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Oct
20
comment Linear algebra: diagonalisation of antisymmetrisation
In that case I will make it a proper answer!
Oct
20
revised Linear algebra: diagonalisation of antisymmetrisation
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Oct
20
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.
Oct
20
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!
Oct
20
revised Solving an equation for X
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Oct
20
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.)
Oct
20
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 $\{v_1,\dotsc,v_k\}$...
Oct
20
revised Question on Showing the dimensions of a Vector Space
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Oct
20
revised Question on Showing the dimensions of a Vector Space
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Oct
20
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!
Oct
20
revised Showing that the image of a homomorphism $d$, with $d^2=0$, is contained in its kernel
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Oct
20
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 $d(y)=x$.
Oct
20
answered Showing that the image of a homomorphism $d$, with $d^2=0$, is contained in its kernel
Oct
20
revised Showing that the image of a homomorphism $d$, with $d^2=0$, is contained in its kernel
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Oct
20
revised Algebraic Geometry: A question about radical ideal
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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 $\phi(e_{G_1})^{-1}$.
Oct
14
revised Prove that the kernel of a group homomorphism $\phi$ is a subgroup and that $\phi$ is injective
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