I don't see/understand the "fact" that :
"We then derived an alternative
formulation using duality. We saw that in this alternative
formulation the data only enters in the form of a \kernel"
K = XX^T"
What do we mean by the data enters only through this form in the dual formulation
Hi,
This means that to solve the dual problem we only need to know the kernel \( K = X X^T \). This might seem trivial and useless but it is very relevant when we want to use a feature map \( \Phi \) to map our data to a more complicated space. In this case to solve the dual we just need to know the associated kernel (which is now \( K = \Phi(X) \Phi(X)^T \) ). It turns out that such a matrix can be efficiently computed without at any time calculating \( \Phi(X) \) (which could be huge or even infinite. For more details please refer to "The Kernel Trick" section in lecture 7b.
Best,
Scott
Kernel
Hey,
I don't see/understand the "fact" that :
"We then derived an alternative
formulation using duality. We saw that in this alternative
formulation the data only enters in the form of a \kernel"
K = XX^T"
What do we mean by the data enters only through this form in the dual formulation
Thanks for your reply
Hi,
This means that to solve the dual problem we only need to know the kernel \( K = X X^T \). This might seem trivial and useless but it is very relevant when we want to use a feature map \( \Phi \) to map our data to a more complicated space. In this case to solve the dual we just need to know the associated kernel (which is now \( K = \Phi(X) \Phi(X)^T \) ). It turns out that such a matrix can be efficiently computed without at any time calculating \( \Phi(X) \) (which could be huge or even infinite. For more details please refer to "The Kernel Trick" section in lecture 7b.
Best,
Scott
Looking into it now thanks for the quick reply !
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