8,041 reputation
1643
bio website people.bath.ac.uk/mdp33
location Bath, United Kingdom
age 25
visits member for 2 years, 6 months
seen Aug 11 at 15:14

Postgraduate student at the University of Bath, UK, studying geometry and representation theory. Currently thinking about cluster algebras and related objects.


May
6
answered Notation in modulo groups
May
1
comment Prove by induction that $\;A^n = PD^nP^{-1}$
I don't know if you get notified of this automatically - in case not, I've edited my answer to comment on your attempted solution.
May
1
revised Prove by induction that $\;A^n = PD^nP^{-1}$
added 1687 characters in body
May
1
comment What is a “unique” mirror line of symmetry?
@user14056 If you edit your post, above the edit box (where the formatting options are), there's an add image button. The phrase "2 are unique" suggests to me that either something is badly wrong here, or the word unique is being used in an extremely counterintuitive way! But this depends on the context...
May
1
comment What is a “unique” mirror line of symmetry?
What is the difference between a "mirror line" and a "mirror line of symmetry"? It sounds to me like they should both mean the same thing, and an equilateral triangle indeed has three of them, not one.
May
1
answered Prove by induction that $\;A^n = PD^nP^{-1}$
May
1
comment Prove by induction that $\;A^n = PD^nP^{-1}$
Do you understand what proof by induction is? (I hope that doesn't sound sarcastic, it's a genuine question!). You need to prove that if $A^{k-1}=PD^{k-1}P^{-1}$ for some $k$, then $A^k=PD^kP^{-1}$.
Apr
29
comment About closure under +
Extra aside for the OP: if $u\in\mathbb{R}^3$, then $u$ cannot be equal to the set $\{(1,0,0),(0,1,0),(0,0,1)\}$, but it could be a member of this set. There are also lots of pairs of linearly independent vectors that are not subsets of this three element set, like $(1,-1,0)$ and $(0,1,-1)$.
Apr
29
comment About closure under +
@MJD I've seen UG textbooks use "closed under scalar multiplication" for a subset $S$ of a vector space over $K$ to mean $av\in S$ for all $v\in S$ and $a\in K$, although I agree this doesn't exactly fit with the usual meaning of "closed". That seems to be the idea here. I agree that the $+$ sign should be a $\cup$ to make the answer correct.
Apr
29
answered Why is this proof complete if only one condition is satisfied?
Apr
25
revised How many orbits are there in the group action $A\colon 3\mathbb{Z} \to\mathbb{Z}_{6}$ with action given by $(3n,m)=(3n+m)\bmod6$.
added 239 characters in body
Apr
25
answered How many orbits are there in the group action $A\colon 3\mathbb{Z} \to\mathbb{Z}_{6}$ with action given by $(3n,m)=(3n+m)\bmod6$.
Apr
25
comment How many orbits are there in the group action $A\colon 3\mathbb{Z} \to\mathbb{Z}_{6}$ with action given by $(3n,m)=(3n+m)\bmod6$.
Great - I will summarise some comments in an answer.
Apr
25
comment How many orbits are there in the group action $A\colon 3\mathbb{Z} \to\mathbb{Z}_{6}$ with action given by $(3n,m)=(3n+m)\bmod6$.
Yes. They certainly aren't all the orbits, because they don't cover $\mathbb{Z}_6$, and in fact none of those $4$ sets is an orbit.
Apr
25
comment How many orbits are there in the group action $A\colon 3\mathbb{Z} \to\mathbb{Z}_{6}$ with action given by $(3n,m)=(3n+m)\bmod6$.
Actually neither is correct - I should probably have said that - I was trying to work out where your misunderstanding is. I'm still not sure I understand your method. An orbit is a subset of the space being acting on, in this case $\mathbb{Z}_6$, and the orbits collectively must partition the space. To find an orbit, you have a to pick an element $m$ of $\mathbb{Z}_6$, and act on it by every element of $3\mathbb{Z}$; in this case this means the orbit of $m$ is $\{3n+m\bmod 6:n\in\mathbb{Z}\}$ - does that make things clearer?
Apr
25
comment How many orbits are there in the group action $A\colon 3\mathbb{Z} \to\mathbb{Z}_{6}$ with action given by $(3n,m)=(3n+m)\bmod6$.
Also, are you saying you think that $\{0,2,4\}$ is an orbit? Or that each of $\{0\}$, $\{2\}$ and $\{4\}$ is an orbit?
Apr
25
comment How many orbits are there in the group action $A\colon 3\mathbb{Z} \to\mathbb{Z}_{6}$ with action given by $(3n,m)=(3n+m)\bmod6$.
Just a small comment on the title - an action of $3\mathbb{Z}$ on $\mathbb{Z}_6$ isn't a map $3\mathbb{Z}\to\mathbb{Z}_6$. You could write it as a map $3\mathbb{Z}\times\mathbb{Z}_6\to\mathbb{Z}_6$ though; and then you probably want to write $(3n,m)\mapsto(3n+m)\bmod6$, rather than using $=$.
Apr
25
revised How many orbits are there in the group action $A\colon 3\mathbb{Z} \to\mathbb{Z}_{6}$ with action given by $(3n,m)=(3n+m)\bmod6$.
added 128 characters in body; edited title
Apr
24
revised What is a co-dimension?
added 47 characters in body
Apr
24
answered What is a co-dimension?