I have given a rotating matrix. and I am asked to give "the representation of the new basis vectors in regard of the old ones. I only have the rotation matrix and no more information. How can I proceed?
To find an eigenvalue, you find where $\lambda I-A$ has determinant zero. This means it has a nontrivial kernel (it sends some nonzero vectors to zero). So solving $(\lambda I-A)v=0$ should give you a nonzero $v$ for such a $\lambda$
Also: the definition of an eigenvector is a nonzero vector $v$ such that there exists a scalar $\lambda$ such that $Av=\lambda v$. Why is "nonzero" included in the definition?
Because otherwise the zero vector $v=0$ would always be an eigenvector! And in fact it would would with any eigenvalue $\lambda$, since $A0=\lambda0=0$.
It doesn't tell us anything about $A$, and it has all eigenvalues, so we exclude it from our definition for convenience.
If $M/L\cong A$ and $N/K\cong B$, is $M\oplus N/L\oplus K \cong A\oplus B$ for $R$-modules, $R$ commutative
I certainly have a map $A\oplus B$ to $\operatorname{coker}(L\oplus K\to M\oplus N)$
Really I want to know if I can direct sum two composition series, term wise, to see that the function that tells me the number of times a specific simple module appears in a composition series is additive with respect to the direct sum
There is a formula for calculating slope (Regression coefficient), b1, for the following regression line:
y= b0 + b1 xi + ei (alternatively y'(predicted)=b0 + b1 * x); which is
b1=(∑(xi-Ẋ) * (yi-Ῡ)) / (∑ ((xi- Ẋ) ^ 2)) ---- (formula-A)
source: https://stattrek.com/statistics/measurement-s...
Can we say anything about the cycle type of an element $g\in S_n$, when $g=hr$ and we know the cycle types of $h,r\in S_n$?
As an application, say we identify $\Bbb Z_2\times \Bbb Z_2\cong \{(1),(12),(34),(12)(34)\}\subset S_4$, and we want to find out which elements are sent where when we take $S_4/(\Bbb Z_2\times \Bbb Z_2)\cong S_3$
I imagine this would be easy if I knew how the cycle types behaved under multiplication
Although, I imagine the answer is 'they don't behave well at all, since we have things like $(1234)(12)=(134)(2)$ and $(1234)(13)=(14)(23)$'
Actually, given that transpositions generate the group, such a thing is hopeless
Nvm, my confusion with $S_4/V_4\cong S_3$ stems from the fact that my choice of $V_4\subset S_4$ isn't normal.
$\{(1),(12)(34),(13)(24),(14)(23)\}$ is the correct subgroup
@Alessandro since you're in the room and I know you know a thing or two about measures; if I have euclidean vector spaces $V, W$, an isomorphism $\varphi : V \to W$ and a measure $\mu_V$ on $V$, can I "transport" the measure on $V$ to a measure on $W$? I assume the isomorphism "changes" the measure by a factor of the determinant of $\varphi$ (notice I'm handwaving like crazy here)
I could find a class function for a finite group $G$, such that it is orthogonal to all but one of the characters, and it is unit length, and still satisfies $\chi(id)>0$, and yet is still not the character of a representation right?
Let $F_1 , F_2 : \mathbb{R^2} \rightarrow \mathbb{R}$ be functions defined by
$F_1 (x_1,x_2 )=-x_2/(x_1^2+x_2^2)$ and $F_2 (x_1,x_2 )=x_1/(x_1^2+x_2^2)$
Then
(i) $∂F_1/∂x_2=∂F_2/∂x_1.$
(ii) there exist a function $f: \mathbb{R^2}-{(0,0)}\rightarrow\mathbb{R}$ such that $∂f/∂x_1=F_1$ an...
For a successive natural numbers $n$ starting with $n=2$ how many values of $k\in\Bbb N^+$ solve the prime counting function equation? I made a sequence of the number of values of $k$ that solve the equation for each $n,$ from $n=2,$ to $n=16.$
$f(x)=\pi(x)\pi(n-x)=k.$
Here's what I tried: when...
not quite, I'm comparing $\pi(x)$ to the number of $k$ that satisfy $\pi(x)\pi(n-x)=k$ for each $n$ where $n$ is positive natural numbers starting at $2$
Let $G$ be a graph, and let $s_{G,n}$ be the number of spanning forests with $n$ components (so $s_{G,1}$ is the number of spanning trees for example and $s_{G,|G|}$ is just $1$ because it's a discrete graph)
@Alessandro fair :P I think I was being silly; I have the measure "vol" on an $n$-dim. euclidean vector space and the claim was that it doesn't depend on the choice of basis
It's handwaved, it just says "the Lebesgue measure on $\Bbb R^n$ gives us a measure $\operatorname{vol}$ on $V$"
well not hand waved
just not explicitly defined
anyway the "reason" stated was that orthogonal matrices have determinant $\pm 1$ so I just wondered if a different linear transformation would dump a factor of the determinant out front
Are there rotation matrices with determinant greater than 1? Can the matrix associated with change to polar coordinates be seen as a rotation about the origin despite the determinant not equaling 1?
yes, the 5 and 4 are touching.... hmmm. Thx for trying. I have a question I hoped to post, but wanted to offer what I've tried. And can waste far more time just trying to show my own work with math jax. I know jpg images are frowned upon.
@AlessandroCodenotti Every compactification of $X$ corresponds to a unitisation of $C_0(X)$, and im pretty sure every unitisation of $C_0(X)$ is contained a sub-algebra of $C_b(X)$
$C_b(X)$ is the multiplier algebra of $C_0(X)$ (a special kind of unitsation) and $C_b(X)$ is also equal to $C(\beta X)$ as an example
whereas the algebra generated by $C_0(X)\cup\{1\}$ in $C_b(X)$ (this is called adjoining a unit) is the same as the algebra of $C(X^*)$ wehre $X^*$ is the one-point compactification
thats the procedure of adjoining a unit to a $C^*$ algebra, which for $C_0(X)$ is the same as the subalgebra of $C_b(X)$ genereated by $C_0(X)$ and the constant function $1$
If $A$ is a $C^*$-algebra then a $C^*$-algebra $\tilde A$ is a unitsation of $A$ if $\tilde A$ is unital and $A$ is big in $\tilde A$ (I think that should be given by $A$ being a maximal ideal, but im not entirely sure)
Ok I wasn't sure about the relationship between $A$ and $\bar{A}$
I think there should some correspondence between subalgebras of $C_b(X)$ containing the constant functions and separating points (or maybe some other adjectives...) and compactifications
that makes sense, if it seperates points I expect Stone-Weierstraß to tell me it contains all of $C_0(X)$ and if it contains the constant functions its unital
So given a unitisation $A$ of $C_0(X)$ the way to get a compactification of $X$ out of it so to look at its spectrum with the weak* topology I suppose?
yes, the thing that prevents the spectrum of $C_0(X)$ from being closed is that $0$ may be a limit point of characters. But the algebra is unital then $\omega(1)=1$ for all characters and $0$ cannot be weak* limit point, hence the characters are closed in weak* + compact since they are bounded
thats a question of extending characters, the extension of a charcter to the constant functions is clear, and I'm not sure how it extends to the other "extra functions"
(cont) I think at that point it needs to be clear what exactly a unitisation is. ie what $A$ being big in $\tilde A$ means
here is what i found about the correct condition: The essentialness is captured by stipulating that every nonzero ideal in the unitization intersects 𝐴 nontrivially. This is equivalent to the condition that 𝑏𝐴={0} implies 𝑏=0
from that i can read off that extension of characters is unique if it exists
Hmm but wait, so I guess that $x\in X$ should correspond to the character of $C_0(X)$ which evaluates at $x$, right? And then we want to extend that to the unitisation?
(Do you know a good reference to read about those things? It's turning out to be surprisingly hard to find a book containing this constructions)
It's clear to me that gives an injection from $X$ into the spectrum if we're working with a subalgebra of $C_b(X)$, the fact that the image of this injection is dense is still not clear to me
there are some super advanced $C^*$algebra books and notes around but I never made much progress reading them (Garth Warner's notes is the one I made the most in, they are nice)
From Jeff's related answer:
This is implemented on http://math.stackexchange.com -- you can check it out there. It will never be on Stack Overflow, though, as it is an extremely heavy dependency.
MathJax is a client-side solution, but it uses a relatively large amount of bandwidth/time to l...
Another reason to avoid having it by default is that you can do some goofy stuff in mathjax, like having $\huge{\text{huge text}}$
so you can imagine the shenanigans that can produce, and why not being able to disable would be an issue
@Semiclassical If the matrix associated with the change of coordinates to polar doesn't represent a rotation about the origin, what does the author mean in the last sentence to the solution 1a (see attached picture)?
hello peoplz. how does the following look like ? the polynom $1-X^2$ I know that the set of all polynoms $ \mathbb{R}[X]$ is basically all $k_0 . x^0 + k_1 . x^1...………$
I am not sure I understand what $1-X^2$ represents.