In particular I'm most interested in the Galois field with 2 elements, binary matrices
But I'm very interested in the Galois field with general prime order
The matrices in question are square
And the order can be assumed to be too large to feasibly search numerically
Er.. when I say "multiplicative order" I'm generalizing the notion to matrices, what I exactly mean to find is the period of the function $f(k) = \mathbf{A}^k$ where $\mathbf{A}$ is the matrix, and $k$ is an integer exponent
Over $\text{GF}(p)$ it's clearly less than or equal to $p^{n^2}$ for an $n\times n$ matrix, but that bound is pretty useless to me :/
Since it's uniformly random, I can even make some guesses about the cycle length, but I'm pretty stumped on finding an exact result
A k-form is a map V^k->F which is linear in each individual argument when the others are held constant. That's how multilinearity works. It's equivalent to specifying a linear map V^{\otimes k}->F.
but anyways I think I get it. the reason that there's no "basis-independent" definiton is because thinking of a form as a map V^k -> F is entirely the wrong way of looking at it
I think I made a nice dent in the problem: When $\mathbf{A}$ is a binary $n\times n$ matrix, the least exponent $k$ which takes $\mathbf{A}^k$ to the identity matrix must divide $\sqrt{2^{n^2-n}} \prod_{i=1}^n (2^i - 1)$
That's pretty much already enough to make a reasonable computer search
@sequence see in Wikipedia Divisor function. Then see the case $\alpha=0$, that is $\tau(n)=\sigma_0(n)$, a notation for the number of divisors of an integer $n>1$, where $\tau(1)=1$, and for example since $1,2,3,$ and $6$ are the divisors of $6$ then $\tau(6)=\sigma_0(6)=4$. See your case $\alpha=0$ in the section Series relations. Is not required a response of this comment, good luck.
Puedo deducir cosas tipo dimensiones, porque la dimension de la suma de subespacios es la dimension de la interseccion + dim de la suma O porque el rango + nulidad de una transformacion lineal es la dimension de V Otras cosas sobre la base dual, como la existencia y unicidad
O porque si una transformacion es isomorfica entonces su nucleo es {0}
cosas simples pero antes no podia deducir de donde salian
I want to find the rank of the matrix $\begin{pmatrix}0&1&1\\ 0&0&1\\ 0&0&0\end{pmatrix}$, but I am not really sure what to do in this case where the first element is zero. Could you give me a hint?
We have the tableau $\begin{pmatrix} \left.\begin{matrix} 1 & 0 & \alpha \\ 0 & 1 & \beta \\ 0 & 0 & 0 \end{matrix}\right|\begin{matrix} c\\ d\\ 0 \end{matrix} \end{pmatrix}$
Since there is a zero-row, do we conclude that the column vectors are linearly dependent? Or to check the linearly (in)dependence of the column vectors do we have to check if there is a zero-column?
Ah ok... The number of linearly independent row- and column vectors is the same. And from the tableau we get that there are 2 linearly independent row- and column vectors, or not? Therefore the dimension of the the vector space spanned by the solumn vectors is $2$, right? And this is also equal to the rank of the matrix, right? @DHMO
e.g. Doing some calculus exercise, and then suddenly I think about vector spaces because a multivariable function has 4 entries, which reminded me of 4 dimensions, which then remind me of R^4
What is a geometric interpretation of all these information? Do we get that two column vectors are either a multiple of each other or they are on the same line? Or is there also an other interpretation? @DHMO
> As scary as it sounds, Daily Routine is suspected to have countably infinite number of parameters, since the procedural nature of Daily Routine means the nodes occur in discrete numbers thus there exist a bijective mapping of them to the natural numbers
Background context: Consider Daily Routine to have some kind of digraph structure
I'm currently revising a course on graph theory that I took earlier this year.
While thinking about planar graphs, I noticed that a finite planar graph corresponds to a (finite) polygonisation of the Euclidean plane (or whichever surface you're working with). By considering, for example a full t...
Search for that because I initially think "discrete number" is ambigurious
Therefore in summary: My brain=Given a proposition, which each word or each phrase is considered a node. There is nonzero probability that a random number of branches follow for each node. This process can be repeated indefinitely with nonzero probability
I only found "Adjunction space" for recollement in the context of topology, not sure about differential equations unless the solution space is equipped with a topology
no it is just joining or gluing the solutions at a point x0 so that the joining have some requested regularity. usually I use wikipedia to find this, but I can't seem to find the wording this time.
I've just obtained the prime factors of a 19728 digit integer :3
Or concretely, 10x the length of a paranoid RSA key
(the number is actually $2^{256}$-smooth *(actually actually (2^{217}$, to be exact), so it's not groundbreaking, but it sure feels good since it's for an app)
the smallest factor is $2$ and the largest factor is $160619474372352289412737508720216839225805656328990879953332340439$, the rest look Poisson distributed
@user90369 many thanks for your feedback, I didn't know such question from the literature but now with your feedback I have more idea. Many thanks then for your attention and help.
Let $X \sim \mathcal{N}(\mu,\sigma)$, i.e. $X$ is a random variable with distribution
$$
p_X(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{-\left(\frac{x-\mu}{\sqrt{2}\sigma}\right)^2}
$$
I haven't touched probability theory for a while it is then possible I'm mistaking something. I want to find the dist...
f is arbitrary (from $S^1$ to $S^1) I am trying to show that $S^1$ is not simply connected by assuming it is nullhomotopic then out of the homotopy deriving a contradiction from the fact that I made a homotopy from the identity on f to the constant function.
@Secret here's my current (flawed) understanding: an open set is such that it is (contains) a neighbourhood of every point; a neighbourhood is which contains an open set containing that point...
Just had another AT lecture, we discussed properties of fundamental groups, the homomorphism induced by a continuous map, they're preserved by homotopy equivalences ans homeomorphisms, stuff like that. It's funny how much we can say even without knowing how to compute any of them
@DHMO but the neighborhood of a and b is {a,b,c} since {a,b} is open and is contained in {a,b,c} by the definition of neighbourhood you provided above?
[That irrational set problem] I don't see how that induction can fail except it is weird that for each iteration (by picking a ratioanl z between x and y where x > y, and then iterate this process indefinitely to get to the result of dense) it means the open sets that can be unioned to form the irrataionals are becoming shorter in size yet have the same cardinality, and they are always countably many of them
so logically, those open intervals can get arbitrarily small. I suspect that might be how they screw up the induction proof
[Recap] Fact 1: Irrationals are not a countable union of open intervals Proof: as mentioned above using countablity of rationals
Fact 2: By induction, the complement of union of closed intervals is a union of open intervals -> Hint: This does not hold for irrationals. figure out why
It's also easy to convince yourself that the CW complex given by a circle, with a disk attached along the boundary by the map $z \mapsto z^n$, has fundamental group Z/n.
Amusingly, the fact that RP^2 has fundamental group Z/2 has some implications in physics. (I specifically have in mind topological defects in liquid crystals.)
well for general point set topology,I guess the fatness of nieghbourhoods is guarenteed by it properly contains an open set, so open sets are sort of something like a generalisation of a $\epsilon$ ball
One rather irritating thing: As far as I know, there aren't any experimental realizations of topological defects classified by nonabelian homotopy groups.
I'm so confused, i'm doing a simulation that runs some system along a gradient of an error function, independently of the random seed the error function first gets small and then goes up again to some stationary value
How can you pick closed intervals (the singletons) if the rationals are not even well ordered
Let P() be the proposition P(n+1) will be completely out of the window because there is no next one.
Now suppose we use that cantor zigzag mapping (forgot its actual name) to index the rationals so as to impose a well ordering(?), then the open intervals that result by taking the complement of say the singleton union [1/2,1/2] U [1/3,1/3] will be the reals with 2 holes. Now moving onto the next step of the induction, say [1/2,1/2] U [1/3,1/3] U [2,2] then you still get something with 2 holes beca…
Ok, I don't really see any wrong logic here. The induction looks fine and thus the complement of union of closed intervals gives the union of open intervals
Then I am really stuck. Given how we can enumerate the rationals, thus avoiding the issue of well ordering and infinite descending sequences, and that so far by putting in singletons one by on I still not seeing anything weird popping out in terms of the logic, I don't really know what is happening and how it fail
Actually the $\omega$-th case should look like this: $$\bigcup_{i\in \Bbb{N}} \{q_i\}$$ and then take its complement, but the problem is I don't know how to prove this since this obviously does not have the same expression as the base case and the successor case (what is said to be the inductive case above)
What is the difference between discrete data and continuous data?
Let $X \sim \mathcal{N}(\mu,\sigma)$, i.e. $X$ is a random variable with distribution
$$
p_X(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{-\left(\frac{x-\mu}{\sqrt{2}\sigma}\right)^2}
$$
I haven't touched probability theory for a while it is then possible I'm mistaking something. I want to find the dist...
