i$ such that $s = sumlimits_{i=1}^malpha_iu_i + varphi$, then, since the $v_i$ form a basis, $varphi$ must be of the form $theta + sumlimits_{i=1}^mgamma_iv_i$ for some scalars $gamma_i$, so $left|varphiright|^2_2 = left|sumlimits_{i=1}^mgamma_iv_i+sumlimits_{i=m+1}^nbeta_iv_iright|^2_2 = sumlimits_{i=1}^m|gamma_i|^2+sumlimits_{i=m+1}^n|beta_i|^2$ by orthonormality of the $v_i$, which is strictly greater than $sumlimits_{i=m+1}^n|b_i|^2 = |theta|_2^2$ (with the strictness since we have equality only when $|gamma_i| = 0$ for all $i$, but in that case, $theta = varphi$, a contradiction). Thus, our chosen $alpha_i$ are those that minimise the associated $theta$, as required.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks! I was thinking differentiation and your solution is obviously smarter and more native to the field of LA. Now there is a little challenge to implement it in code, but it shouldn't be too hard.
            $endgroup$
            – Denis Kulagin
            Dec 29 '18 at 8:07








          • 1




            $begingroup$
            I've just noticed two minor errors in there (firstly: I'm assuming that the $alpha_i$ are linearly independent: if not, just take a maximal linearly independent subset first; and secondly, there are a whole bunch of $^2$s missing in the proof. Neither changes the result at all, but I've fixed them.
            $endgroup$
            – user3482749
            Dec 29 '18 at 9:58
















          1












          $begingroup$

          First, if the $u_i$ are not linearly independent, replace them with a maximal linearly independent subset (apply the sifting algorithm).



          Since the $L_2$ norm comes from an inner product $langle -,-rangle$, we can orthonormalise $(u_1,ldots,u_m)$ by the Gram-Schmidt process into $(v_1,ldots,v_m)$. Extend this to an orthonormal basis of the whole space (this is easy: extend it by the standard basis vectors, sift, and Gram-Schmidt whatever's left: if you're doing this calculation lots of times and need it very efficient, the extended basis doesn't actually need to be orthonormal for an actual implementation) $(v_1,ldots,v_n)$. Then there are unique coefficients $beta_1,ldots,beta_n$ such that $s = sumlimits_{i=1}^n beta_i v_i$. Choose $theta = sumlimits_{i=m+1}^nbeta_iv_i$. Then $s - theta = sumlimits_{i=1}^mbeta_iv_i$ lies in the linear span of the $v_i$, which is exactly the linear span of the $u_i$, so there are unique $alpha_1,ldots,alpha_m$ such that $s-theta = sumlimits_{i=1}^malpha_iu_i$ (and obtaining these from the $beta_i$ is easy, since we get the transition matrix from the $u_i$ to the first $m$ of the $v_i$ out of the Gram-Schmidt process for free).



          Now, $|theta|_2^2 = sumlimits_{i=m+1}^n|beta_i|^2$, since the $v_i$ are orthonormal. Further, if there is some $varphineqtheta$ and some other choice of $alpha_i$ such that $s = sumlimits_{i=1}^malpha_iu_i + varphi$, then, since the $v_i$ form a basis, $varphi$ must be of the form $theta + sumlimits_{i=1}^mgamma_iv_i$ for some scalars $gamma_i$, so $left|varphiright|^2_2 = left|sumlimits_{i=1}^mgamma_iv_i+sumlimits_{i=m+1}^nbeta_iv_iright|^2_2 = sumlimits_{i=1}^m|gamma_i|^2+sumlimits_{i=m+1}^n|beta_i|^2$ by orthonormality of the $v_i$, which is strictly greater than $sumlimits_{i=m+1}^n|b_i|^2 = |theta|_2^2$ (with the strictness since we have equality only when $|gamma_i| = 0$ for all $i$, but in that case, $theta = varphi$, a contradiction). Thus, our chosen $alpha_i$ are those that minimise the associated $theta$, as required.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks! I was thinking differentiation and your solution is obviously smarter and more native to the field of LA. Now there is a little challenge to implement it in code, but it shouldn't be too hard.
            $endgroup$
            – Denis Kulagin
            Dec 29 '18 at 8:07








          • 1




            $begingroup$
            I've just noticed two minor errors in there (firstly: I'm assuming that the $alpha_i$ are linearly independent: if not, just take a maximal linearly independent subset first; and secondly, there are a whole bunch of $^2$s missing in the proof. Neither changes the result at all, but I've fixed them.
            $endgroup$
            – user3482749
            Dec 29 '18 at 9:58














          1












          1








          1





          $begingroup$

          First, if the $u_i$ are not linearly independent, replace them with a maximal linearly independent subset (apply the sifting algorithm).



          Since the $L_2$ norm comes from an inner product $langle -,-rangle$, we can orthonormalise $(u_1,ldots,u_m)$ by the Gram-Schmidt process into $(v_1,ldots,v_m)$. Extend this to an orthonormal basis of the whole space (this is easy: extend it by the standard basis vectors, sift, and Gram-Schmidt whatever's left: if you're doing this calculation lots of times and need it very efficient, the extended basis doesn't actually need to be orthonormal for an actual implementation) $(v_1,ldots,v_n)$. Then there are unique coefficients $beta_1,ldots,beta_n$ such that $s = sumlimits_{i=1}^n beta_i v_i$. Choose $theta = sumlimits_{i=m+1}^nbeta_iv_i$. Then $s - theta = sumlimits_{i=1}^mbeta_iv_i$ lies in the linear span of the $v_i$, which is exactly the linear span of the $u_i$, so there are unique $alpha_1,ldots,alpha_m$ such that $s-theta = sumlimits_{i=1}^malpha_iu_i$ (and obtaining these from the $beta_i$ is easy, since we get the transition matrix from the $u_i$ to the first $m$ of the $v_i$ out of the Gram-Schmidt process for free).



          Now, $|theta|_2^2 = sumlimits_{i=m+1}^n|beta_i|^2$, since the $v_i$ are orthonormal. Further, if there is some $varphineqtheta$ and some other choice of $alpha_i$ such that $s = sumlimits_{i=1}^malpha_iu_i + varphi$, then, since the $v_i$ form a basis, $varphi$ must be of the form $theta + sumlimits_{i=1}^mgamma_iv_i$ for some scalars $gamma_i$, so $left|varphiright|^2_2 = left|sumlimits_{i=1}^mgamma_iv_i+sumlimits_{i=m+1}^nbeta_iv_iright|^2_2 = sumlimits_{i=1}^m|gamma_i|^2+sumlimits_{i=m+1}^n|beta_i|^2$ by orthonormality of the $v_i$, which is strictly greater than $sumlimits_{i=m+1}^n|b_i|^2 = |theta|_2^2$ (with the strictness since we have equality only when $|gamma_i| = 0$ for all $i$, but in that case, $theta = varphi$, a contradiction). Thus, our chosen $alpha_i$ are those that minimise the associated $theta$, as required.






          share|cite|improve this answer











          $endgroup$



          First, if the $u_i$ are not linearly independent, replace them with a maximal linearly independent subset (apply the sifting algorithm).



          Since the $L_2$ norm comes from an inner product $langle -,-rangle$, we can orthonormalise $(u_1,ldots,u_m)$ by the Gram-Schmidt process into $(v_1,ldots,v_m)$. Extend this to an orthonormal basis of the whole space (this is easy: extend it by the standard basis vectors, sift, and Gram-Schmidt whatever's left: if you're doing this calculation lots of times and need it very efficient, the extended basis doesn't actually need to be orthonormal for an actual implementation) $(v_1,ldots,v_n)$. Then there are unique coefficients $beta_1,ldots,beta_n$ such that $s = sumlimits_{i=1}^n beta_i v_i$. Choose $theta = sumlimits_{i=m+1}^nbeta_iv_i$. Then $s - theta = sumlimits_{i=1}^mbeta_iv_i$ lies in the linear span of the $v_i$, which is exactly the linear span of the $u_i$, so there are unique $alpha_1,ldots,alpha_m$ such that $s-theta = sumlimits_{i=1}^malpha_iu_i$ (and obtaining these from the $beta_i$ is easy, since we get the transition matrix from the $u_i$ to the first $m$ of the $v_i$ out of the Gram-Schmidt process for free).



          Now, $|theta|_2^2 = sumlimits_{i=m+1}^n|beta_i|^2$, since the $v_i$ are orthonormal. Further, if there is some $varphineqtheta$ and some other choice of $alpha_i$ such that $s = sumlimits_{i=1}^malpha_iu_i + varphi$, then, since the $v_i$ form a basis, $varphi$ must be of the form $theta + sumlimits_{i=1}^mgamma_iv_i$ for some scalars $gamma_i$, so $left|varphiright|^2_2 = left|sumlimits_{i=1}^mgamma_iv_i+sumlimits_{i=m+1}^nbeta_iv_iright|^2_2 = sumlimits_{i=1}^m|gamma_i|^2+sumlimits_{i=m+1}^n|beta_i|^2$ by orthonormality of the $v_i$, which is strictly greater than $sumlimits_{i=m+1}^n|b_i|^2 = |theta|_2^2$ (with the strictness since we have equality only when $|gamma_i| = 0$ for all $i$, but in that case, $theta = varphi$, a contradiction). Thus, our chosen $alpha_i$ are those that minimise the associated $theta$, as required.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 29 '18 at 9:59

























          answered Dec 28 '18 at 22:29









          user3482749user3482749

          4,3291119




          4,3291119












          • $begingroup$
            Thanks! I was thinking differentiation and your solution is obviously smarter and more native to the field of LA. Now there is a little challenge to implement it in code, but it shouldn't be too hard.
            $endgroup$
            – Denis Kulagin
            Dec 29 '18 at 8:07








          • 1




            $begingroup$
            I've just noticed two minor errors in there (firstly: I'm assuming that the $alpha_i$ are linearly independent: if not, just take a maximal linearly independent subset first; and secondly, there are a whole bunch of $^2$s missing in the proof. Neither changes the result at all, but I've fixed them.
            $endgroup$
            – user3482749
            Dec 29 '18 at 9:58


















          • $begingroup$
            Thanks! I was thinking differentiation and your solution is obviously smarter and more native to the field of LA. Now there is a little challenge to implement it in code, but it shouldn't be too hard.
            $endgroup$
            – Denis Kulagin
            Dec 29 '18 at 8:07








          • 1




            $begingroup$
            I've just noticed two minor errors in there (firstly: I'm assuming that the $alpha_i$ are linearly independent: if not, just take a maximal linearly independent subset first; and secondly, there are a whole bunch of $^2$s missing in the proof. Neither changes the result at all, but I've fixed them.
            $endgroup$
            – user3482749
            Dec 29 '18 at 9:58
















          $begingroup$
          Thanks! I was thinking differentiation and your solution is obviously smarter and more native to the field of LA. Now there is a little challenge to implement it in code, but it shouldn't be too hard.
          $endgroup$
          – Denis Kulagin
          Dec 29 '18 at 8:07






          $begingroup$
          Thanks! I was thinking differentiation and your solution is obviously smarter and more native to the field of LA. Now there is a little challenge to implement it in code, but it shouldn't be too hard.
          $endgroup$
          – Denis Kulagin
          Dec 29 '18 at 8:07






          1




          1




          $begingroup$
          I've just noticed two minor errors in there (firstly: I'm assuming that the $alpha_i$ are linearly independent: if not, just take a maximal linearly independent subset first; and secondly, there are a whole bunch of $^2$s missing in the proof. Neither changes the result at all, but I've fixed them.
          $endgroup$
          – user3482749
          Dec 29 '18 at 9:58




          $begingroup$
          I've just noticed two minor errors in there (firstly: I'm assuming that the $alpha_i$ are linearly independent: if not, just take a maximal linearly independent subset first; and secondly, there are a whole bunch of $^2$s missing in the proof. Neither changes the result at all, but I've fixed them.
          $endgroup$
          – user3482749
          Dec 29 '18 at 9:58


















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