# Matt Pressland

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bio website people.bath.ac.uk/mdp33 location Bath, United Kingdom age 25 member for 2 years, 6 months seen Aug 11 at 15:14 profile views 810

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

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 Mar5 comment Does associativity justify $(f^{-1}gf)(f^{-1}hf) = f^{-1}gff^{-1}hf$? @foo1899 I guess my language possibly isn't clear. I mean that if you have no associativity, you don't know whether $fgh$ is supposed to mean $(fg)h$ or $f(gh)$, and it might matter. Hence you can't write $fgh$ (or any product of three elements) without establishing a convention about which is the correct reading. Mar5 comment Does associativity justify $(f^{-1}gf)(f^{-1}hf) = f^{-1}gff^{-1}hf$? @foo1899 It also isn't a priori defined: the fact that we can unambiguously write expressions consisting of the product of more than 2 elements without brackets is a consequence of associativity. (Also, SE should let you post an answer, so if you can't, this is a bug.) Mar5 comment Does associativity justify $(f^{-1}gf)(f^{-1}hf) = f^{-1}gff^{-1}hf$? (5) isn't really a well-defined statement because $fgh$ isn't a priori defined; $fgh$ may be defined to be equal to $(fg)h$, which is equal to $f(gh)$ by (2). This definition would be dangerous without associativity. The statement you want (where you drop all the brackets) follows by repeated application of associativity, but this may be a little fiddly to do in practice. (It would probably help to start by putting more brackets in, so that the elements all occur in bracketed pairs). Mar4 revised Do you know to find the sum of the series? Turned attached image into MathJax Mar4 revised Rigorous proof that any linear map between vector spaces can be represented by a matrix added 5 characters in body Mar4 comment A dense subset of a finite group The definition gives you the meaning of dense - the closure is $G$. The finiteness isn't relevant, this is purely topological. A subset $S$ of a topological space $X$ is dense if $\overline{S}=X$. Here you are taking the closure in the Zariski topology. Mar3 revised For the following function, determine whether it is a homomorphism. added 190 characters in body Mar3 comment For the following function, determine whether it is a homomorphism. OK - now you should check the definition of homomorphism. Take matrices $\left(\begin{smallmatrix}a&0\\b&a\end{smallmatrix}\right)$ and $\left(\begin{smallmatrix}c&0\\d&c\end{smallmatrix}\right)$, and multiply them together. If $\varphi$ takes the result to $\frac{b}{a}+\frac{d}{c}$, then you do have a homomorphism. I hope this last statement is clear, please ask if not! Mar3 comment For the following function, determine whether it is a homomorphism. You need to define $G$ and $R$ for us. If you already have some thoughts on the problem, it would be good to hear them as well. Feb26 comment Name of a certain type of rings They certainly exist (see $\mathbb{Z}$, or any finite ring), but I don't know of any name. Feb26 reviewed Reopen Abstract Algebra (FUNCTIONS) Feb25 comment Proof that a certain subset of the reals is not a ring It's not immediately clear why there isn't some $y\in\mathbb{Z}$ such that $y\sin{y}=-x\sin{x}$ though (I agree that this is unlikely). Feb25 comment How to prove the equation |xy|=|x||y| if we assume x and y are real numbers by using analysis. The left-hand side won't equal the right hand side in general if you remove the absolute values. But the idea of removing the absolute values is a good one, and the hint tells you how to do this - $|x|$ is always equal to $x$ or $-x$ depending on whether $x\geq0$ or $x<0$. So making a choice of sign for both $x$ and $y$ lets you take the absolute values away, and there are four possible pairs of sign choices... Feb19 comment What does it meet for an ideal to meet as set? Do you have more context? It would usually mean that they have non-empty intersection. Feb19 comment How is an open set defined without referring to any distance function in a topology? Technically you could call them closed sets, but that would then be very confusing, because the open sets in a metric space would be closed sets of the topology defined by the metric. And nobody wants that. Feb19 comment How is an open set defined without referring to any distance function in a topology? This has definitely come up before but I can't find it. The elements of the topology are simply called the open sets, that's it. The definition of an open set in this context is "an element of the topology". The name is the same as in the metric spaces sense because it is a generalization; a metric defines a topology in which the open sets are those open in the metric sense. Feb17 comment Short question about 'multilinearity' of determinant Yes - this also (sort of) explains the name of the property. You can think of the determinant as a function of $n$ vectors (in two different ways) by making these vectors the rows or columns of a matrix and taking the determinant. This function is then linear in each of its $n$ arguments. Feb17 awarded Yearling Feb13 awarded Nice Answer Jan31 answered Simple questions about isomorphisms between vector spaces