# 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 811

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

 Aug8 comment Proof of uniqueness of identity element of addition of vector space @rschwieb True - but of course at that point the statement $0=0'$ has been proved, which was the point. I guess I meant "which you never use in obtaining the result that you actually wanted". Jul29 comment Corollary to Smith normal form @mvcouwen This isn't even generally true for operators on vector spaces - not every linear operator is diagonalizable. The best you can do with a linear map $A\colon V\to V$ ($V$ finite dimensional), if you only allow yourself to choose one basis for $V$, is to pick a basis so that the matrix of $A$ is in Jordan normal form. Jul25 comment How do I turn my verbal argument into something formal in [Real Analysis]? (proving every compact set is bounded) I think you mean "If not every compact set on a metric space is bounded...". Jul25 comment How do I turn my verbal argument into something formal in [Real Analysis]? (proving every compact set is bounded) To attempt to answer the second part of the question - if the statements you make in the proof were correct, and the exercise occurs in a context in which they had been proved earlier, then I would say your proof is sufficiently formal as it is. Jul23 comment Differential function as a linear map. For part 2), you've described the set slightly inefficiently, because $0$ is itself a constant, so it isn't a special case. You have proved that this set is contained in the kernel; why is it the whole thing? Jul23 comment Differential function as a linear map. This is in many ways the root of your problem; you need to check that $T(f)$ is linear in $f$, so if $f$ and $g$ are two differentiable functions, then $T(f+g)=T(f)+T(g)$ (and something similar for scalars). You shouldn't try to check linearity in the arguments of the functions $f$ and $g$, because it won't hold, as you point out. Said differently, you want to check that $T(f+g)(x)=T(f)(x)+T(g)(x)$, not that $T(f)(x+y)=T(f)(x)+T(f)(y)$ - this second statement says that $T(f)$ is linear, rather than that $T$ is linear, and this won't be true for general $f$. Jul21 comment Repeated Irreducible Representations in a representation If you forget about representation theory for a minute - an $n$-dimensional vector space over $K$ decomposes as $V=V_1\oplus\dotsb\oplus V_n$, where each $V_i$ is isomorphic to $K$. So all of the summands are isomorphic to each other as vector spaces, despite being different subspaces of $V$. Given this, it shouldn't be so hard to believe that you can have a representation $W=W_1\oplus W_2$ with $W_1$ and $W_2$ isomorphic representations, even though $W_1$ and $W_2$ are as disjoint as possible inside $W$. Jul18 comment Proof needed for this exercise from “Linear Algebra Done Right” @Hobbit6094 Careful - you can't have $\operatorname{Null}(ST)\subseteq\operatorname{Null}(S)$ because the first is a subspace of $U$ and the second is a subspace of $V$. But what you actually proved is that $T(\operatorname{Null}(ST))\subseteq\operatorname{Null}(S)$, which, if you think it through carefully and use the Rank-Nullity theorem, is enough to get what you want. Jul11 comment If $A$ is compact and connected, then is $\Bbb R^2\setminus A$ connected? It's not really a stupid question! But the lesson should probably be to always try out some examples first - Heine-Borel makes it fairly easy to generate lots of them. (Of course, if you didn't have a good handle on what it means to be compact, then generating example might be more problematic, but it seems that you do.) Jul11 comment How Max plus algebra is different from conventional algebra? In a sense, your tag provides an answer to the question - the max-plus algebra gives us a neat way to tropicalize a variety, and then there are various theorems relating properties of the original variety to properties of the tropical variety, which are potentially easier to compute because of the combinatorial nature of the object. I believe there are other applications of max-plus not directly to do with tropical geometry, but I don't know what they are. Jul10 comment writing papers: definition in word or formula? This question might be too opinion-based. I don't have any real preference between the three options. If the set had a more complicated definition, then I would probably be more strongly against C) and more strongly in favour of A). The aim should always be clarity, but this can be extremely subjective. Jun26 comment Origin in homogeneous coordinates @HaraldHanche-Olsen Standard that may be, but canonical it is not! ;-) Jun26 comment Origin in homogeneous coordinates To elaborate a little - the origin of a vector space is special because it is the additive identity under the group structure. But once you projectivise the vector space, the group structure disappears - and there is no "special" line in the vector space to play the role of the origin in projective space. Jun26 comment Origin in homogeneous coordinates There is no origin in projective space! Jun26 comment Building quotient rings @o2genum I'm not sure this is a concrete question - you already have described the rings! If you want to describe them in another form, you need to be quite specific about what that form should be. Asking whether the first two are isomorphic to $\mathbb{Z}_n$ is a perfectly good question, but "what ring is $\mathbb{Z}[\sqrt{-2}]/(2)$ isomorphic to?" is too open-ended (unless there's a "standard" ring it's isomorphic to that I don't immediately see). Jun19 comment Is it true that every eigenvalue has at least one eigenvector? You could in theory define an eigenvalue to be a root of the characteristic polynomial of $f$. Michael's answer provides half of the proof that these two definitions are equivalent. Jun17 comment Is the intersection of dense sets dense? @freebird The complement of a nowhere dense set is always dense - indeed, one definition of a nowhere dense is that the complement of its closure is dense. But the complement of a dense set is not necessarily nowhere dense, as in the example in your question. So you while you can't prove that the intersection of two arbitrary dense sets is dense (because this is false!) the sets $A'$ and $B'$ in your setup have the stronger property that their complements $A$ and $B$ are nowhere dense. I agree with Ittay that this is probably not the best way to think about the problem though. Jun5 comment Solving matrices I don't think I entirely understand your question - but I will try to say something useful. The question you link to asks you to calculate the determinant of the matrix, which you can do without thinking at all about a linear system that the matrix might represent. Jun4 comment Questions for first year students at the University. Some pedantry - I would argue that $1+2+3+\dotsb+100$ already is written as a sum. Maybe you mean "using $\Sigma$-notation" or something like that? May29 comment Indecomposable quiver representations Literally, no, because there are several representations with the same dimension - for example, all the vertex simple representations have dimension $1$. But maybe you mean "dimension vector"?