Let $\textbf{A} \in M_{n\times n}(\mathbb{C})$ be a Matrix and $\textbf{A}^m=\textbf{I}$ where $m\neq \infty$. Show (with Jordan normal form) that $\textbf{A}$ is diagonalizable. I have no clue how to do that...
Let $A$ be a complex matrix such that $A^n = I$, show that $A$ is diagonalisable.  How do I do this?    I would hazard a guess that since $A^n = I$ that means that $A^n$ is obvious diagonal, and I suppose that a diagonal matrix can only be formed by the square of another diagonal matrix, but I am not sure if that is the case. Also I think this neglects the complex cases.  Is there a Jordan block argument that I should use?
The only 1-manifolds are $\mathbb R$ and $S^1$ I recall having heard somewhere that the only 1-manifolds (second countable, Hausdorff, connected spaces locally homeomorphic to $\mathbb R$) are $\mathbb R$ and $S^1$. Is this true? If so, is there a reasonably elementary proof? Prove that the multiplicative inverse of $a$ modulo $m$ exists if and only if $a$ and $m$ are coprime. Prove that the multiplicative inverse of $a$ modulo $m$ exists if and only if $a$ and $m$ are coprime.  Can someone help me with this? Drush hangs when trying to bootstrap Drupal What is the difference between 'sudo' and 'su -c' Hint on how to prove the identity $\sum_{0 \leq k \leq n } \binom{k}{m} = \binom{n + 1}{m + 1}$ Is the use of a hyphen between "non" and an adjective strictly necessary? Do I need to put a "-" between "non" and an adjective? As an example in physics we say "a non isolated photon", "non tight photon"... The context is very formal (paper publications and similar). Is there a general rule? Are there some differences between countries? Changing Unsubscribe reason through SSJS I noticed that you can unsubscribe a user , is it possible to use SSJS to change the unsubscribe reason?  I notice that doing it through ? why is that?  Thanks Where are the NetworkManager logs? A previous poster on the regular forums (  ) said /var/log/daemon.log, but I don't seem to have such a file. Do I need to enable logging? Forward local port or socket file to remote socket file Quick question - I run two linux boxes, one my own desktop and the other my VPS. For security reasons on the VPS end I opted for socket connections to MySQL (/var/run/mysqld/mysql.sock). I know I can tunnel like this: ssh -L 3307:127.0.0.1:3306 user@site.com if I set up the remote sql server to listen on some port, but what I want to know is can I do something like: ssh -L /path/to/myremotesqlserver.sock:/var/run/mysqld/mysql.sock thereby tunnelling two sockets, as opposed to two ports?  A perfectly acceptable solution would also be to forward a local port to the remote socket file, but where possible I'm trying not to have tcp servers running on the remote box.  (and yes, I know tcp would be easier). I wish to have long and short forms of command line options invoked using my shell script.  I know that getopts can be used, but like in Perl, I have not been able to do the same with shell.  Any ideas on how this can be done, so that I can use options like:  ./shell.sh --copyfile abc.pl /tmp/ ./shell.sh -c abc.pl /tmp/   In the above, both the commands mean the same thing to my shell, but using getopts, I have not been able to implement these? Application launchers in '/usr/share/applications' not working in Nautilus We had a couple of examples where we were supposed to calculate the  (CT), but we never actually talked about a condition that decides whether a transformation is a canonical one or not.   Let me give you an example: We had the transformation: $$P=q \cdot \cot(p), \qquad Q=\ln \left(\frac{\sin(p)}{q}\right).$$  How do I see whether this transformation is a canonical one or not?   You don't have to carry out the full calculation, but maybe you can give me a hint what I need to show here? Every infinite subset of a countable set is countable. Here is the proof I tried to weave while trying to prove this theorem:     Theorem. Every infinite subset of a countable set is countable.      Proof. Let $A$ be a countable set and $E\subset A$ be infinite. Then $A\thicksim\mathbb{N}$. This implies that there is a sequence $\left\{x_{n}\right\}_{n\in\mathbb{N}}$ where $x_{n}\in A$. Construct a sequence $\left\{n_{k}\right\}_{k\in\mathbb{N}}$ where $x_{n_{k}}\in A$, $n_{1}$ is the smallest positive integer such that $x_{n_{1}}\in A$, and $n_{k+1}&gt;n_{k}$. Then      $$ E=\bigcup_{k=1}^{\infty}x_{n_{k}}, $$      and $E$ is countable, because $E\thicksim\mathbb{N}$. $\blacksquare$   Is it convincing? I want to install selenium webdriver in my Ubuntu 16.04 system for python The feeling of always wanting newer possessions? I am looking for a word that describes the emotion or feeling of an intrinsic want of a new and better possession. Covetous or envious seem a bit too focused on the other other person's possession of the desired object whereas I am looking for something that is defined by the internal thoughts we have when seeing an advertisement for the latest iPhone or life-changing new service.    but I am wondering if there is a word with less emphasis on the second entity? Otherwise, would I covet new things from Amazon or Apple? Can I be envious of corporations?  For example, I just bought a new car and now, looking at my slightly out of date television, I feel the need to upgrade my T.V. with a newer, shiny one from Amazon. Would I be coveting that T.V. which Amazon sells?     I saw the ad for the latest iPhone, the iPhone 3000, and feel that my current phone is inadequate. I am a(n) ______ person, since I always desire something better than what I currently possess. If I was a caveman with a sharpened stick, I'd want your sharper, pointier stick.   I feel like there is probably some niche German word for exactly this, but I'm unsure it exists in English. Thanks! Best way to remove an event handler in jQuery? I have an input type="image". This acts like the cell notes in Microsoft Excel. If someone enters a number into the text box that this input-image is paired with, I setup an event handler for the input-image. Then when the user clicks the image, they get a little popup to add some notes to the data.  My problem is that when a user enters a zero into the text box, I need to disable the input-image's event handler. I have tried the following, but to no avail.  $('#myimage').click(function { return false; }); As the title say, I want to prove that  $C_c^{\infty}(\mathbb R^n)$ is dense in $W^{k,p}(\mathbb R^n)$  i.e.  $\displaystyle{ W^{k,p} (\mathbb R^n) = W_0^{k,p}(\mathbb R^n)  \quad (\star)}$. In a book I found that in order to prove it, we need the following claim:  Claim: Let $\displaystyle{\zeta \in C^{\infty} ([0, \infty)) }$ such that $\zeta (t)  =1 $ for $ t \leq 1 $ and $\displaystyle{ \zeta (t) =0 }$ for $ t \geq 2$. For $R&gt;0$ and  $x \in \mathbb R^n$ define $\displaystyle{ \eta_R (x) = \zeta \left( \frac{|x|}{R} \right)}$. If $ \displaystyle{ u \in W^{k,p} ( \mathbb R^n) }$ with $1 \leq p &lt; \infty $ then, $\displaystyle{ \eta_R u \to u }$ in $W^{k,p}(\mathbb R^n) $ as $ R \to \infty$.  If I prove the claim then the $(\star)$ follows using mollifiers. But unfortuently I have stuck, with the proof of the claim.  Any ideas? EEA Family Permit: No appointments are currently available for this location My Austrian husband and I will be travelling to the UK in April 2019 so I've applied for the EEA (just like before). I can't really wait with my application till January (3 months until the trip) because we'll be abroad in February and I need my passport back for that.  Upon filling out the application I saw some appointments on Dec, 27 that I couldn't book because we'll be on vacation in another country that week.  For the last three days, there have been no appointments available at all, I keep getting the message "No appointments are currently available for this location" (that's for Vienna).  I have contacted the visa4uk support per Email. They told me to use the approved Internet browser and closed the ticket. I replied to their Email asking for further help since my issue wasn't of technical nature (I use an approved browser), but they don't seem to reply anymore.  So how can the situation with unavailable appointments be resolved?  13.12.2018 Edit after comment: My national residence permit is not the Article 10 card. Answer appears automatically converted as a comment From official :     Due to a head injury, Cyclops is unable to shut off his optic blasts   at will and must therefore wear ruby quartz lenses to block the beams.   From the same page:     Cyclops possesses the mutant ability to project a beam of heatless   ruby-colored concussive force from his eyes, which act as   inter-dimensional apertures between this universe and another.   When his eyes are always blocked by ruby-colored beam of concussive force, how does he see?