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Kernel function with a feature space equipped with an inner product that is not the dot product


Inner product space computationThe complex inner product spacedot product vs inner product?Implicit feature space of Power KernelWhy is $mathbbR^3$ not the Reproducing Kernel Hilbert Space defined by kernel $k(x,y)=(x_1y_1+x_2y_2)^2$Hilbert space with an additional semi-inner product structureComplexification of real inner product spaces and how the inner product extends to a complex spaceLinear regression with feature representation confusion - is design matrix column space the feature space?Semi-inner product structure in complex Hilbert spacesSimple (?) Quesiton on Inner Product in Reproducing Kernel Hilbert Space













0












$begingroup$


Premise:




A function $K: mathbb R^d times mathbb R^d to mathbb R$ is called
a kernel function on $mathbbR^d$ if there exists a Hilbert space
$mathcalH$ and a map $phi: mathbb R^d to mathcalH$ such that
for any $mathbf x, mathbf yin mathbbR^d$: beginequation
labeleq:kerdef K(mathbf x,mathbf y) = langle phi(mathbf x),
phi(mathbf y) rangle_mathcalH, endequation
where $
langlecdot,cdotrangle$
is an inner product.




I have recently noticed that under this definition $ langlecdot,cdotrangle$ is not necessary the standard inner product (dot product).



I have thought of the following example.
Consider the degree-two polynomial kernel on $mathbb R^2$:
beginequation
K(mathbf x,mathbf y) = (mathbf xcdot mathbf y)^2.
endequation

Then, the following is a valid feature map for this kernel:
beginequation
phi(mathbf x) = (2mathbf x_1 mathbf x_1,~ 2mathbf x_1 mathbf x_2,~ 2mathbf x_2 mathbf x_1,~ 2mathbf x_2 mathbf x_2 ),
endequation

considering the feature space $mathcal H=mathbb R^4$, with the custom inner product:
beginequation
langlemathbf x, mathbf y rangle = fracmathbf xcdot mathbf y4.
endequation



Proof:
beginalign
langle phi(mathbf x), phi(mathbf y) rangle_mathcalH &= (2mathbf x_1 mathbf x_1,~ 2mathbf x_1 mathbf x_2,~ 2mathbf x_2 mathbf x_1,~ 2mathbf x_2 mathbf x_2 ) cdot (2mathbf y_1 mathbf y_1,~ 2mathbf y_1 mathbf y_2,~ 2mathbf y_2 mathbf y_1,~ 2mathbf y_2 mathbf y_2 )/4\
&= (mathbf x cdot mathbf y)^2 \&= K(mathbf x, mathbf y)
endalign



Question:



Can you provide a less trivial example of kernel function, feature map and feature space, with an inner product that is not the dot product?










share|cite|improve this question









$endgroup$
















    0












    $begingroup$


    Premise:




    A function $K: mathbb R^d times mathbb R^d to mathbb R$ is called
    a kernel function on $mathbbR^d$ if there exists a Hilbert space
    $mathcalH$ and a map $phi: mathbb R^d to mathcalH$ such that
    for any $mathbf x, mathbf yin mathbbR^d$: beginequation
    labeleq:kerdef K(mathbf x,mathbf y) = langle phi(mathbf x),
    phi(mathbf y) rangle_mathcalH, endequation
    where $
    langlecdot,cdotrangle$
    is an inner product.




    I have recently noticed that under this definition $ langlecdot,cdotrangle$ is not necessary the standard inner product (dot product).



    I have thought of the following example.
    Consider the degree-two polynomial kernel on $mathbb R^2$:
    beginequation
    K(mathbf x,mathbf y) = (mathbf xcdot mathbf y)^2.
    endequation

    Then, the following is a valid feature map for this kernel:
    beginequation
    phi(mathbf x) = (2mathbf x_1 mathbf x_1,~ 2mathbf x_1 mathbf x_2,~ 2mathbf x_2 mathbf x_1,~ 2mathbf x_2 mathbf x_2 ),
    endequation

    considering the feature space $mathcal H=mathbb R^4$, with the custom inner product:
    beginequation
    langlemathbf x, mathbf y rangle = fracmathbf xcdot mathbf y4.
    endequation



    Proof:
    beginalign
    langle phi(mathbf x), phi(mathbf y) rangle_mathcalH &= (2mathbf x_1 mathbf x_1,~ 2mathbf x_1 mathbf x_2,~ 2mathbf x_2 mathbf x_1,~ 2mathbf x_2 mathbf x_2 ) cdot (2mathbf y_1 mathbf y_1,~ 2mathbf y_1 mathbf y_2,~ 2mathbf y_2 mathbf y_1,~ 2mathbf y_2 mathbf y_2 )/4\
    &= (mathbf x cdot mathbf y)^2 \&= K(mathbf x, mathbf y)
    endalign



    Question:



    Can you provide a less trivial example of kernel function, feature map and feature space, with an inner product that is not the dot product?










    share|cite|improve this question









    $endgroup$














      0












      0








      0





      $begingroup$


      Premise:




      A function $K: mathbb R^d times mathbb R^d to mathbb R$ is called
      a kernel function on $mathbbR^d$ if there exists a Hilbert space
      $mathcalH$ and a map $phi: mathbb R^d to mathcalH$ such that
      for any $mathbf x, mathbf yin mathbbR^d$: beginequation
      labeleq:kerdef K(mathbf x,mathbf y) = langle phi(mathbf x),
      phi(mathbf y) rangle_mathcalH, endequation
      where $
      langlecdot,cdotrangle$
      is an inner product.




      I have recently noticed that under this definition $ langlecdot,cdotrangle$ is not necessary the standard inner product (dot product).



      I have thought of the following example.
      Consider the degree-two polynomial kernel on $mathbb R^2$:
      beginequation
      K(mathbf x,mathbf y) = (mathbf xcdot mathbf y)^2.
      endequation

      Then, the following is a valid feature map for this kernel:
      beginequation
      phi(mathbf x) = (2mathbf x_1 mathbf x_1,~ 2mathbf x_1 mathbf x_2,~ 2mathbf x_2 mathbf x_1,~ 2mathbf x_2 mathbf x_2 ),
      endequation

      considering the feature space $mathcal H=mathbb R^4$, with the custom inner product:
      beginequation
      langlemathbf x, mathbf y rangle = fracmathbf xcdot mathbf y4.
      endequation



      Proof:
      beginalign
      langle phi(mathbf x), phi(mathbf y) rangle_mathcalH &= (2mathbf x_1 mathbf x_1,~ 2mathbf x_1 mathbf x_2,~ 2mathbf x_2 mathbf x_1,~ 2mathbf x_2 mathbf x_2 ) cdot (2mathbf y_1 mathbf y_1,~ 2mathbf y_1 mathbf y_2,~ 2mathbf y_2 mathbf y_1,~ 2mathbf y_2 mathbf y_2 )/4\
      &= (mathbf x cdot mathbf y)^2 \&= K(mathbf x, mathbf y)
      endalign



      Question:



      Can you provide a less trivial example of kernel function, feature map and feature space, with an inner product that is not the dot product?










      share|cite|improve this question









      $endgroup$




      Premise:




      A function $K: mathbb R^d times mathbb R^d to mathbb R$ is called
      a kernel function on $mathbbR^d$ if there exists a Hilbert space
      $mathcalH$ and a map $phi: mathbb R^d to mathcalH$ such that
      for any $mathbf x, mathbf yin mathbbR^d$: beginequation
      labeleq:kerdef K(mathbf x,mathbf y) = langle phi(mathbf x),
      phi(mathbf y) rangle_mathcalH, endequation
      where $
      langlecdot,cdotrangle$
      is an inner product.




      I have recently noticed that under this definition $ langlecdot,cdotrangle$ is not necessary the standard inner product (dot product).



      I have thought of the following example.
      Consider the degree-two polynomial kernel on $mathbb R^2$:
      beginequation
      K(mathbf x,mathbf y) = (mathbf xcdot mathbf y)^2.
      endequation

      Then, the following is a valid feature map for this kernel:
      beginequation
      phi(mathbf x) = (2mathbf x_1 mathbf x_1,~ 2mathbf x_1 mathbf x_2,~ 2mathbf x_2 mathbf x_1,~ 2mathbf x_2 mathbf x_2 ),
      endequation

      considering the feature space $mathcal H=mathbb R^4$, with the custom inner product:
      beginequation
      langlemathbf x, mathbf y rangle = fracmathbf xcdot mathbf y4.
      endequation



      Proof:
      beginalign
      langle phi(mathbf x), phi(mathbf y) rangle_mathcalH &= (2mathbf x_1 mathbf x_1,~ 2mathbf x_1 mathbf x_2,~ 2mathbf x_2 mathbf x_1,~ 2mathbf x_2 mathbf x_2 ) cdot (2mathbf y_1 mathbf y_1,~ 2mathbf y_1 mathbf y_2,~ 2mathbf y_2 mathbf y_1,~ 2mathbf y_2 mathbf y_2 )/4\
      &= (mathbf x cdot mathbf y)^2 \&= K(mathbf x, mathbf y)
      endalign



      Question:



      Can you provide a less trivial example of kernel function, feature map and feature space, with an inner product that is not the dot product?







      hilbert-spaces inner-product-space machine-learning reproducing-kernel-hilbert-spaces






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Mar 22 at 9:06









      Daniel LópezDaniel López

      1185




      1185




















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