# Matt Pressland

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bio website people.bath.ac.uk/mdp33 location Bath, United Kingdom age 25 member for 2 years, 6 months seen Aug 11 at 15:14 profile views 811

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

# 1,916 Actions

 Apr24 comment A proof of $x^TAx=\mathrm{tr}(Axx^T)$ @user89987 Yes, that is correct - unfortunately I only noticed after the time limit for editing comments had elapsed. Apr23 comment A proof of $x^TAx=\mathrm{tr}(Axx^T)$ @StevenStadnicki No problem, it happens to us all...hence me finding it very easy to believe I had got it wrong! Apr23 comment A proof of $x^TAx=\mathrm{tr}(Axx^T)$ @StevenStadnicki Maybe I'm just having a brain-freeze; surely if you check the two maps I wrote above (correcting the typo in the first one) agree on a basis, then they are equal by bilinearity, and in particular they are equal when the two inputs are the same. I can't see what I'm missing... The difference between this and what the OP said is that you have to check all $n^2$ pairs of basis vectors, not just the $n$ pairs consisting of the same basis vector twice (which I agree is not enough). This is the (longer) argument in Martin's answer. Apr23 comment A proof of $x^TAx=\mathrm{tr}(Axx^T)$ The maps aren't bilinear either; that requires them to take two arguments, but they only take one. The maps $(x,y)\mapsto x^TAx$ and $(x,y)\mapsto\operatorname{tr}(Ayx^T)$ are bilinear though, so you could apply your argument to those, and then specialise to $x=y$. Apr23 answered Matrix representations of Transformation with change of basis (Fraleigh Beauregard) Apr23 comment Matrix representations of Transformation with change of basis (Fraleigh Beauregard) OK - that makes more sense. (In my previous comment, $R_{B,B''}$ should just say $R_{B''}$.) Apr23 comment Matrix representations of Transformation with change of basis (Fraleigh Beauregard) Oh, that's probably not a useful way of thinking about it (although it is true). The procedure is to write $T(v_j)=\sum_{i=1}^n\lambda_{ij}v_i$ for some constants $\lambda_j$, for each $v_j$ in the basis $B$, and then the $j$-th column of $R_B$ consists of the $\lambda_{ij}$s. So each column should tell you the action of the transformation on a single basis vector. I think there might be a typo in the second example; $B$ and $B'$ are the same, and $R_B$ isn't correct. This matrix is actually $R_{B,B''}$, where $B''=(x^2,x,1)$. Apr23 comment Matrix representations of Transformation with change of basis (Fraleigh Beauregard) The $R_B$ and $R_{B'}$ could be written more fully as $R_{B,B}$ and $R_{B',B'}$ to match the notation from before - this time the transformation is from a vector space to itself, so it makes more sense to only choose one basis. (I don't know if this is what you meant when you said you weren't sure what $R_B$ and $R_{B'}$ represent, but maybe it's helpful to say anyway!). Apr23 comment Matrix representations of Transformation with change of basis (Fraleigh Beauregard) I don't see what you mean - it looks vertical in the second example; the third entry in the first column of the augmented part is $0$, which is the third component of $T([1,1,0])$, not the first component of $T([0,1,1])$. The $R_{B,B'}$ are matrices of a linear transformation with respect to the bases $B$ and $B'$, I'm not aware of a shorter name than that. Apr22 comment Permutations expressed as product of transpositions Ah, you made a different mistake that gives the same result as doing the transpositions backwards - on the third arrow you should be swapping the numbers $1$ and $4$ over, wherever they appear, not the first position with the fourth. (I was drawing my way of doing this calculation; put $1$ in on the right and see where it goes - it first gets mapped to $4$ (by $(1,4)$), then $4$ is mapped to $6$ by $(4,6)$, and finally $6$ is mapped to $2$ by $(2,6)$). Apr22 comment Permutations expressed as product of transpositions I'm not sure I understand your first question - but for the second one, you're just reading the transpositions in the wrong order. Your convention for $\sigma_1\circ\sigma_2$ is that $\sigma_2$ is applied first, so with the transpositions you need to start from $(3,6)$ and read from right to left. So $1\mapsto 4\mapsto 6\mapsto 2$ etc. Apr22 answered What does $f: 2^{\mathcal{S}}\rightarrow\,\mathbb{R}$ mean? Apr17 revised Why we can consider both modules as modules over $R_{(p)}$? (Bruns and Herzog, Theorem 1.5.9) added 4 characters in body; edited title Apr14 comment Representation of $GL_2$ on $K^2$ If you follow what they say completely literally, you will need to invert all the matrices, which will make the expressions messier. Apr14 comment Representation of $GL_2$ on $K^2$ In what I said, the induced action was a right action because the action on $K^2$ was a left action - in what you say two comments ago, the action $*$ on the domain is a right action. What you say one comment ago is correct. In terms of what to do about your problem, you will have to use the context of the problem to work out what they want. I wouldn't be surprised if they intended the action on $K[x,y]$ described in my answer, but this isn't consistent with the action induced by the standard left action on $K^2$. Apr11 comment Representation of $GL_2$ on $K^2$ Even more precisely; a left action of $G$ is "the same" as a right action of the opposite group $G^{\mathrm{op}}$, which has the same underlying set, but the operation is defined by $g\times_{\mathrm{op}}h=h\times g$, where $\times$ is the multiplication in $G$. But $G$ and $G^{\mathrm{op}}$ are isomorphic via $g\mapsto g^{-1}$, hence the relation between left and right actions in the comment above. Apr11 comment Representation of $GL_2$ on $K^2$ Not quite - you also need an inverse in the second expression, else you have the problem in my earlier comment. I'm not sure I would call them "exactly the same" either. A more precise statement is that each left action $\cdot$ induces a right action $*$ by defining $x*g=g^{-1}\cdot x$ (and vice versa) and the actions $g\cdot x(a,b)=g^{-1}(a,b)$ and $x\cdot g(a,b)=(a,b)g^{-1}$ are related in this way. Apr11 comment Representation of $GL_2$ on $K^2$ No - you have a left action on $K^2$, so it induces a right action on $K[x,y]$; if you had a right action it would induce a left action. Either way, they can't be on the same side - in what you just wrote, you would have $(x\cdot g)\cdot h(a,b)=x\cdot g((a,b)\cdot h)=x((a,b)\cdot hg)$, which isn't $x\cdot gh(a,b)$. You either have to have the actions on opposite sides, or introduce an inverse. Apr11 comment Representation of $GL_2$ on $K^2$ You extend the definition so that evaluation is a ring homomorphism - so $(2xy^2)(a,b)=2ab^2$. Moreover, $(2xy^2\cdot g)(a,b)$ is $2(g_{11}a+g_{12}b)(g_{21}a+g_{22}b)^2$, so $2xy^2\cdot g=2(g_{11}x+g_{12}y)(g_{21}x+g_{22}y)^2$. Apr11 answered Representation of $GL_2$ on $K^2$