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 Jan 11 comment Group Duality with respect to Generators and Relations (1/2) I think my real point is: in order for such a definition to be mathematically meaningful, its conception needs to be mathematically well-motivated. You can't just say "I want duals to exist; where are they?". Better questions: what do you want $G^D$ to do? Or how do you want it to look relative to $G$? I see no mathematical reason to swap the generators and relations of a group (and no guarantee that it could be made to work). My geometric proposal was just something slightly more well-motivated (but, if you work through the details, you'll see it's also not very interesting). Jan 10 comment Group Duality with respect to Generators and Relations Suppose $G = \langle x | x^2 \rangle$. What does your proposed object $G^D$ mean? (But, if you want your 'dual' group to be the group of symmetries of the respective 'dual' polyhedron - which seems like quite a reasonable request to me - then the right definition is probably simpler than you think.) Jan 10 comment How to prove that a set spans a plane @Pow: aha, okay - take an element (x,y,z) on your plane. Supposing $c \neq 0$, we can more helpfully write this as (x,y,$\frac{-ax-by}{c}$). Now it suffices for you to find r and s (in terms of a, b, c, x and y) such that r(b,-a,0) + s(0,c,-b) = (x,y,$\frac{-ax-by}{c}$) - shouldn't be too hard. (Then of course you have to deal with the case $c = 0$...) :) Jan 10 answered How to prove that a set spans a plane Sep 10 awarded Yearling Jan 25 comment What's the difference between $\mathbb{R}^2$ and the complex plane? (You might also find it helpful to note that the "natural" multiplication on $\mathbb{R}^2$, namely $(u,v)\cdot (x,y) = (ux, vy)$, does not agree with the natural multiplication on $\mathbb{C}$ for any choice of $g\in \mathrm{GL}_2(\mathbb{R})$.) Jan 25 comment What's the difference between $\mathbb{R}^2$ and the complex plane? @laovultai: Let $f$ be the map $\mathbb{R}^2\to \mathbb{C}$, $(a,b) \mapsto a+ib$, and let $g$ be any element of $\mathrm{GL}_2(\mathbb{R})$. Then $f\circ g$ is an isomorphism $\mathbb{R}^2\to \mathbb{C}$ as $\mathbb{R}$-vector spaces. Now simply "pull back" the multiplication from $\mathbb{C}$ to $\mathbb{R}^2$ along the map $f\circ g$ (e.g. when $g$ is the identity map, the multiplication inherited is $(u,v)\cdot (x,y)=(ux−vy,uy+vx)$), and you get an isomorphism of rings (or, equivalently, $\mathbb{C}$-vector spaces). Does that answer your question? Jan 17 comment How should I understand $R[x]/(f)$ for a ring $R$? f should be irreducible, otherwise it is not true that R[alpha] = R[x]/f. Jan 17 awarded Custodian Jan 17 reviewed Approve Optimization Word Problem Jan 17 answered If $X$ and $Y$ are objects of $\mathrm{Set}$, is there any reason not to regard $\mathrm{Hom}(X,Y)$ as an object, too? Jan 17 comment What's the difference between $\mathbb{R}^2$ and the complex plane? @laovultai: Because they're not "equal". They're isomorphic, but in order to prove that, I have to choose an isomorphism. I chose the "obvious" one, $(a,b) \mapsto a+ib$. (There are lots more, e.g. $(a,b) \mapsto 2a + b - 5ia$, or $(a,b) \mapsto ia - b$.) Dec 22 awarded Nice Answer Sep 10 awarded Yearling Aug 2 comment Surjection/Injection in Product of Linear Transformation Also, why not just pick an explicit example? The first example I think of is $$S = \begin{pmatrix} 1&0&0\\0&1&0\\0&0&1\\0&0&0\end{pmatrix}, T = \begin{pmatrix} 1&0&0&0\\0&1&0&0\\0&0&1&0\end{pmatrix}.$$ What properties does this have? Aug 2 comment Surjection/Injection in Product of Linear Transformation "Say, S(abc)=abcd, also S(abc)=pqrs." But this can never happen! Functions don't work like this. If you plug in an input, you get one output. Jul 29 comment Should we simplify or not for function domain? Do you mean "what is the domain of $f$"? Personally, if I was being very strict, I would say that the definition of $f$ was "take $x$, square it, and then divide the result by $x$", which always gives you $x$ back except when $x = 0$, when the calculation doesn't make sense. So the domain is the set of non-zero real numbers. (In practice, I am never this strict, except when I am teaching students how to be this strict.) Jul 29 awarded Citizen Patrol Jul 28 comment linear map vs operator: raised to power Your example illustrates the point nicely, but of course, it's slightly worse than that: even if $V\cong W$, there's no obvious way of defining $T^2$. (Matrices are evil notation in that they hide the implicitly chosen bases. The OP might want to consider a map from $V$, the space of real polynomials of degree at most 4, and $W$, the space of real symmetric 3*3 matrices of trace 0. Both are $\cong \mathbb{R}^5$, and there are clearly lots of nice linear maps between them, but there's no obvious way of applying one twice.) Jul 28 comment linear map vs operator: raised to power What does $T^2$ mean, when $T: V\to W$? If I try to plug in a vector $v\in V$, I get a little confused: $T^2 v = T(T(v))$, but $T(v)\in W$, so I can't apply $T$ again.