8,129 reputation
1643
bio website people.bath.ac.uk/mdp33
location Bath, United Kingdom
age 25
visits member for 2 years, 7 months
seen Sep 12 at 15:53

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


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?
Apr
24
comment A proof of $x^TAx=\mathrm{tr}(Axx^T)$
@user89987 Yes, that is correct - unfortunately I only noticed after the time limit for editing comments had elapsed.
Apr
23
comment A proof of $x^TAx=\mathrm{tr}(Axx^T)$
@StevenStadnicki No problem, it happens to us all...hence me finding it very easy to believe I had got it wrong!
Apr
23
comment A proof of $x^TAx=\mathrm{tr}(Axx^T)$
@StevenStadnicki Maybe I'm just having a brain-freeze; surely if you check the two maps I wrote above (correcting the typo in the first one) agree on a basis, then they are equal by bilinearity, and in particular they are equal when the two inputs are the same. I can't see what I'm missing... The difference between this and what the OP said is that you have to check all $n^2$ pairs of basis vectors, not just the $n$ pairs consisting of the same basis vector twice (which I agree is not enough). This is the (longer) argument in Martin's answer.
Apr
23
comment A proof of $x^TAx=\mathrm{tr}(Axx^T)$
The maps aren't bilinear either; that requires them to take two arguments, but they only take one. The maps $(x,y)\mapsto x^TAx$ and $(x,y)\mapsto\operatorname{tr}(Ayx^T)$ are bilinear though, so you could apply your argument to those, and then specialise to $x=y$.
Apr
23
answered Matrix representations of Transformation with change of basis (Fraleigh Beauregard)