# Matt Pressland

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bio website people.bath.ac.uk/mdp33 location Bath, United Kingdom age 25 member for 2 years, 10 months seen 7 hours ago profile views 924

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

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 Oct22 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. Oct22 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.) Oct22 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). Oct22 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. Oct22 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. Oct21 revised The ideal for image of Segre embedding edited body; edited title Oct21 revised Unnecessary Elements in the Tensor Product? added 2 characters in body Oct21 revised Is an ideal also a normal subgroup? added 7 characters in body Oct21 revised If $K_X$ is not $\mathbb Q$-Cartier then it is not nef deleted 1 character in body; edited title Oct21 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. Oct21 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. Oct21 revised Calculating the images of transformations of matrices added 53 characters in body Oct21 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. Oct21 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 $3(1,0,0)-1(1,1,0)+(0,-1,1)=(2,0,1)\ne(3,-1,1)$. Oct20 answered What does “$\mathbb{F^n}$ is a vector space over $\mathbb{F}$” mean? Oct20 revised Vector spaces and direct sums added 27 characters in body Oct20 answered Linear algebra: diagonalisation of antisymmetrisation Oct20 comment Linear algebra: diagonalisation of antisymmetrisation In that case I will make it a proper answer! Oct20 revised Linear algebra: diagonalisation of antisymmetrisation edited body Oct20 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.