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bio website people.bath.ac.uk/mdp33
location Bath, United Kingdom
age 24
visits member for 2 years, 5 months
seen 8 hours ago

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


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)
Apr
23
comment Matrix representations of Transformation with change of basis (Fraleigh Beauregard)
OK - that makes more sense. (In my previous comment, $R_{B,B''}$ should just say $R_{B''}$.)
Apr
23
comment Matrix representations of Transformation with change of basis (Fraleigh Beauregard)
Oh, that's probably not a useful way of thinking about it (although it is true). The procedure is to write $T(v_j)=\sum_{i=1}^n\lambda_{ij}v_i$ for some constants $\lambda_j$, for each $v_j$ in the basis $B$, and then the $j$-th column of $R_B$ consists of the $\lambda_{ij}$s. So each column should tell you the action of the transformation on a single basis vector. I think there might be a typo in the second example; $B$ and $B'$ are the same, and $R_B$ isn't correct. This matrix is actually $R_{B,B''}$, where $B''=(x^2,x,1)$.
Apr
23
comment Matrix representations of Transformation with change of basis (Fraleigh Beauregard)
The $R_B$ and $R_{B'}$ could be written more fully as $R_{B,B}$ and $R_{B',B'}$ to match the notation from before - this time the transformation is from a vector space to itself, so it makes more sense to only choose one basis. (I don't know if this is what you meant when you said you weren't sure what $R_B$ and $R_{B'}$ represent, but maybe it's helpful to say anyway!).
Apr
23
comment Matrix representations of Transformation with change of basis (Fraleigh Beauregard)
I don't see what you mean - it looks vertical in the second example; the third entry in the first column of the augmented part is $0$, which is the third component of $T([1,1,0])$, not the first component of $T([0,1,1])$. The $R_{B,B'}$ are matrices of a linear transformation with respect to the bases $B$ and $B'$, I'm not aware of a shorter name than that.
Apr
22
comment Permutations expressed as product of transpositions
Ah, you made a different mistake that gives the same result as doing the transpositions backwards - on the third arrow you should be swapping the numbers $1$ and $4$ over, wherever they appear, not the first position with the fourth. (I was drawing my way of doing this calculation; put $1$ in on the right and see where it goes - it first gets mapped to $4$ (by $(1,4)$), then $4$ is mapped to $6$ by $(4,6)$, and finally $6$ is mapped to $2$ by $(2,6)$).
Apr
22
comment Permutations expressed as product of transpositions
I'm not sure I understand your first question - but for the second one, you're just reading the transpositions in the wrong order. Your convention for $\sigma_1\circ\sigma_2$ is that $\sigma_2$ is applied first, so with the transpositions you need to start from $(3,6)$ and read from right to left. So $1\mapsto 4\mapsto 6\mapsto 2$ etc.
Apr
22
answered What does $f: 2^{\mathcal{S}}\rightarrow\,\mathbb{R}$ mean?