User dan fodor - MathOverflow

Answer by Dan Fodor for Intersection of an uncountable number of sets.

2012-05-07T16:48:14Z

Not that I know of necessarily , other than calculating it . To see how this can be the case , consider the complement of $E_i$ . It has measure 0 .Now the union of the complements (or the complement of $\cap_{i\in \mathcal{I}} E_i$) can be a very general set considering we have uncountable $i$'s . In particular , for $[0,1]$ with the usual topology , it can be an any subset . To see this , let $S$ be a subset of $[0,1]$ , such that the complement of $S$ , $C_S$ is uncountable , and let $b :[0,1] \rightarrow C_S $ be a bijection . Define $E_i$ , $i \in [0,1] $ , to be $[0,1] - { b(i) } $ . Now $\cap_{i\in [0,1] } E_i$ is $S$ . Notice that if $\Omega$ is countable then you can not obtain non-measurable sets by intersecting $E_i$ . The general condition for this to happen is that the union of an uncountable ($R$ , the cardinality of the continuum ) number of the subsets with measure $0$ has measure $0$ .If this doesn't happen , there will be $E_i$'s such that the intersection is a non-measurable set . In general , for a triplet $(\Omega, \mathcal{F},\mathbb{P})$ , if there exists an uncountable subset $J \subseteq F $ , such that the elements of $J$ are pairwise disjoint , $\cup J= \Omega $ and the probability space $(J,M(J) ,\mathbb{P})$ (where $M(J)$ is the set of measurable subsets of the $powerset$ of $J$ )is isomorphic to $([0,1],M([0,1]) , \mathbb{P} _{ [ 0,1 ] } )$ , then there exist $E_i$ for $(\Omega, \mathcal{F},\mathbb{P})$ such that the union is non-measurable . In general , the condition for non-masurable unions of such sets not to exist for $(\Omega, \mathcal{F},\mathbb{P})$ is that $P(E_s)$ = 1 , where $E_s$ is the set of elementary events with non-zero probability .

Answer by Dan Fodor for Eigenvectors of a certain big upper triangular matrix

2012-05-07T11:32:06Z

Might be a wild intuition , I'd say the eigenvalues are the entries of the first row , and that the eigenvector coresponding to the $nth$ eigenvalue ,$k$ is made by adjoining a column of zeroes to the eigenvector coresponding to the eigenvector coresponding to the same eigenvalue for the first $n*n$ minor of the matrix .

Example :

for eigenvaue $1$ we take the matrix $ \begin{pmatrix} 1 \end{pmatrix}$ ,the eigenvetor corresponding to $1 $ is $\begin{pmatrix} 1 \end{pmatrix}$ , so we obtain $\begin{pmatrix} 1 \cr 0 \cr 0 \cr 0 \cr 0 \cr \vdots \end{pmatrix} $ as the first eigevector .

for eigenvaue $1/2$ we take the matrix $ \begin{pmatrix} 1& 1/2 \cr 0 & 1/2\end{pmatrix}$ ,the eigenvetor corresponding to $1/2$ is $\begin{pmatrix} 1 \cr -1\end{pmatrix}$ , so we obtain $\begin{pmatrix} 1 \cr -1 \cr 0 \cr 0 \cr 0 \cr \vdots \end{pmatrix} $ as the second eigevector .

The eigenvector coresponding to $1/8$ for $ \begin{pmatrix} 1& 1/2 & 1/8 \cr 0 & 1/2 & 1/4 \cr 0 & 0 & 1/8 \end{pmatrix}$ is $\begin{pmatrix} 5 \cr -14 \cr 21 \cr \end{pmatrix}$, you get the ideea .Also , the eigenvectors span the entire space , ie if a possibly infinite (but convergent) sum of eigenvectors is $\vec 0$ then the coefficients of those vectors are $0$ .

Here is an explicit formula for the eigenvectors :first select $M_n$ , the $n*n$ truncation of the matrix and calculate $M_n - I*v_n $ , the nt'h eigenvalue . Example : for n=3 , we obtain \begin{pmatrix} 7/8 & 1/2 & 1/8 \cr 0 & 3/8 & 1/4 \cr 0 & 0 & 0 \end{pmatrix} . Now let $S$ be the $(n-1)*(n-1)$ truncation of that , ie \begin{pmatrix} 7/8 & 1/2 \cr 0 & 3/8 & \end{pmatrix} Calculate $S^{-1}$ = \begin{pmatrix} 8/7 & -32/21 \cr 0 & 8/3 & \end{pmatrix} , now multiply $S^{-1}$ with the truncation of the last column of $M_n$ , \begin{pmatrix} 1/8 \cr 1/4 \end{pmatrix} You obtain \begin{pmatrix} -5/21 \cr 2/3 \cr \end{pmatrix} . Concatenating $-1$ to that , you obtain \begin{pmatrix} -5/21 \cr 2/3 \cr -1 \end{pmatrix} , the third eigenvector ,or the nt'h eigenvector in the general case .

Answer by Dan Fodor for Cramer's rule for eigenvectors

2012-05-07T09:45:00Z

http://en.wikipedia.org/wiki/Jordan_normal_form

http://www.wolframalpha.com/input/?i=jordan+normal+form+calculator

Let $J$ be the Jordan normal form of a matrix $A$ (that is , $A=PJP^{-1}$ ) . Then the $V_{An}$ eigenvectors of $A$ can be written as $PV_{Jn}$ , where $V_{Jn}$ are the eigenvectors of $J$ . Although calculating $J$ and $P$ can be rather complicated , calculating the eigenvectors of $J$ is trivial .

Also recomend reading this

http://www.cs.berkeley.edu/~wkahan/MathH110/DownDets.pdf

At least to me it seems like the best tutorial on linear algebra out there .(edit:I did not intend for this to sound condescending or anything , I recommend that paper to anyone who does linear algebra)

Answer by Dan Fodor for Vector "product" diagonalization

2012-05-06T15:37:53Z

Assuming you want $(A+B) ⊙ V = A⊙V + B⊙V$ (if not things get a lot more complicated , you might as well ignore the vector structure having no way to link the addition and product) , the product , being bi-linear , can be described as a tensor (essentially n square matrices $M_p$, n*n each , such that if $V_1⊙V_2 =C $ , then $< V_1 | M_k | V2 > = C_k$ , the k'th component of $C$ . ) . I like to imagine them as a cube .

A⊙V = V⊙A means that each of these matrices will be symmetric . The second constraint is $v\odot v_0=v$ . Linear algebra ensures we can chose a convenient form for $v_0$ , so let $v_0= (1,0,0 ,0...) $ . We have $< v_0 | M_k | B > =B_k $ , the k'th component of $B$ . This constrains the first column of $M_k$ to be (0,0..,0,1,0..0) where the k'th component is 1. By symmetry , the first row is also constrained . So the only freedom we have is choosing the parameters of $n \hspace{5 mm} (n-1)*(n-1)$ symmetric matrices , in total $n*n*(n-1)$ parameters for a fixed $v_0$ . Since $v_0$ can be chosen arbitrarily from a space of n dimentions , we have $(n + n*n*(n+1))$ degrees of freedom for the product operator . Hope that helps .

Help with Kelvin–Stokes theorem

2012-05-06T13:48:42Z

Consider f:R^3 -> R^3 f(x,y,z) = (-y,x,0) . Consider a circle in the x0y plane with the radius 1 , and center in (0,0,0) . The curl of f is (0,0,2) everywhere , so integral of (curl f) over the surface of the circle is (0,0,2pi ) . The integral of f over the boundary of the circle is (0,0,0) . According to the theorem , they should be equal , so I assume I'm doing something horribly wrong in understanding or calculating integrals .

Hall polynomial when the subgroup is cyclic?

2012-04-26T07:45:50Z

Does anyone know the formula for a Hall polynomial $g_{u,v}^{\lambda}(p)$ when $v$ is the type of cyclic subgroup (ie. $v=(v_{1})$ ) . http://en.wikipedia.org/wiki/Hall_algebra

I was hoping this particular case would be simple enough to describe .