Let $w = \{w(i,j)\}_{1 \leq i,j \leq m}$ be an $m \times m$ symmetric matrix with non-negative real entries such that $w(i,j)=0$ if and only if $i=j$. Show that $$d(i,j)=\min\left\{\sum_{j=0}^{k-1}w(i_j,i_{j+1}) \mid k \geq 1, i_0=i, i_k=j,i_j \in \{1,2,...,m\}\right\}$$ is a metric on $\{1,...,m...
Fair enough. I can't say I understood very much of your question either. Also, I don't really understand a lot of stuff related to Mandelbrot set or fractals. But rendering/coloring them is addictive, even without understanding everything
@AlessandroCodenotti So, for example, the function f(x) = 1/2x is a non-expansion, because, given two numbers x1 and x2, then |1/2*x1 - 1/2*x2| <= |x1 - x2|. For instance, x1=6 and x2=9, then |3 - 4.5| <= |-3| <=> 1.5 <= 3
For which of the subspaces $U$ and $V$ of $\textbf{R}^4$ is the sum $U+V$ direct? $U=\{\textbf{x}\in\textbf{R}^4: x_1+x_2+x_3+x_4=0\}$ and $V=\{\textbf{x}\in\textbf{R}^4: x_1-x_2+x_3-x_4=0\}$.
Consider the following theorem: Let $G$ be a locally compact abelian group, and suppose that the Gelfand homomorphism $\Gamma : L^{1}(G) \to C_{0}(\hat{G})$ given by $f \mapsto \hat{f}$ (Fourier transform, I think) is surjective, then $G$ must be discrete.
My question is, is the converse true? If $G$ is discrete, does it follow that the Gelfand transform is surjective?
Alright, yesterday I posted that incorrect result, and my mistake was thinking that thinking about subgroups of order 2 would be the same as thinking about all subgroups. After reviewing, what I posted actually does hold when $n=2$, because $S_3\cong GL_2(\mathbb{F}_2)$.
If I wanted to motivate the $n=3$ case by thinking about permutations on subgroups, I'm kind of coming up short. I know there are seven subgroups of order 4 in this group, and there are six automorphisms on the group those subgroups are isomorphic to, so multiplication gives 42 automorphisms which leaves at least one group of order 4 fixed, but there are 168 automorphisms, so I'm missing a factor of 4, and I don't know where to find that factor.
@SubhasisBiswas Please do not ping people like this. You did it to me earlier. You can post your question, of course, but do not annoy people with pings.
@YOUSEFY Did you try clearing out the denominator? I don't know what you mean by "calculate", though.
Because n is the number of vertices in graph and W is the maximum weight of a tour in the graph (a tour is a walk from a vertex and ending the same vertex such that you visit all vertices exactly once) and the weight of each edge is either 1 or 2. Thus, W is maximum will be 2n.
There's a computation I've been using a bunch lately, and I'm trying to make sure I actually understand it
If I want to project onto the 1D subspace in the direction of unit vector $u$, that's easy: The projector is $P_u=uu^\top$
More generally, if I have a subspace $U$with orthonormal basis $u_1,u_2,\ldots,u_n$, then the projector onto $U$ is just the sum of the projectors onto the basis vectors: $P_U=P_{u_1}+P_{u_2}+\cdots +P_{u_n}$
That's familiar enough to me.
What I wasn't familiar with was how to write the projector if the basis isn't orthonormal
to be precise, I've been interested in the history of regression/correlation
and wanted to get a geometrical perspective on how Yule (one of the early workers in the subject) derived one of his results (specifically, an inequality on correlation coefficients)
The geometric story you get is simple enough, happily: $\|x-p\|^2\geq 0$
That's all he needs to derive the inequality. So that's cute.
@Semiclassic: I've commented before that a lot of the statistical stuff has very pretty linear algebraic (geometric) interpretations, but the statisticians obscure/hide those.
For example, the fact that the algebraic sum of the errors when you do least squares is $0$, is immediate from the fact that one of the columns of your matrix is all $1$'s. @Semiclassic
and another interprets the partial correlation coefficient as the angle you get when you project two vectors onto a plane. (this one shows up in Jacobson's thesis, though I don't remember if he cited Fisher)
(though one of the late papers of Pearson they cite is how I got the idea for the spherical law of cosines)
"R.A. Fisher (1 890-1962) used geometry throughout his work on distribution theory but his treatment of correlation is of most interest here. In his paper on the exact distribution of the correlation coefficient he (1915, p. 509) wrote, “The five quantities [associated with the bivariate normal] have . . . an exceedingly beautiful interpretation in generalised space.” The n observations on a pair of variables can be represented by two points P and Q in n-dimensional space and the correlation coefficient interpreted as the cosine of the angle between OP and OQ."
So I just pulled out my algebra book. The spherical law of cosines I did with a bunch of cross products and dot products (and considering the dual triangle).
"Later Fisher (1924, p. 330) observed that when there is a third vector corresponding to point R, the partial correlation of the original vectors is the “cosine of the angle between the projections of OP and OQ upon the region perpendicular to OR”. This is a geometric interpretation of residuals and of (8.2) above."
So yeah, Fisher definitely had that
@TedShifrin One of the points of that paper, though, is how long it took that geometric perspective to make headway into statistical regression
When proving if a sum of two subspaces $U,V$ of a linear space $W$ is a direct sum, is it sufficient to prove that the intersection of $U$ with $V$ is the zero set (hence $U+V$ is a direct sum?), or does one also have to prove the equality $W=U+V$? If yes to the latter, how would one go about doing that?
"The treatise on the trigonometry of correlation that Pearson (1916, p. 237) thought “greatly to be desired” never materialised. Kendall’s (1961) Course in the Geometry of n Dimensions is a partial offering but it appeared just as a new approach was taking off. This drew on the Hilbert space theory developed in the early part of the century and assembled by Stone (1932): Birkhoff & Kreysig (1984) and Dieudonne (1 98 1) discuss the development. The “projections” that invaded least squares in the 1960s (see Section 13) owed more to Stone (pp. 70-5) and his idempotent operators than to Fisher."
I am trusting this author's reading on the situation, of course
I have been trying to find an elegant proof of this mathb.in/36275 I was able to show that it is true if mathb.in/36276 is true. I can't figure out an elegant way to prove either. Does anyone have any hints?
At the least, you can choose digits in the following way. Multiply $X$ by 3 to get a number between $0$ and $3$. It'll either by closer to 0 or to 2; pick the closer one
$\{0,2\}^{\Bbb N}$ is a Cantor space, if you can prove that $f$ is continuous and injective then its image will also be a Cantor space since the domain is compact and the codomain is Hausdorff
"For each $n \in \mathbb{N}^+$, let $X_n = \{0, 2\}$. Let $X = \Pi_{n \in \mathbb{N}^+} X_n$. Define a function $f: X \to [0, 1]$ by setting $$f(x) = \sum_{n = 1}^{\infty} \frac{x_n}{3^n}.$$ Prove that $f$ is one-to-one."
@Semiclassical I think I am going to post this as a math stack exchange question. I felt like I was missing something quick and obvious which is why I posted it in the chat
@psa: Everything perpendicular to $u$ projects to $0$, so all the vectors that project to $p=cu$ will be the line perpendicular to $u$ passing through $p$.
@Semiclassical Referring to me? I can see why you'd think that, but that is a direct consequence of "subtracting" and seeing where they differ as @TedShifrin suggested..
@William: Note that I specified that the $x$ and $y$ disagreed for the first time in a certain place. I.e., $x_i = y_i$ for $1\le i\le N-1$ and $x_N\ne y_N$.
Start over: let $\mathbf{u}$ and $\mathbf{v}$ be perpendicular unit vectors [with u = (1,0) and v = (0,1)], and $\mathbf{r}$ be the vector of interest. Then if the scalar projection of r on u is $a=|r|\cos{\theta}$ where $\theta$ is the angle between $\mathbf{r}$ and $\mathbf{u}$ and $\mathbf{r}$ is in the first quadrant, then the scalar projection of $\mathbf{r}$ on $\mathbf{v}$ is $b=|r|\cos(\pi/2 - \theta)$ where the argument of the cosines is measured in radians.
eh, geometrically is probably easier and then I was thinking I could come up with a set of linear equations which I could show only has one solution which would indicate uniqueness
geometrically would be easier just because I can see where I'm going
Okay... I would keep trying this but I have been thinking about this for way too long... haha... Could you tell me what you had in mind for the lower bound @TedShifrin? I give up...
Ha, this is actually from a multivariable calculus class but we're basically just doing introduction to planes/boundaries/neighbourhoods/vectors now, which is what this question is from.
@TedShifrin Its $\frac{2}{3^6} - \frac{2}{3^9}$ I see in this case that it is non 0, the problem is that there can be infinite terms like this and I can't guarantee that one term won't equal all of the others...
So, if we have a point $[x:y:z:w]$ satisfying those equations... Well, let's say $x=0$. Then $y^2 = xz = 0$, so $y=0$. But then $z^2 = yw = 0$. So that's just the point $[0:0:0:1]$
Otherwise $x=1$. Then $z=y^2$ and $w = wx = yz = y^3$
So yeah this should be our twisted cubic, it's a copy of the affine twisted cubic plus one point at infinity
By the way, my ultimate challenge to you will be to figure out the normal bundle of the twisted cubic in $\Bbb P^3$. One of my all-time favorite exercises.
@William: No, worst case scenario is that they'll be different for all $i\ge 7$.
And the signs all are the same to make you most worried.
So let's say you have $yw = z^2$ and $xz = y^2$. If $x=0$ we're still stuck with the point $[0:0:0:1]$, and if $x=1$, well $z=y^2$, and $yw = z^2 = y^4$. So the question here becomes whether $w$ could be something other than $y^3$
Now let's try $yz = xw$ and $xz=y^2$. If $x=1$ we have $[1:y:y^2:y^3]$. If $x=0$, then $y=0$, but then $z$ and $w$ can do anything, so we have a copy of $\mathbb{P}^1$ at infinity
The closure of the affine cubic includes one of those points at infinity. So the intersection of the two surfaces is the twisted cubic union the line at infinity.
@TedShifrin I want to come up with an intuitive way of specifying a trivialization of the normal bundle of an orientable knot (in an oriented 3-manifold) via a non-vanishing section.
I originally thought of just using the tangent vector to complete the section to a basis of the tangent space of the ambient space.
That is, if $K\subset M$ is the knot in the 3-manifold, then at each point $p\in K$ a section $v\colon K\to NK$ specifies a 2-frame of $NK$ via completing $(K'(p),v_p)$ to a basis of $T_pM$, where $K'(p)$ is the oriented tangent vector of the knot.
@anakhro: Any time you have an oriented 2-dimensional vector space, you can always complete a non-zero vector to a basis. But finding a (smooth nonvanishing) section in the first place is the challenge.
@TedShifrin So the non-vanishing section does indeed imply triviality of NK, but isn't any easier to find than a trivialization of NK in the first place.
@TedShifrin I guess if I had a Seifert surface $S$ for $K$ which I knew were orientable, then I'd have a non-vanishing section for $NS$ which I could use as $v$.
@TedShifrin So, my thoughts so far: by a similar argument, there should be a geometric argument applied to v so that we have a line orthogonal to v. Now all that's left to show is that these orthogonal lines cannot be parallel, which will lead to a necessary condition that there is (by some Euclidean axiom) one meeting place for these non-parallel straight lines. This means that the vector r we picked to get a scalar projection of r on u to be a and r on v to be b must be unique.
@anakhro: I think you can get it out of Sard's Theorem.
@nbro: You just told me that if $1\le q\le 5$, then $5=-5$.
@psa: Yes, that's the gist of the argument the answer to your question gave. But the two lines are parallel if and only if the original vectors are (nonzero) scalar multiples of one another.
@TedShifrin I didn't say this (I think). I wanted to say that q is maximised at a if -q is minimized at a. In other words, if q has a maximum at a, then this corresponds to a minimum of -q
