8,514 reputation
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bio website people.bath.ac.uk/mdp33
location Bath, United Kingdom
age 25
visits member for 2 years, 8 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.


2d
comment Isomorphism vs Bijection in Real Analysis
An isomorphism (in general) is not a structure-preserving bijection, but a structure-preserving map with a structure-preserving inverse - there are situations in which a structure-preserving map is a set-theoretic bijection, but the inverse is not structure-preserving, and other situations in which "bijection" is not defined, but "structure-preserving map" still is. More relevantly to your question - the answer is probably isometry (i.e. it preserves distance) or homeomorphism (it preserves openness of sets) but it depends what you're doing exactly.
2d
revised Linear algebra: diagonalisation of antisymmetrisation
edited body
2d
comment Linear algebra: diagonalisation of antisymmetrisation
I agree with your concern - providing you are only allowed to use a single basis for the space of matrices, this is impossible, as there are no eigenvectors with eigenvalue $1$. However, you could replace the $1$ by a $2$, or redefine $f(A)=\frac{1}{2}(A-A^t)$, and then you're fine.
2d
comment Unity of a subring of $\mathbb Z_{10}$
The calculations you do are consistent with $[6]$ being a multiplicative identity, and indeed it is - I'm not sure where you think you're making a mistake!
2d
revised Solving an equation for X
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2d
comment Proof vector x + ⃗y = ⃗x + ⃗z then ⃗y = ⃗z
Do you know how you would prove this for numbers? Look at the vector space axioms. (Presumably the sentence "Show that $S$ is a vector space" shouldn't be part of the question, as $S$ doesn't appear anywhere else.)
2d
reviewed Approve suggested edit on Is the polynomial $X^{32} + 1$ irreducible?
2d
comment Prove that the linear map of the basis $V$ is a spanning set of the image of $f$
You don't need $f$ to be injective (at least on $V$) - it isn't in general. If $f(v)=0$ then $v\in\ker{V}$, so $v$ is in the span of $\{v_1,\dotsc,v_k\}$...
2d
revised Question on Showing the dimensions of a Vector Space
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2d
revised Question on Showing the dimensions of a Vector Space
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2d
comment Showing that the image of a homomorphism $d$, with $d^2=0$, is contained in its kernel
@DanielRust Ah, of course - not quite sure what I was thinking when I wrote that!
2d
revised Showing that the image of a homomorphism $d$, with $d^2=0$, is contained in its kernel
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2d
comment Showing that the image of a homomorphism $d$, with $d^2=0$, is contained in its kernel
Small typo - in your first line, you mean $d(y)=x$.
2d
answered Showing that the image of a homomorphism $d$, with $d^2=0$, is contained in its kernel
2d
revised Showing that the image of a homomorphism $d$, with $d^2=0$, is contained in its kernel
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2d
revised Algebraic Geometry: A question about radical ideal
added 72 characters in body; edited title
Oct
16
answered Can we say anything about the unit of a $k$-algebra $A$ in terms of the unit $1\in k$?
Oct
14
comment Why is a projective subspace itself a projective variety?
Pick a basis $w_1,\dotsc,w_n$ of $W$ and extend it to a basis $w_1,\dotsc,w_{n+1}$ of $K^{n+1}$. Let $\lambda_1,\dotsc,\lambda_{n+1}$ be the corresponding coordinate functions; then in these coordinates the equation of $W$ is $\lambda_{n+1}=0$. You can then change coordinates into something else, like the standard basis of $K^{n+1}$, if you like.
Oct
14
comment Prove that the kernel of a group homomorphism $\phi$ is a subgroup and that $\phi$ is injective
I'm not talking about injectivity at the moment - the important statement, which I sketched the proof of in my previous comment, is that for any homomorphism $\phi$ (not necessarily injective), $\phi(e_{G_1})=e_{G_2}$. You need to use this fact a lot in both part a) and b) of the exercise - you already have used it in your first two comments on this answer! For example, if $\phi$ is injective, and $\phi(g)=e_{G_2}$, then $\phi(g)=\phi(e_{G_1})$, so $g=e_{G_1}$ - this proves one direction of part b.
Oct
14
comment Prove that the kernel of a group homomorphism $\phi$ is a subgroup and that $\phi$ is injective
You should prove that $\phi(e_{G_1})=e_{G_2}$ - in the courses I took this would always be proved before setting the exercise you are doing, and in fact you have already used this fact. If you haven't seen a proof of it, then you can use that $\phi(e_{G_1})\phi(e_{G_1})=\phi(e_{G_1}^2)=\phi(e_{G_1})$, and now multiply by $\phi(e_{G_1})^{-1}$.