If $A$ is a  square matrix of size $n$  with complex entries and is nilpotent , then I can show that all the eigenvalues of $A^k$ , for any $k$ , is $0$ , so $Tr(A^k)=0 , \forall k \ge 1$ . Now conversely if $A $ is a square matrix of size $n$ with complex entries such that $Tr(A^k)=0 , \forall k \ge 1$ , then is it true that $A$ is nilpotent ?
Let $A$ be an $n\times n$ complex nilpotent matrix. Then we know that because all eigenvalues of $A$ must be $0$, it follows that $\text{tr}(A^n)=0$ for all positive integers $n$.  What I would like to show is the converse, that is,      if $\text{tr}(A^n)=0$ for all positive integers $n$, then $A$ is nilpotent.   I tried to show that $0$ must be an eigenvalue of $A$, then try to show that all other eigenvalues must be equal to 0. However, I am stuck at the point where I need to show that $\det(A)=0$.  May I know of the approach to show that $A$ is nilpotent?
Salesforce query on profile not working on user language change I have a query on profile like  profilename = [Select name from profile where id=:userinfo.getUserprofileid ].name  Now I have a check in  VF page  like  profilename  == 'System Administrator' . this works fine if the logged in user language is english. If the user language is different then profile name will be  'System admin' in that language . So how to prevent this beahviour?   Also  is there any particular field in profile which I can use  to identify the profile is system admin? so that I can have check on that field instead of checking for name which is causing the problem, for example I think modifyall filed is only checked for system admin. Restoring Windows XP PATH variable I installed an application that screwed my PATH variable. Now I cannot run anything from the command prompt. Is there any way I can restore PATH variable? How do you prove that $\sum\limits_{n=0}^{\infty}{2n \choose n}x^n=(1-4x)^{-1/2}$? 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According to tests (Galileo from Pro7, AFAIR) it has also the best coverage. However, calls to other countries are very expensive.  For to second criteria, Lebara and Mobilka are interesting. The calls are very cheap, but there are problems with signal strength in many locations, where T-Mobile has no problem. The voice quality, at least in Lebara, is also limited. They both have mobile internet with unlimited transfer, but after reaching the high speed limit, the transfer is reduced to 64 kbps, which is too slow.  I've found out that O2 has foreign calls at moderate prices, but I haven't yet checked for details. There are so many other options, making it is hard to compare them all without spending the whole weekend on it. So, if someone has found a better option than having 2 SIM cards and constantly switching them, this answer would be greatly appreciated. I want to learn more about hopf algebras but I am having trouble finding a down to earth introduction to the subject with lots of motivation and examples. My algebra knowledge ranges from Dummit and Foote, Atiyah-Macdonald commutative algebra, and Hartshorne level algebraic geometry.  What are good places to look at for a hopf algebra beginner? My main motivation is for studying groupoid schemes from an algebraic point of view. When will my phone get the Android 2.3 update (Gingerbread)? Can I have multiple primary keys in a single table? Can I have multiple primary keys in a single table? I have ubuntu mate. I have recently changed my password using the passwd command. I know now that I have made a mistake because I also have my home directory encrypted with the previous password.  Consequently after enter my new password at login screen it return me back to the login screen.  Can someone please help me?  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A topological space is extremally disconnected iff every two disjoint open sets have disjoint closures Show that for any topological space $X$ the following are equivalent:         $X$ is extremally disconnected   Every two disjoint open sets in $X$ have disjoint closures.      My attempt at a solution  Assume every two disjoint open sets in $X$ have disjoint closures. Let A open in $X$, $A$ and $X \setminus \overline{A} $  disjoint open set $\overline{A} \cap (\overline{X \setminus \overline{A}} )= \emptyset$,  $(\overline{X \setminus \overline{A}} ) \subset X \setminus \overline{A}$, which implies that $X \setminus \overline{A}$ is closed in $X$. Thus $\overline{A}$ is open in $X$. So we obtain $X$ is extremally disconnected.  Otherwise Let $X$ is extremally disconnected. Then the closure of every open set is open.  So $X$ is completely separated. Let $A$, $B$ two disjoint open sets.  Then there exist a continuous function $g:X \rightarrow [0,1]$ such that  $g(A)=0$, $g(B)=1$  How can I continue? Any group of index 2 is normal I'm trying to understand the proof that any group of index $2$ is normal.   Let $G$ be a group and $H$ a subgroup of index $2$.  I understand that if $x\in H$ then $xH=H=Hx$ since $H$ is a subgroup. However, I don't understand why if $x\in G\setminus H$ then $xH=G\setminus H.$  I know there are other proofs of this on this site but none of them specifically address this question. Let $f$ be entire and suppose that $\text{Im} f\ge0$. Show that $\text{Im} f$ is constant. How do you show that if $f$ is entire and if $\text{Im} f\ge0$, then $\text{Im} f$ is constant? Is this even true in the first place?