8,899 reputation
1646
bio website people.bath.ac.uk/mdp33
location Bath, United Kingdom
age 25
visits member for 2 years, 10 months
seen 21 hours ago

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


Sep
27
revised Largest number on multiplying with itself gives the same number as last digits of the product
latex formatting, making it easier to understand
Sep
27
comment Show that Aut(Q) is isomorphic to S4
Presumably $S_4$ is the symmetric group on four elements? You might want to explain $\operatorname{Aut}(Q)$, because I can think of a few things it could mean. (Perhaps there is an "obvious" interpretation to people more familiar with quaternions, and I think I can guess from the fact it should be isomorphic to $S_4$, but it would be nice to clarify). You should also say what you have already tried, or be more specific about what you don't understand.
Sep
27
reviewed No Action Needed having trouble with recognizing the dimension
Sep
26
reviewed Close Orders of group elements
Sep
26
comment non linear transformation that satisfies $T(cx) = cT(x)$
@ChristophPegel Haha, as well as getting the answer in faster, you got this comment in while I was editing!
Sep
26
answered non linear transformation that satisfies $T(cx) = cT(x)$
Sep
26
comment What is the first proof that you've done using induction?
Interestingly, when I did this, we did it the other way round, stating the well-ordering principle axiomatically, and using it to prove that induction works.
Sep
26
comment What is the first proof that you've done using induction?
I concur with Danny, although we started from 0. ;)
Sep
26
comment Can abstract nonsense be helpful here?
I think it's even a good viewpoint to have for simpler things like linear algebra sometimes. I never really understood why the fact that $V\cong (V^*)^*$ for finite-dimensional $V$ was interesting or important when I was an undergraduate (after all, $V\cong V^*$, right?). But a little explanation of how the double dual functor is naturally isomorphic to the identity (ideally without using the words "category", "functor" or "naturally isomorphic") can be helpful - the diagram is clear enough without the general theory. I tried this on some undergrads and it seemed to make things clearer.
Sep
26
answered linear transformation's geometric meaning
Sep
26
comment How to measure “linear dependence” of more than two vectors?
The angle seems like too much work for just two vectors - if two vectors are linearly dependent, then one is a multiple of the other. If you have $n$ vectors of length $n$, then they can be put as the columns of a square matrix and the determinant is zero if and only if there is a dependency. (Of course if you have more than $n$ vectors of length $n$, they're automatically dependent).
Sep
25
reviewed Leave Open Time and work issue
Sep
25
reviewed Close Open subgroups of $\mathbb{R}$
Sep
25
reviewed Reject Deriving the approximation formula
Sep
25
reviewed Edit Category of fractions: transitivity and cancellation property
Sep
25
revised Category of fractions: transitivity and cancellation property
changed title
Sep
25
revised Write the following statements in symbolic form using either a universal quantifier or existential quantifier
added 87 characters in body
Sep
25
comment Write the following statements in symbolic form using either a universal quantifier or existential quantifier
@BrianM.Scott I agree - I wonder if somebody saw the list format and assumed it was a verbatim copy of an exercise. At any rate, I vote not to close (although I don't know if people here pay attention to such comments as they do on Mathoverflow).
Sep
25
comment Showing a vectorspace equals a span of polynomials?
You will need to tell us what $P_{2,3}$ is. Homogeneous polynomials of degree 3 in 2 variables?
Sep
25
comment Propositional Logic - induction
@user18921 Fair point. I think I now feel that the rewording of "closed" to "on hold", and the descriptive message explaining what can be done to have the question reopened make this sort of thing gentle enough, but it is hard to imagine exactly how the average new user will respond to it.