First of all, thanks for sending us our scanned and graded sheet.
There is a question for which I disagree with the solution though. This is the question regarding kernels, where we have to tell which kernel is not valid.
I believe the \(K(x, x') = xA^Tx'\) kernel is not valid when \(A\) is PSD. We need A to be symmetric as well. In case we were using complex numbers, I would agree, but it is said that the entries are real valued, hence there are PSD matrix that are not symmetric. For instance, taking A := [[1, 0], [1, 1]] (sorry for the python notation), we can see that A is PSD as it has eigenvalue 1 only, but it does not yield a valid kernel, since the result is not symmetric in x, x'.
I personally agree with you and that's also why I did not check the "all kernels are valid" answer. Also another reason why A must be symmetric. The natural high dimensional mapping for $$K$$ would be
yes we're investigating if we might allow both answers as correct (both if you assumed A symmetric or not).
PSD is often assumed to include symmetric, but you're right that we should have stated this explicitly in the question. we'll keep you guys updated
Both of your remarks are indeed correct. The source of the confusion comes from the definition of a positive semi-definite matrix. The most common definition is: a psd matrix is a symmetric matrix with all positive eigenvalues (for example: https://en.wikipedia.org/wiki/Definite_symmetric_matrix or https://www.math.brown.edu/streil/papers/LADW/LADW_2017-09-04.pdf). Hence the notion of symmetry is very often included in the definition. However there does exist some literature (though not seen in class) in which the definition of positive definiteness is extended to non-symmetric and non-hermitian matrices.
Nonetheless to avoid such source of confusion, which was obviously not intended, the symmetry precision should / could have been added. It will be up to Martin and Nicolas to decide whether to accept both answers as correct or not.
Best.
Misunderstanding wrt MCQ Kernel question
Hello,
First of all, thanks for sending us our scanned and graded sheet.
There is a question for which I disagree with the solution though. This is the question regarding kernels, where we have to tell which kernel is not valid.
I believe the \(K(x, x') = xA^Tx'\) kernel is not valid when \(A\) is PSD. We need A to be symmetric as well. In case we were using complex numbers, I would agree, but it is said that the entries are real valued, hence there are PSD matrix that are not symmetric. For instance, taking A := [[1, 0], [1, 1]] (sorry for the python notation), we can see that A is PSD as it has eigenvalue 1 only, but it does not yield a valid kernel, since the result is not symmetric in x, x'.
Do you see my point ?
4
I personally agree with you and that's also why I did not check the "all kernels are valid" answer. Also another reason why A must be symmetric. The natural high dimensional mapping for $$K$$ would be
$$K(x,x') = x(PDP^{T})^Tx'=xP^{-1}^TD^{1/2}D^{1/2}P^Tx' = (D^{1/2}P^{-1}x^T)^T(D^{1/2}P^{T}x^T)$$
Thus
$$\phi(x) = (D^{1/2}P^{T}x^T)$$
However we can not have an orthogonal decomposition with \(P^{-1} = P^T\) of A unless A is symmetric.
yes we're investigating if we might allow both answers as correct (both if you assumed A symmetric or not).
PSD is often assumed to include symmetric, but you're right that we should have stated this explicitly in the question. we'll keep you guys updated
5
Thank you, that is nice of you to consider that option
Both of your remarks are indeed correct. The source of the confusion comes from the definition of a positive semi-definite matrix. The most common definition is: a psd matrix is a symmetric matrix with all positive eigenvalues (for example: https://en.wikipedia.org/wiki/Definite_symmetric_matrix or https://www.math.brown.edu/streil/papers/LADW/LADW_2017-09-04.pdf). Hence the notion of symmetry is very often included in the definition. However there does exist some literature (though not seen in class) in which the definition of positive definiteness is extended to non-symmetric and non-hermitian matrices.
Nonetheless to avoid such source of confusion, which was obviously not intended, the symmetry precision should / could have been added. It will be up to Martin and Nicolas to decide whether to accept both answers as correct or not.
Best.
1
hi all
we decided to count both possible answers as correct:
the reason A is also counted correct now is that we didn't specify if A psd is also symmetric or not (we should have).
so everyone who ticked A gets 2.5 more points. there are very minor impacts to the grades only, and only upwards.
we will not contact people individually, but remain assured that everyone who ticked A now got the additional points
4
Add comment