7,706 reputation
1642
bio website people.bath.ac.uk/mdp33
location Bath, United Kingdom
age 24
visits member for 2 years, 2 months
seen 11 hours ago

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


Jan
30
revised Understanding Isomorphism on Binary Operation
added 24 characters in body
Jan
28
comment “Easy” (maybe not) question about dual spaces (Lineal Algebra).
I would argue it is better to say "is" (if you can) than "is isomorphic to". This is quite a subtle issue; when dealing with vector spaces, you don't often see genuine equality! Certainly $V$ is not equal to $(V^*)^*$, but the two are more than just isomorphic, because there is a unique "best" way of identifying them. A lot of people (maybe unfortunately) will say that one vector space "is" another one (without the word equal) when they are naturally isomorphic, but not when they are only unnaturally isomorphic.
Jan
28
comment “Easy” (maybe not) question about dual spaces (Lineal Algebra).
A little bit about the last part; a finite dimensional vector space $V$ is isomorphic to the dual of any vector space $W$ with the same dimension as $V$, but this should not be thought of as an identification because there are lots of choices of isomorphism. However, $V$ is isomorphic to $(V^*)^*$ in a natural way, that involves no choices, and so it is more reasonable to treat this as an identification and say that $V$ is a dual space (it's the dual of its dual).
Jan
28
comment Understanding of a formula with matrix summation
$T$ has to be the length of the vector $e$, and the number of vectors $x_i$. As tabstop says, it is not a matrix. If your problem is with dividing a matrix by $T$; you do this by dividing each entry by $T$.
Jan
28
revised spectrum of a ring
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Jan
28
comment spectrum of a ring
Do you mean the part i) $\implies$ iii)?
Jan
28
reviewed Leave Closed Let V be a complex vector space. If $\langle T(v),v\rangle\in \mathbb R$ for every $v \in V$, then is T self adjoint?
Jan
27
comment Solve the following equation with combinatorics…
@user123499 In relation to the comment you asked for an explanation for - if $r>n$, there are $0$ ways of choosing $r$ things from a set of $n$. If you're question is related to a polynomial expansion, it's unlikely you care about this, and instead you want $n$ such that the two numbers are equal and non-zero. It might be worth editing your question to include the exercise you are trying to solve.
Jan
27
comment Solve the following equation with combinatorics…
@SpamIAm So do I - using the C-notation, the original post (before my edit) said that the 3 and the 12 should be at the top - it's possible this means one should read $\mathrm{C}^3_n$ as $\binom{3}{n}$; I don't remember. Either way, my edit to the actual question reflects what the OP described.
Jan
27
revised Solve the following equation with combinatorics…
added 13 characters in body
Jan
27
comment Solve the following equation with combinatorics…
Your question is a little unclear; do you want to find $n$ such that $\binom{3}{n}=\binom{12}{n}$? (Or perhaps you prefer the notation $\mathrm{C}^3_n=\mathrm{C}^{12}_n$.)
Jan
27
revised Quotient of projective indecomposable modules by their radicals
added 2 characters in body; edited tags; edited title
Jan
27
comment Quotient of projective indecomposable modules by their radicals
This looks like an exercise from a book - which is fine, but it would be nice to know which book (Auslander-Reiten-Smalø?), and the number of the exercise, so that people answering might be able to refer to numbered theorems from the book, and can see what tools you're supposed to have. I think it would also be helpful to say a little bit about what you've tried or already know. (It looks to me like $Y$ should be the projective cover of $X$ as a $\Lambda$-module, but I haven't thought extremely hard about this - maybe you already have and it doesn't work).
Jan
27
answered Can prime quadruplets be adjacent
Jan
24
answered Conceptual query for finding eigen values during change of basis
Jan
22
reviewed Reject suggested edit on How to get reverse number mathematically?
Jan
21
comment Solve this system by rewriting in row-echelon form $ \left\{ \begin{aligned} x+y+z&=6 \\ 2x-y+z&=3 \\ 3x-z&=0 \end{aligned} \right. $
It is acceptable to perform more than one operation (because it's valid to perform each operation on any system of linear equations). You are also in a position to check the algebra yourself - the last set of equations is comparatively simple to solve (you have one equation in $z$ only, so solve it, and substitute the value into the equation only involving $y$ and $z$ etc.) Then you can put the three values back into the original equation and check you have a solution (it looks like you will find that you don't, because of the error spotted by David Mitra).
Jan
21
comment Notation for quotient ring
@rschwieb Ha, sorry, missed that. I read your whole comment once, and then when I wanted to remind myself what you said about equivalence relations I started again from the @. Oops.
Jan
21
comment Notation for quotient ring
@rschwieb I suppose one could also backwards engineer a possible root of the notation that way as well, with each $I$ defining an equivalence relation on $R$ by $x\sim y$ iff $x-y\in I$, so that $R/I=R/{\sim}$.
Jan
21
comment Relation between direct sum and tensor product
Your question might need additional clarity - every object is the direct sum of just itself. In the Lie group setting, at least when the group is simple, the category of representations is Krull-Schmidt, so every object is a direct sum of indecomposables, rather than this being something special about tensor products. Same for the category of finite dimensional vector spaces (although of course what you have written is not in general the decomposition into a sum of indecomposables).