Let's take the sequence of naturals at or above two: $2, 3, 4, \dots$ and cross out just the primes $2,3$ and all their multiples: $\not{2}, \not{3}, \not{4}, 5, \not{6}, 7, \not{8}, ...$. Well if you keep going on paper, there appears a pattern, namely $\bullet\bullet\bullet \circ \bullet \cir...
Rule 3: If you asked your question on the main site, please don't post it on the chat. It will get a lot of exposure without that. If you do choose to post it, title is a better format; it is compact and easier on the eyes.
Does anyone know where I can find a complete set of the Lectures in Geometry series by Mikhail Postnikov? My university library has the second volume (and ONLY the second volume) and I really enjoy it. But I'd like to start reading the series from the beginning. I'd prefer the English translation, though I can read a bit of Russian, so if someone knows where to get the original Russian set that'd work too.
@DiscipleofBarney Can you help me prove that the product of two permutation matrices is a permutation matrix? I have been trying, but I can't generalise from the first row
By "permutation form", you mean elements of the symmetric group? Because proving the symmetric group is closed (which is equivalent to this) is not hard.
@Incurrence I personally prefer to think of matrix multiplication as linear combinations of rows, but anyways. $(AB)_{ij}= \sum_{r=1}^n A_{ir}B_{rj}$, when is that term nonzero? only when we have at some point the both parts are 1 in the sum
@Incurrence If you're intent on using matrix multiplication, here's an idea: AB = [Ab_1 Ab_2 ... Ab_n], where b_i is the ith column of B. Ab_i is just the ith column of A. Because each b_i has a 1 in a different row (by definition), each Ab_i is a different column of A. So all you've done is rearranged the columns of A. That won't change the fact that each row and column of A has a single 1 in it and 0s elsewhere.
So it's still a permutation matrix. With some thought, the above could probably be made rigorous.
Look at this edit: math.stackexchange.com/review/suggested-edits/398027, first it changes the meaning and on top of that its like a two year old post (with a perfectly good title), and then for some reason it gets accepted
Indeed @AlexWertheim. I'm doing an REU with Dr. Robert Rumely at UGA. As I understand, we're going to be looking at arithmetic dynamics in $\mathbb{Q}_p$, whatever that means.
We're going to have a crash course in the area once finals are over.
$P_1 = \begin{pmatrix}1&2&\cdots&n\\\pi(1)&\pi(2)&\cdots&\pi(n)\end{pmatrix}\to M_\pi$ Where $M$ is a permutation matrix with $1$'s at positions only $i,j=\pi(i)$
I want to show this is homomorphic, which I can do easily with any given example
But I don't know how to compose $P_1 = \begin{pmatrix}1&2&\cdots&n\\\pi_1(1)&\pi_1(2)&\cdots&\pi_1(n)\end{pmatrix}, P_2 = \begin{pmatrix}1&2&\cdots&n\\\pi_2(1)&\pi_2(2)&\cdots&\pi_2(n)\end{pmatrix}$
Given that I am not sure how to say which element of $\pi_1(i)$ will go into which mapping in $P_2$
Basically there is no nice way with that notation, but since this is a permutation you can have the top part of $P_2$ to be $\pi_i(i)$'s and then the bottom rows to be $ \pi_2 ( \pi_1 (i))$'s @Incurrence
And showing this map is surjective seems really strange, pretty much I just take a $M_\pi$ and say it has values $1$ at all $i,\pi(i)$ and zeroes elsewhere, and then I can just make my $P\in S_n$ from this
Is that acceptable? It pretty much just comes from the definition
I am not sure which, but I think I have to show that permutation matrices are a subset(or maybe a subgroup) of the normaliser of the diagonal matrices group
So $N_G(T)=\{g\in G| gTg^{-1} =T\}$
So all the elements that don't change any diagonal matrix(??)
Oh wait no
I misunderstood
$T$ is the subgroup of diagonal matrices
$G=GL_n(\mathbb{R})$
So I need to find all $g\in G$ that don't change $T$ under conjugation