I feel like B should be positive definite to make K positive definite, since K being positive definite means we can have the critical point be a minimum. So, if we have a positive definite matrix in R^3, then in R^4, it would also be positive definite?
jacheng
I tried plugging in the matrix from K into x^TBx and got x^2(a^2+ab+ac) + y^2(ab+b^2+bc) + z^2(ac + bc + c^2). So all the ^2 terms are positive, but couldn't it still end up being negative if for example a^2 + ab + ac < 0?
Heisenberg
@jacheng cuz it's square <x,p> which can be written as (ax+by+cz)^2
aylu
Consider arbitrary uTKu. K can be expressed vvT (or sum thereof) so we can write it as uTvvTu or (uTv)(uTv)T which is just the dot product of a vector with itself so it has to be positive unless u and v are perpendicular in which case it is 0.
Azure
So it's actually not positive-definite, but rather positive-semidefinite.
0x484884
Maybe this is a slightly circular argument, but distance should never be negative so if this correctly computes the distance then it has to be positive semidefinite. I also think that it would be positive semidefinite if there's a points where all the planes intersect, otherwise it would be positive definite.
I feel like B should be positive definite to make K positive definite, since K being positive definite means we can have the critical point be a minimum. So, if we have a positive definite matrix in R^3, then in R^4, it would also be positive definite?
I tried plugging in the matrix from K into x^TBx and got x^2(a^2+ab+ac) + y^2(ab+b^2+bc) + z^2(ac + bc + c^2). So all the ^2 terms are positive, but couldn't it still end up being negative if for example a^2 + ab + ac < 0?
@jacheng cuz it's square <x,p> which can be written as (ax+by+cz)^2
Consider arbitrary uTKu. K can be expressed vvT (or sum thereof) so we can write it as uTvvTu or (uTv)(uTv)T which is just the dot product of a vector with itself so it has to be positive unless u and v are perpendicular in which case it is 0.
So it's actually not positive-definite, but rather positive-semidefinite.
Maybe this is a slightly circular argument, but distance should never be negative so if this correctly computes the distance then it has to be positive semidefinite. I also think that it would be positive semidefinite if there's a points where all the planes intersect, otherwise it would be positive definite.