Proving that a function f is a kernel.
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The theory states that $$f(x,y)\ text{with} x text{and} y in R^n$$ in order to be a valid kernel, beside being symmetric, has to be an inner product in a suitable space.
Is this latter requirement equivalent to claim that the following Matrix:
$$ begin{bmatrix}
f(x,x) & f(x,y) \
f(y,x) & f(y,y)
end{bmatrix} $$
is semidefinite positive for any pair x and y?
If not, why?
inner-product-space machine-learning
asked Jan 6 at 17:14
Tommaso BendinelliTommaso Bendinelli
14110
$endgroup$
|
show 3 more comments
$begingroup$
The theory states that $$f(x,y)\ text{with} x text{and} y in R^n$$ in order to be a valid kernel, beside being symmetric, has to be an inner product in a suitable space.
Is this latter requirement equivalent to claim that the following Matrix:
$$ begin{bmatrix}
f(x,x) & f(x,y) \
f(y,x) & f(y,y)
end{bmatrix} $$
is semidefinite positive for any pair x and y?
If not, why?
inner-product-space machine-learning
asked Jan 6 at 17:14
Tommaso BendinelliTommaso Bendinelli
14110
$endgroup$
$begingroup$
I think you need to check this for an arbitrary number of datapoints. You are checking for two datapoints. I'm not sure if this is enough.
$endgroup$
– Kenny Wong
Jan 6 at 21:08
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Some examples here: stats.stackexchange.com/questions/35634/…
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– Kenny Wong
Jan 6 at 21:14
$begingroup$
But why do when need to check for an arbitrary number of points?
$endgroup$
– Tommaso Bendinelli
Jan 6 at 22:16
$begingroup$
Because that's Mercer's theorem tells us to do!
$endgroup$
– Kenny Wong
Jan 6 at 22:16
$begingroup$
ahh still not clear
$endgroup$
– Tommaso Bendinelli
Jan 6 at 22:45
|
show 3 more comments
$begingroup$
The theory states that $$f(x,y)\ text{with} x text{and} y in R^n$$ in order to be a valid kernel, beside being symmetric, has to be an inner product in a suitable space.
Is this latter requirement equivalent to claim that the following Matrix:
$$ begin{bmatrix}
f(x,x) & f(x,y) \
f(y,x) & f(y,y)
end{bmatrix} $$
is semidefinite positive for any pair x and y?
If not, why?
inner-product-space machine-learning
asked Jan 6 at 17:14
Tommaso BendinelliTommaso Bendinelli
14110
$endgroup$
The theory states that $$f(x,y)\ text{with} x text{and} y in R^n$$ in order to be a valid kernel, beside being symmetric, has to be an inner product in a suitable space.
Is this latter requirement equivalent to claim that the following Matrix:
$$ begin{bmatrix}
f(x,x) & f(x,y) \
f(y,x) & f(y,y)
end{bmatrix} $$
is semidefinite positive for any pair x and y?
If not, why?
inner-product-space machine-learning
inner-product-space machine-learning
asked Jan 6 at 17:14
Tommaso BendinelliTommaso Bendinelli
14110
asked Jan 6 at 17:14
Tommaso BendinelliTommaso Bendinelli
14110
edited Jan 6 at 17:55
Tommaso Bendinelli
asked Jan 6 at 17:14
Tommaso BendinelliTommaso Bendinelli
14110
asked Jan 6 at 17:14
Tommaso BendinelliTommaso Bendinelli
14110
asked Jan 6 at 17:14
Tommaso BendinelliTommaso Bendinelli
14110
14110
$begingroup$
I think you need to check this for an arbitrary number of datapoints. You are checking for two datapoints. I'm not sure if this is enough.
$endgroup$
– Kenny Wong
Jan 6 at 21:08
$begingroup$
Some examples here: stats.stackexchange.com/questions/35634/…
$endgroup$
– Kenny Wong
Jan 6 at 21:14
$begingroup$
But why do when need to check for an arbitrary number of points?
$endgroup$
– Tommaso Bendinelli
Jan 6 at 22:16
$begingroup$
Because that's Mercer's theorem tells us to do!
$endgroup$
– Kenny Wong
Jan 6 at 22:16
$begingroup$
ahh still not clear
$endgroup$
– Tommaso Bendinelli
Jan 6 at 22:45
|
show 3 more comments
$begingroup$
I think you need to check this for an arbitrary number of datapoints. You are checking for two datapoints. I'm not sure if this is enough.
$endgroup$
– Kenny Wong
Jan 6 at 21:08
$begingroup$
Some examples here: stats.stackexchange.com/questions/35634/…
$endgroup$
– Kenny Wong
Jan 6 at 21:14
$begingroup$
But why do when need to check for an arbitrary number of points?
$endgroup$
– Tommaso Bendinelli
Jan 6 at 22:16
$begingroup$
Because that's Mercer's theorem tells us to do!
$endgroup$
– Kenny Wong
Jan 6 at 22:16
$begingroup$
ahh still not clear
$endgroup$
– Tommaso Bendinelli
Jan 6 at 22:45
$begingroup$
I think you need to check this for an arbitrary number of datapoints. You are checking for two datapoints. I'm not sure if this is enough.
$endgroup$
– Kenny Wong
Jan 6 at 21:08
$begingroup$
I think you need to check this for an arbitrary number of datapoints. You are checking for two datapoints. I'm not sure if this is enough.
$endgroup$
– Kenny Wong
Jan 6 at 21:08
$begingroup$
Some examples here: stats.stackexchange.com/questions/35634/…
$endgroup$
– Kenny Wong
Jan 6 at 21:14
$begingroup$
Some examples here: stats.stackexchange.com/questions/35634/…
$endgroup$
– Kenny Wong
Jan 6 at 21:14
$begingroup$
But why do when need to check for an arbitrary number of points?
$endgroup$
– Tommaso Bendinelli
Jan 6 at 22:16
$begingroup$
But why do when need to check for an arbitrary number of points?
$endgroup$
– Tommaso Bendinelli
Jan 6 at 22:16
$begingroup$
Because that's Mercer's theorem tells us to do!
$endgroup$
– Kenny Wong
Jan 6 at 22:16
$begingroup$
Because that's Mercer's theorem tells us to do!
$endgroup$
– Kenny Wong
Jan 6 at 22:16
$begingroup$
ahh still not clear
$endgroup$
– Tommaso Bendinelli
Jan 6 at 22:45
$begingroup$
ahh still not clear
$endgroup$
– Tommaso Bendinelli
Jan 6 at 22:45
|
show 3 more comments
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$begingroup$
I think you need to check this for an arbitrary number of datapoints. You are checking for two datapoints. I'm not sure if this is enough.
$endgroup$
– Kenny Wong
Jan 6 at 21:08
$begingroup$
Some examples here: stats.stackexchange.com/questions/35634/…
$endgroup$
– Kenny Wong
Jan 6 at 21:14
$begingroup$
But why do when need to check for an arbitrary number of points?
$endgroup$
– Tommaso Bendinelli
Jan 6 at 22:16
$begingroup$
Because that's Mercer's theorem tells us to do!
$endgroup$
– Kenny Wong
Jan 6 at 22:16
$begingroup$
ahh still not clear
$endgroup$
– Tommaso Bendinelli
Jan 6 at 22:45