Linearly independence after feature embedding












0












$begingroup$


Problem



If feature vectors $mathbf{x}_1, mathbf{x}_2,cdots,mathbf{x}_m$ are linearly independent, argue whether or not their embedding $psi(mathbf{x}_1), psi(mathbf{x}_2),cdots,psi(mathbf{x}_m)$ are linearly independent.



Some Thoughts



The complication here is that the choice of $psi(cdot)$ is not random and often requires that the resulting $psi(mathbf{x}_i)$'s being linearly separable (or we could always do $psi:mathbf{x}_imapsto mathbf{0}$, but this is hardly useful). I am wondering whether this requirement could guarantee that the mapping keeps the linearly independence of original vectors $mathbf{x}_1, mathbf{x}_2,cdots,mathbf{x}_m$.






linear-algebra machine-learning







share|cite|improve this question









$endgroup$












  • $begingroup$
    Could you tell us more precisely what you mean by "requires that the resulting $psi(mathbf{x}_i)$'s being linearly separable"?
    $endgroup$
    – Federico
    Dec 3 '18 at 18:01










  • $begingroup$
    @Federico My bad. Suppose $mathbf{x}_i$'s belongs to two classes, then in original space they are not separable by a hyperplane. But after transformation, i.e. $psi(mathbf{x}_i)$'s, they could be separated by a hyperplane.
    $endgroup$
    – Mr.Robot
    Dec 3 '18 at 18:30






  • 1




    $begingroup$
    Well then it's even more severe that what I wrote. For example, all points from the first class could be mapped to the same vector $v$ and all points from the second class to another vector $w$. From the point of view of machine learning, this is quite a perfect situation, but for sure the images are not linearly independent.
    $endgroup$
    – Federico
    Dec 3 '18 at 18:34






  • 1




    $begingroup$
    It seems to me that the linear independence of the images is not really relevant for the purpose of ML, but I might be wrong.
    $endgroup$
    – Federico
    Dec 3 '18 at 18:35










  • $begingroup$
    @Federico Thank you. I think your example is sufficient for my purpose. Since polynomial and RBF mapping are often used, I have always been struggling with these more complicated mappings but ignore the most straightforward case.
    $endgroup$
    – Mr.Robot
    Dec 3 '18 at 19:09
















0












$begingroup$


Problem



If feature vectors $mathbf{x}_1, mathbf{x}_2,cdots,mathbf{x}_m$ are linearly independent, argue whether or not their embedding $psi(mathbf{x}_1), psi(mathbf{x}_2),cdots,psi(mathbf{x}_m)$ are linearly independent.



Some Thoughts



The complication here is that the choice of $psi(cdot)$ is not random and often requires that the resulting $psi(mathbf{x}_i)$'s being linearly separable (or we could always do $psi:mathbf{x}_imapsto mathbf{0}$, but this is hardly useful). I am wondering whether this requirement could guarantee that the mapping keeps the linearly independence of original vectors $mathbf{x}_1, mathbf{x}_2,cdots,mathbf{x}_m$.






linear-algebra machine-learning







share|cite|improve this question









$endgroup$












  • $begingroup$
    Could you tell us more precisely what you mean by "requires that the resulting $psi(mathbf{x}_i)$'s being linearly separable"?
    $endgroup$
    – Federico
    Dec 3 '18 at 18:01










  • $begingroup$
    @Federico My bad. Suppose $mathbf{x}_i$'s belongs to two classes, then in original space they are not separable by a hyperplane. But after transformation, i.e. $psi(mathbf{x}_i)$'s, they could be separated by a hyperplane.
    $endgroup$
    – Mr.Robot
    Dec 3 '18 at 18:30






  • 1




    $begingroup$
    Well then it's even more severe that what I wrote. For example, all points from the first class could be mapped to the same vector $v$ and all points from the second class to another vector $w$. From the point of view of machine learning, this is quite a perfect situation, but for sure the images are not linearly independent.
    $endgroup$
    – Federico
    Dec 3 '18 at 18:34






  • 1




    $begingroup$
    It seems to me that the linear independence of the images is not really relevant for the purpose of ML, but I might be wrong.
    $endgroup$
    – Federico
    Dec 3 '18 at 18:35










  • $begingroup$
    @Federico Thank you. I think your example is sufficient for my purpose. Since polynomial and RBF mapping are often used, I have always been struggling with these more complicated mappings but ignore the most straightforward case.
    $endgroup$
    – Mr.Robot
    Dec 3 '18 at 19:09














0












0








0





$begingroup$


Problem



If feature vectors $mathbf{x}_1, mathbf{x}_2,cdots,mathbf{x}_m$ are linearly independent, argue whether or not their embedding $psi(mathbf{x}_1), psi(mathbf{x}_2),cdots,psi(mathbf{x}_m)$ are linearly independent.



Some Thoughts



The complication here is that the choice of $psi(cdot)$ is not random and often requires that the resulting $psi(mathbf{x}_i)$'s being linearly separable (or we could always do $psi:mathbf{x}_imapsto mathbf{0}$, but this is hardly useful). I am wondering whether this requirement could guarantee that the mapping keeps the linearly independence of original vectors $mathbf{x}_1, mathbf{x}_2,cdots,mathbf{x}_m$.






linear-algebra machine-learning







share|cite|improve this question









$endgroup$




Problem



If feature vectors $mathbf{x}_1, mathbf{x}_2,cdots,mathbf{x}_m$ are linearly independent, argue whether or not their embedding $psi(mathbf{x}_1), psi(mathbf{x}_2),cdots,psi(mathbf{x}_m)$ are linearly independent.



Some Thoughts



The complication here is that the choice of $psi(cdot)$ is not random and often requires that the resulting $psi(mathbf{x}_i)$'s being linearly separable (or we could always do $psi:mathbf{x}_imapsto mathbf{0}$, but this is hardly useful). I am wondering whether this requirement could guarantee that the mapping keeps the linearly independence of original vectors $mathbf{x}_1, mathbf{x}_2,cdots,mathbf{x}_m$.






linear-algebra machine-learning




linear-algebra machine-learning






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 3 '18 at 17:52









Mr.RobotMr.Robot

35019




35019












  • $begingroup$
    Could you tell us more precisely what you mean by "requires that the resulting $psi(mathbf{x}_i)$'s being linearly separable"?
    $endgroup$
    – Federico
    Dec 3 '18 at 18:01










  • $begingroup$
    @Federico My bad. Suppose $mathbf{x}_i$'s belongs to two classes, then in original space they are not separable by a hyperplane. But after transformation, i.e. $psi(mathbf{x}_i)$'s, they could be separated by a hyperplane.
    $endgroup$
    – Mr.Robot
    Dec 3 '18 at 18:30






  • 1




    $begingroup$
    Well then it's even more severe that what I wrote. For example, all points from the first class could be mapped to the same vector $v$ and all points from the second class to another vector $w$. From the point of view of machine learning, this is quite a perfect situation, but for sure the images are not linearly independent.
    $endgroup$
    – Federico
    Dec 3 '18 at 18:34






  • 1




    $begingroup$
    It seems to me that the linear independence of the images is not really relevant for the purpose of ML, but I might be wrong.
    $endgroup$
    – Federico
    Dec 3 '18 at 18:35










  • $begingroup$
    @Federico Thank you. I think your example is sufficient for my purpose. Since polynomial and RBF mapping are often used, I have always been struggling with these more complicated mappings but ignore the most straightforward case.
    $endgroup$
    – Mr.Robot
    Dec 3 '18 at 19:09


















  • $begingroup$
    Could you tell us more precisely what you mean by "requires that the resulting $psi(mathbf{x}_i)$'s being linearly separable"?
    $endgroup$
    – Federico
    Dec 3 '18 at 18:01










  • $begingroup$
    @Federico My bad. Suppose $mathbf{x}_i$'s belongs to two classes, then in original space they are not separable by a hyperplane. But after transformation, i.e. $psi(mathbf{x}_i)$'s, they could be separated by a hyperplane.
    $endgroup$
    – Mr.Robot
    Dec 3 '18 at 18:30






  • 1




    $begingroup$
    Well then it's even more severe that what I wrote. For example, all points from the first class could be mapped to the same vector $v$ and all points from the second class to another vector $w$. From the point of view of machine learning, this is quite a perfect situation, but for sure the images are not linearly independent.
    $endgroup$
    – Federico
    Dec 3 '18 at 18:34






  • 1




    $begingroup$
    It seems to me that the linear independence of the images is not really relevant for the purpose of ML, but I might be wrong.
    $endgroup$
    – Federico
    Dec 3 '18 at 18:35










  • $begingroup$
    @Federico Thank you. I think your example is sufficient for my purpose. Since polynomial and RBF mapping are often used, I have always been struggling with these more complicated mappings but ignore the most straightforward case.
    $endgroup$
    – Mr.Robot
    Dec 3 '18 at 19:09
















$begingroup$
Could you tell us more precisely what you mean by "requires that the resulting $psi(mathbf{x}_i)$'s being linearly separable"?
$endgroup$
– Federico
Dec 3 '18 at 18:01




$begingroup$
Could you tell us more precisely what you mean by "requires that the resulting $psi(mathbf{x}_i)$'s being linearly separable"?
$endgroup$
– Federico
Dec 3 '18 at 18:01












$begingroup$
@Federico My bad. Suppose $mathbf{x}_i$'s belongs to two classes, then in original space they are not separable by a hyperplane. But after transformation, i.e. $psi(mathbf{x}_i)$'s, they could be separated by a hyperplane.
$endgroup$
– Mr.Robot
Dec 3 '18 at 18:30




$begingroup$
@Federico My bad. Suppose $mathbf{x}_i$'s belongs to two classes, then in original space they are not separable by a hyperplane. But after transformation, i.e. $psi(mathbf{x}_i)$'s, they could be separated by a hyperplane.
$endgroup$
– Mr.Robot
Dec 3 '18 at 18:30




1




1




$begingroup$
Well then it's even more severe that what I wrote. For example, all points from the first class could be mapped to the same vector $v$ and all points from the second class to another vector $w$. From the point of view of machine learning, this is quite a perfect situation, but for sure the images are not linearly independent.
$endgroup$
– Federico
Dec 3 '18 at 18:34




$begingroup$
Well then it's even more severe that what I wrote. For example, all points from the first class could be mapped to the same vector $v$ and all points from the second class to another vector $w$. From the point of view of machine learning, this is quite a perfect situation, but for sure the images are not linearly independent.
$endgroup$
– Federico
Dec 3 '18 at 18:34




1




1




$begingroup$
It seems to me that the linear independence of the images is not really relevant for the purpose of ML, but I might be wrong.
$endgroup$
– Federico
Dec 3 '18 at 18:35




$begingroup$
It seems to me that the linear independence of the images is not really relevant for the purpose of ML, but I might be wrong.
$endgroup$
– Federico
Dec 3 '18 at 18:35












$begingroup$
@Federico Thank you. I think your example is sufficient for my purpose. Since polynomial and RBF mapping are often used, I have always been struggling with these more complicated mappings but ignore the most straightforward case.
$endgroup$
– Mr.Robot
Dec 3 '18 at 19:09




$begingroup$
@Federico Thank you. I think your example is sufficient for my purpose. Since polynomial and RBF mapping are often used, I have always been struggling with these more complicated mappings but ignore the most straightforward case.
$endgroup$
– Mr.Robot
Dec 3 '18 at 19:09










1 Answer
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$begingroup$

It depends on what you mean by "requires that the resulting $psi(mathbf{x}_i)$'s being linearly separable". If you mean that any of the points $psi(mathbf{x}_i)$ can be linearly separated from the others, then the answer is false.



For instance, the points $(0,0)$, $(1,0)$ and $(0,1)$ are linearly separable in this sense, but are not linearly independent.






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    $begingroup$

    It depends on what you mean by "requires that the resulting $psi(mathbf{x}_i)$'s being linearly separable". If you mean that any of the points $psi(mathbf{x}_i)$ can be linearly separated from the others, then the answer is false.



    For instance, the points $(0,0)$, $(1,0)$ and $(0,1)$ are linearly separable in this sense, but are not linearly independent.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      It depends on what you mean by "requires that the resulting $psi(mathbf{x}_i)$'s being linearly separable". If you mean that any of the points $psi(mathbf{x}_i)$ can be linearly separated from the others, then the answer is false.



      For instance, the points $(0,0)$, $(1,0)$ and $(0,1)$ are linearly separable in this sense, but are not linearly independent.






      share|cite|improve this answer









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        0





        $begingroup$

        It depends on what you mean by "requires that the resulting $psi(mathbf{x}_i)$'s being linearly separable". If you mean that any of the points $psi(mathbf{x}_i)$ can be linearly separated from the others, then the answer is false.



        For instance, the points $(0,0)$, $(1,0)$ and $(0,1)$ are linearly separable in this sense, but are not linearly independent.






        share|cite|improve this answer









        $endgroup$



        It depends on what you mean by "requires that the resulting $psi(mathbf{x}_i)$'s being linearly separable". If you mean that any of the points $psi(mathbf{x}_i)$ can be linearly separated from the others, then the answer is false.



        For instance, the points $(0,0)$, $(1,0)$ and $(0,1)$ are linearly separable in this sense, but are not linearly independent.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 '18 at 18:00









        FedericoFederico

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