If anyone could clear why this is true, I will appreciate it a lot. Why is (u, (v,w1)w1)=($\bar(v,w1)$(u,w1)) and not (v,w1)(u,w1) ((the first one has over-bar above it, not sure it got it right with math jax))
I didn't read/see any rule stating it is so, and they are just using it.. I can't figure out why it is true.
( , ) is not necesserily the standart multipication
Precisely, I understand what the manipulation needs to be - I also saw that's the one they used in the solution. But I can't figure out wht it is true.
Well, unless they gave you some random form with a rule or something and said "Show this is Hermitian," then I'd assume they just told you it's Hermitian.
In which case, a Hermitian form is linear in the second variable, but Hermitian symmetric in the first variable.
Given inner prduct space V (with some unknown inner multipication ( , ) ), let $w_1$, $w_2$ be two vectors in V so that ||$w_1$||=||$w_2$||=1, (w1,w2)=0. Let T:V-->V Be a linear transformation, so that: Tv=v-2(v,$w_1$)$w_1$-2(v,$w_2$)$w_2$
That bar. It's the bar that means, say, Z=a+i-->$\overline{z}$=a-i
I do not know whether it is complex or real, it's just a given field. So I have to assume the 'harsher' rules, so I go by the complex rules for it, for if F=R then (x,y)=(y,x)
but when F is just a given field, then for V (x,y)=$\overline{y,x}$
@robjohn back. While walking I realized I can come up with a HUGE improvement. See the last double integral in the second part before being split into $I$ and $J$. By changing the integration order there, things turn into a piece of case.
@nullgeppetto I think the best way would a change of coordinates so that your distribution becomes the rotationally invariant standard normal distrubution. Then rotate so that your hyperplane is $\{x_n = k\}$. That makes the first $n-1$ integrals trivial and leaves you with $$\int_{-\infty}^k (k-x) e^{-x^2/2}\,dx + \int_k^\infty (x-k)e^{-x^2/2}\,dx.$$
@DanielFischer, then it's the integral of the gaussian pdf by $\mathbf{w}\cdot\mathbf{x}+b$, just like one of my older questions in which you gave the correct answer? :)
Similar, @nullgeppetto. But wait, it's a little more complicated than what I wrote above, because the first change of coordinates doesn't leave distances invariant.
@nullgeppetto First part, change of coordinates so that the axes are eigenvectors of $\Sigma$ is no problem. The somewhat hairy part is the scaling of the axes. Being not so good in geometry, I don't know off-hand how that translates to the distance thing.
@DanielFischer, ok, so if we find out what's happening with the scaling of the axis, then we can compute the mean distance, right? I mean, it's a matter of a scalar factor after all?
@robjohn I also sent an e-mail to a professor from Harvard to ask from an alternative way of getting the first equality for that I needed to work a lot. He didn't answer me back yet.
@nullgeppetto Ah, got it. the distance is $$\frac{\lvert \mathbf{w}\cdot\mathbf{X} + b\rvert}{\lVert \mathbf{w}\rVert}.$$ We can pull out the denominator, so we have an integral $$\int \lvert \mathbf{w}\cdot\mathbf{X}+b\rvert\, f(\mathbf{X})\,d\mathbf{X},$$ and when we transform $\mathbf{X}$, writing $\mathbf{X} = A\mathbf{Y}$, we get $\mathbf{w}\cdot\mathbf{X} = (A^T\mathbf{w})\cdot\mathbf{Y}$.
Hi everybody can someone help me to solve my problem here http://math.stackexchange.com/questions/682765/the-space-spanned-by-the-rows-of-the-kernel-matrix
@DanielFischer, it seems nice, but I need sometime to understand it. I start now! Anyway, thank you very much Daniel! I'll let you know if it works (I am sure it does for sure)!
@DanielFischer, I think it's the same with previous question (except for the denominator), yes? But the limites of the integration are infinite? Am I right?
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