hm, the minimal polynomial of a rank $2$ $n \times n$ matrix A can have one of two forms -- $m_A(\lambda) = \lambda(\lambda - \lambda_1)(\lambda - \lambda_2)$ or $m_A(\lambda) = \lambda(\lambda - \lambda_1)$? Correct?
@Astyx I think if the eigenvalues are the same, then the degree of the jordan block is $1$. Alternatively, the largest ($1 \times 1$) jordan block that corresponds to $\lambda_1$ is $1$.. that's why I omitted that term. I'm not sure if the $\lambda$ has any power, and I think it shouldn't..
This is slightly weird. So I applied for a job as an Android developer, and they contacted me asking me to do a small programming task to test my skills. The task is based on Firebase, and when I went to learn how to use that, I saw a weekend course on it, which basically develops the exact app this company has asked me to do (except for a few features which will be basically trivial to add on).
@MikeMiller try going into machine learning or language theory, or anything else that's not traditional programming. else youre dooming yourself to a life amongst idiots
and for that folk there's an inverse relation between how much they know, and how positively they think of their skill
well yeah but I claim in general there are $3$ distnict forms (which really can be collapsed into one..) $(1)\ m_A(t) = t(t-\lambda_1)(t-\lambda_2)$ $(2)\ m_A(t) = t(t-\lambda_1)$ $(1)\ m_A(t) = t(t-\lambda_1)^2$
@MikeMiller Maybe there could be some set of homeomorphisms on a infinite type surface(say a surface with infinite genus and set of ends a cantor set) by doing some "general Nielson realization"/choosing a nice hyperbolic metric where it has lots of finite order subgroups...?
I guess just for the $t$ term, I am wondering how much contribution to the power do I get from the singular part of the matrix alone. So the power on the characteristic polynomial is going to be $n-2$, but I think the power on the minimal polynomial is going to have to be $1$, once again, contributed to strictly from the singular part of the matrix
and I think I have a proof of that fact..
where, by the way, your matrix falls under category $(2)$, say
Sure, there is a sort of standard/simple construction for surfaces to show that for any finite subgroup there is a surface of genus g containing that subgroup in the homeomorphism group(and a way to make it isometries), it seems like with some modification maybe it could work for infinite type surfaces with enough symmetry in the ends. Compact case would be interesting though(and I would be surprised if there where such a manifold)
The rough idea is to thicken the cayley graph into a surface(i think you need be a little careful, maybe vertices correspond to tori), then you have the group acting on it. It seems possible to generalize this construction to to the "tree of infinite genus" by having the tree parts attached in some way that preserves the action.
Hi all. I need to find a map between the group $\mathbb R^n$ and the group $Diag(n,\mathbb R^+)$ which consists of $n\times n$ diagonal matrices having positive real diagonal entries, so that I can check if it is an isomorphism. Any suggestion?
@LeylaAlkan ok I figured it out. I would point out that the first one is a group under addition and the second one is a group under multiplication. Hint: can you do it for $n=1$?
@MikeMiller ok so today we learnt about degree of a map between sufficiently nice spaces (I have already seen this concept before in homology albeit only for a sphere), and they proved that the antipodal map $A : S^n \to S^n$ has degree $(-1)^{n+1}$, and I don't quite like their proof
So I would probably just take the volume form $\eta := \iota^\ast (\sum_{i=0}^n x_i \mathrm dx^1 \land \cdots \land \widehat{\mathrm dx^i} \land \cdots \land \mathrm dx^n)$
it should be $\eta := \iota^\ast (\sum_{i=0}^n x_i \mathrm dx^0 \land \cdots \land \widehat{\mathrm dx^i} \land \cdots \land \mathrm dx^n)$... then $A^\ast$ reverses the $x_i$ as well as every $\mathrm dx^j$, so there are $n+1$ terms being reversed, so $A^\ast \eta = (-1)^{n+1} \eta$
@LeylaAlkan correct; can you now generalize for $n$?
Well, each entry in the resultant matrix would be an infinite sum. Like for matrices $A$ and $B$, $AB$ would have an $i$-th row and $j$-th column of $\sum_{k=1}^{\infty} a_{ik}b_{kj}$
(Assuming I've got the symbols in the right places there. It looks correct at a glance.)
Yah the closed case is interesting, like you I would be surprised if there was such a manifold. Did you have a proof for the closed case that it can't be of diffeomorphisms @MikeMiller
The idea I had was to consider the action of $S_d$. For some reason I thought I would be able to control the orbit types that arise as $d \to \infty$ (this is the flawed part). If there is some $S_d$ orbit with cardinality $d$, then this gives an injective homomorphism $S_{d-1} \to O(n)$ for $n$ the dimension of the manifold by averaging out a Riemannian metric, by looking at one of the fixed points
But the finite subgroups of $O(n)$ are classified and in particular, $S_d$ does not arise for all $d$
I am confident I saw this in an Andy Putman answer on MO but couldn't find the post
Ultimately I don't see how to control the possible orbit types. The idea should be that you can't have that many free quotients, which forces certain stabilizer types - but I couldn't see how to make this precise
@LeakyNun I showed that $f$ is a homomorphism and it is injective. Now trying to show that it is surjective, and I started by saying take any $y= \begin{pmatrix} \exp x_1 & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & \exp x_n\\ \end{pmatrix} \in Diag(n,\mathbb R^+)$ . But then I don't know how to write $x=(x_1,\cdots , x_n)$ in terms of $y$
@MikeMiller I am not really familiar with transformation groups and Smith theory, but the idea seems to be that if you have an elementary abelian $p$-group rank $k$, and it acts effectively, you can inductively apply some of the Smith-type theory arguments to subgroups(rank k-1 for example, to rank 1) to relate the dimension of the manifold to the dimension of fixed point set and rank of the group
When people say $GL_\infty$ they usually mean $\bigcup_n GL_n$, in which case these are matrices which are supported in the top-left $n \times n$ block (outside of which they are 0, except on the diagonal, where they are 1).
hm, question. If $A$ is $n \times n$ over $\mathbb{C}$, and $rank(A) = 2$, then under the two linearly independent eigenvectors as a basis (and any other vectors not in the eigenspace(s) completing the collection to a basis of $\mathbb{C}^n$), $A \sim \begin{bmatrix} X & 0 \\ 0 & 0 \end{bmatrix}$; where $X$ is $2 \times 2$ is the nonsingular part of $A$?
it is not necessarily true that there are two linearly independent eigenvectors (think of the standard Jordan block [[1, 1], [0, 1]]), but yes, one can always put your matrix in such a form after a change of basis - for instance, by putting it in Jordan normal form
