8,502 reputation
1644
bio website people.bath.ac.uk/mdp33
location Bath, United Kingdom
age 25
visits member for 2 years, 8 months
seen 7 hours ago

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


7h
revised Showing $\mathbb{H}$ is isomorphic to a subring of $M_2(\mathbb{C})$ as $\mathbb{R}$-algebras
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8h
comment Linear Algebra Subspace question
Admittedly I don't know what happened to them many years later...but they were certainly able to solve linear algebra problems comfortably by the end of the course. I might be a little concerned if they were consistently making algebra mistakes like $b_1b_2=0\implies b_1=b_2=0$, but there's no evidence here that that was any more than a one-off error.
8h
comment Linear Algebra Subspace question
The other thing, which John's answer touches on, is that you should be doing things very carefully if you're learning this for the first time. So no saying things like "no limitation"; check explicitly that the zero vector is in the set, if $v$ is in the set then $\lambda v$ is in the set for all $\lambda\in\mathbb{R}$, and that if $v,w$ are in the set, then so is $v+w$. This should be fairly mechanical...what requires a little more creativity is finding a counterexample when one of these statements isn't true.
8h
comment Linear Algebra Subspace question
I think your last comment is a bit too pessimistic - these are the kinds of problems that about half of the students have at the start of the linear algebra course at my university. Most of them get over them after a couple of weeks, and go on to do perfectly well. (Anticipating possible confusion - I didn't downvote your answer.)
8h
comment Linear Algebra Subspace question
The numbers $b_1$, $b_2$ and $b_3$ are not vectors. This seems to be a source of confusion in many of the examples - in b), for example, you are being asked to consider the the set $\{(1,b_2,b_3):b_2,b_3\in\mathbb{R}\}\subset\mathbb{R}^3$, which is indeed a plane (although it doesn't pass through the origin...). It may help to add the tuple $(b_1,b_2,b_3)$ after the word "vectors" in b), c), e) and f).
12h
comment Permutations with forbidden values
Actually, forget that, I should have followed your Wikipedia link - to flag it up for other readers, in this question a permutation of a set is an ordering of the elements, and a permutation of length $r$ (called an $r$-permutation on Wikipedia) is an ordering of any $r$-subset of the original set.
12h
comment Permutations with forbidden values
What notation are you using? In your $n=3$ example, it seems you are just listing all the values of the permutation, but then once you introduce $r$ it seems like you might have switched to cycle notation - except that then $[1,2]=[2,1]$, but you list them separately. I think it would be helpful to clarify the notation, and maybe explain more clearly what a "permutation of length $r$" is.
1d
revised The ideal for image of Segre embedding
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1d
revised Unnecessary Elements in the Tensor Product?
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1d
revised Is an ideal also a normal subgroup?
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1d
revised If $K_X$ is not $\mathbb Q$-Cartier then it is not nef
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1d
comment Can Cayley-Menger Determinant Be Negative?
You mean that you input arbitrary values for $\beta_{ik}$? In that case I imagine you could make it take essentially any value at all, and it no longer has anything particularly to do with volumes. Zero is certainly achievable - for example, you could take all the $\beta_{ik}$s to be zero. Although as I point out in the last comment, it can be zero even when the $\beta_{ik}$s are distances between points.
1d
comment Can Cayley-Menger Determinant Be Negative?
What do you mean by the last sentence? The point is that the formula can't tell if you've input a degenerate shape. If I want to compute the area of a triangle, I input the coordinates of the vertices to the formula. However, I could input three points which all lie on a line, and then the formula would output zero; which is in a sense the area of a degenerate triangle whose vertices are colinear.
1d
revised Calculating the images of transformations of matrices
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1d
comment Calculating the images of transformations of matrices
What did you get for the image? Remember that two different sets of vectors can have the same span, so you might have written down the correct set even if you didn't choose $(1,0,-1)$ and $(0,1,1)$ as a basis.
1d
comment Linear map problem
You need to write $(3,-1,1)$ as a combination of the vectors $(1,0,0)$, $(1,1,0)$ and $(0,-1,1)$, i.e. you need $\lambda_1,\lambda_2,\lambda_3$ such that $(3,-1,1)=\lambda_1(1,0,0)+\lambda_2(1,1,0)+\lambda_3(0,-1,1)$, and then you know how to apply $T$ to this combination. You tried to use $3,-1,1$, but $3(1,0,0)-1(1,1,0)+(0,-1,1)=(2,0,1)\ne(3,-1,1)$.
2d
answered What does “$\mathbb{F^n}$ is a vector space over $\mathbb{F}$” mean?
2d
revised Vector spaces and direct sums
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2d
answered Linear algebra: diagonalisation of antisymmetrisation
2d
comment Linear algebra: diagonalisation of antisymmetrisation
In that case I will make it a proper answer!