Vector decomposition in linear space
$begingroup$
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.
linear-algebra optimization vector-spaces vectors
$endgroup$
add a comment |
$begingroup$
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.
linear-algebra optimization vector-spaces vectors
$endgroup$
add a comment |
$begingroup$
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.
linear-algebra optimization vector-spaces vectors
$endgroup$
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
linear-algebra optimization vector-spaces vectors
edited Dec 28 '18 at 22:11
Denis Kulagin
asked Dec 28 '18 at 22:00
Denis KulaginDenis Kulagin
1454
1454
add a comment |
add a comment |
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.
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
$begingroup$
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
add a comment |
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