8,504 reputation
1644
bio website people.bath.ac.uk/mdp33
location Bath, United Kingdom
age 25
visits member for 2 years, 8 months
seen 9 hours ago

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


9h
answered What does “$\mathbb{F^n}$ is a vector space over $\mathbb{F}$” mean?
10h
revised Vector spaces and direct sums
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11h
answered Linear algebra: diagonalisation of antisymmetrisation
11h
comment Linear algebra: diagonalisation of antisymmetrisation
In that case I will make it a proper answer!
11h
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.
11h
revised Linear algebra: diagonalisation of antisymmetrisation
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11h
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.
12h
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!
13h
revised Solving an equation for X
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13h
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.)
13h
reviewed Approve suggested edit on Is the polynomial $X^{32} + 1$ irreducible?
13h
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 $\{v_1,\dotsc,v_k\}$...
15h
revised Question on Showing the dimensions of a Vector Space
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15h
revised Question on Showing the dimensions of a Vector Space
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15h
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!
15h
revised Showing that the image of a homomorphism $d$, with $d^2=0$, is contained in its kernel
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15h
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 $d(y)=x$.
15h
answered Showing that the image of a homomorphism $d$, with $d^2=0$, is contained in its kernel
15h
revised Showing that the image of a homomorphism $d$, with $d^2=0$, is contained in its kernel
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15h
revised Algebraic Geometry: A question about radical ideal
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