8,041 reputation
1643
bio website people.bath.ac.uk/mdp33
location Bath, United Kingdom
age 25
visits member for 2 years, 6 months
seen Aug 11 at 15:14

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


Aug
18
awarded  abstract-algebra
Aug
8
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".
Aug
8
answered Proof of uniqueness of identity element of addition of vector space
Aug
8
revised Will the Goldbach's conjecture be a law?
added 426 characters in body
Aug
6
revised Atiyah & Macdonald Prop. 2.9
added 24 characters in body
Aug
6
revised differences and similarities between Linear transformations, Linear functionals, Dual Spaces and Isomorphisms
added 834 characters in body
Aug
6
answered differences and similarities between Linear transformations, Linear functionals, Dual Spaces and Isomorphisms
Jul
29
reviewed Approve suggested edit on How to figure out the solution to this equality problem?
Jul
29
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.
Jul
28
revised Find modulus of $z$ given modulus of $(z-3w)/(3-z\overline{w})$
edited body
Jul
28
reviewed Edit suggested edit on Is the coefficient ring $R$ of a group ring $RG$ necessarily projective as an $RG$-module?
Jul
28
revised Is the coefficient ring $R$ of a group ring $RG$ necessarily projective as an $RG$-module?
re-tagging
Jul
25
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...".
Jul
25
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.
Jul
25
revised Logarithmic question
edited title
Jul
23
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?
Jul
23
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 $f$.
Jul
21
revised Subsequences and limit inferior
added 3 characters in body
Jul
21
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 $W$.
Jul
18
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.