Question:  A and B are matrices size $n\times n$  given $AB=BA$ and A has n eigenvalues, prove $B$ is diagonalizable.  I would have written what I tried to do, but It's really nothing worth reading..  Any help would appreciated,  Thanks.
We were asked to prove the following:  Let $ A $ be an $n \times n$ matrix with $n$ distinct real eigenvalues. If $AB=BA$, show that $B$ is diagonalizable.   It was suggested I show that an eigenvector of $A$ is also an eigenvector of $B$. I am both having trouble doing this and failing to see how I would complete the proof after. Any help would be appreciated.  Thanks
Prove that the sum of digits of $(999...9)^{3}$ (cube of integer with $n$ digits $9$) is $18n$ Someone had posted a question on this site as to what would be sum of digits of $999999999999^3$ (twelve $9s$ ) equal to?   I did some computation and found the pattern that sum of digits of  $9^3 = 18$,  $99^3 = 36$,  $999^3 = 54$ and so on. So, I had replied that sum of digits of  $999999999999^3 = 12 \cdot 18 = 216$.  Can anybody help me prove this, that sum of digits of  $(\underbrace{999\dots9}_{n\text{ times}})^3 = 18n.$ Einstein-Hilbert action as an effective field theory I have read the following statement:      "The Einstein-Hilbert action can be seen as an effective field theory: setting $g_{\mu\nu}= \eta_{\mu\nu}+h_{\mu\nu}$, one gets a free Lagrangian for the field $h_{\mu\nu}$ plus interactions."   Now, how is it possible to define all the interaction terms?  EDIT: I ask apologize if my statement was appeared unclear: I have created a  which is more precise; please don't continue to consider seriously this one. What happened to Charles Xavier after X-Men 3? I remember watching X-Men 3 and seeing Charles Xavier get disintegrated by Phoenix, however the post-credits scene and his appearance in later movies indicate that he actually survived. How is that possible? Not sure if this option ever existed, but the new QGIS version hides a lot of things somewhere else and this seems so simple that I'm kind of convinced it was possible to do this before. I have quite a large data set with ship tracks. I now want to analyse individual ship tracks (made from point layer with  Plugin). However, it is so crowded that it is hard to look at individual tracks. I don't want to save each single track as a new layer, so I wonder if it is possible to select one track in the Attribute table and display only that one on the map/hide the others? Show that $g\circ f\equiv 0$ implies $\det Df\equiv 0$, where $g(x_1,...,x_n)=x_1^5+...+x_n^5$ Let $f:\mathbb{R}^n \rightarrow \mathbb{R}^n$ be a smooth function and let $g:\mathbb{R}^n \rightarrow \mathbb{R}$ be defined by $g(x_1,...,x_n)=x_1^5+...+x_n^5$. Suppose $g\circ f\equiv 0$. Show that $\det Df\equiv 0$.  I was going to start the solution from $n=2$ and then use induction on $n$.  So, for $n=2$, $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ is a smooth function and $g:\mathbb{R}^2 \rightarrow \mathbb{R}$ where $g(x_1,x_2)=x_1^5+x_2^5$. Then $Df=\begin{pmatrix}   f_{x_1x_1} &amp; f_{x_1x_2}  \\   f_{x_2x_1} &amp; f_{x_2x_2}   \end{pmatrix}$  But now I am stuck in proving the result. Am I in the correct track? Can anyone plese give me a hint to continue?  Ok so since $g\circ f$ is $0$, the image of $f$ is the line $x_1+x_2=0$ which has no min, max or saddle.   If det$Df(x_0)&gt;0$ and $f_{x_1x_1}(x_0)&gt;0$ OR det$Df(x_0)&gt;0$ and $f_{x_1x_1}(x_0)&lt;0$, we have contradiction.   But if det$Df(x_0)&gt;0$ and $f_{x_1x_1}(x_0)=0$, then det$Df(x_0)=0$ or, det$Df(x_0)&lt;0$(Since $f_{x_1x_2}=f_{x_2x_1}$). So this is impossible.  If det$Df(x_0)&lt;0$ we have saddle and it is impossible.   So finally we conclude that det$Df(x)=0$ for all $x$. Am I correct? Does something like  exist for topological (eg. braided) circuits?   Alternatively, any ideas on how @ is getting  (or something similar)? How to get a vintage/dreamy look like this photo?   It has some yellow color cast and seems like extra blur added in post process? Accessing inherited variable from templated parent class Consider the following code:  template&lt;class T&gt; class Foo { public:   Foo() { a = 1; }  protected:   int a; };  template&lt;class T&gt; class Bar : public Foo&lt;T&gt; { public:   Bar() { b = 4; };    int Perna(int u);  protected:   int b;  };  template&lt;class T&gt; int Bar&lt;T&gt;::Perna(int u)  {    int c = Foo&lt;T&gt;::a * 4; // This works   return (a + b) * u;    // This doesn't }   g++ 3.4.6, 4.3.2 and 4.1.2 give error   test.cpp: In member function `int Bar&lt;T&gt;::Perna(int)': test.cpp:25: error: `a' was not declared in this scope   g++ 2.96 and MSVC 6, 7, 7.1, 8 and 9 accept it, as do (at least) older Intel and SGI c++ compilers.   Do the new Gnu C++ compiler obey the standard or not? If they do, what is the rationale behind that inheriting class is not able to see a protected inherited member variable?  Also, if there's  int A() { return a; }    in Foo, I get error  test.cpp:25: error: there are no arguments to A that depend on a template parameter, so a declaration of A must be available test.cpp:25: error: (if you use -fpermissiveâ, G++ will accept your code, but allowing the use of an undeclared name is deprecated)   when I try to use it in a member function of Bar. I find that curious as well: Bar inherits Foo, so I think it's obvious that A() in scope of Bar is Foo::A(). The Monty Hall problem I was watching the movie 21 yesterday, and in the first 15 minutes or so the main character is in a classroom, being asked a "trick" question (in the sense that the teacher believes that he'll get the wrong answer) which revolves around theoretical probability.  The question goes a little something like this (I'm paraphrasing, but the numbers are all exact):  You're on a game show, and you're given three doors. Behind one of the doors is a brand new car, behind the other two are donkeys. With each door you have a $1/3$ chance of winning. Which door would you pick?  The character picks A, as the odds are all equally in his favor.  The teacher then opens door C, revealing a donkey to be behind there, and asks him if he would like to change his choice. At this point he also explains that most people change their choices out of fear; paranoia; emotion and such.  The character does change his answer to B, but because (according to the movie), the odds are now in favor of door B with a $1/3$ chance of winning if door A is picked and $2/3$ if door B is picked.  What I don't understand is how removing the final door increases the odds of winning if door B is picked only. Surely the split should be 50/50 now, as removal of the final door tells you nothing about the first two?  I assume that I'm wrong; as I'd really like to think that they wouldn't make a movie that's so mathematically incorrect, but I just can't seem to understand why this is the case.  So, if anyone could tell me whether I'm right; or if not explain why, I would be extremely grateful. is there a way to request a new tag be created? Possible Duplicate:          I do not have sufficient reputation on any site to create new tags, but I do on occasion find myself in need of a tag that does not exists.  Is there a feature that allows for a person with sub-1500 reputation to request a new tag? I'm studying for actuarial exams, but I always pick up mathematics books because I like to challenge myself and try to learn new branches.  Recently I've bought Topology by D. Kahn and am finding it difficult.  Here is a problem that I think I'm am answering sufficiently but any help would be great if I am off.  If $d$ is a metric on a set $S$, show that  $$d_1(x,y)=\frac{d(x,y)}{1+d(x,y)}$$ is a metric on $S$.  The conditions for being a metric are $d(X,Y)\ge{0}, d(X,Y)=0$ iff $X=Y$, $d(X,Y)=d(Y,X)$, and $d(X,Y)\le{d(X,Z)+d(Z,Y)}$.  Thus, we simply go axiom by axiom.  1)  Since both $d(x,y)\ge{0}$ and $1+d(x,y)\ge{0},$ it is clear that $d_1(x,y)\ge{0}$.  (Is this a sufficient analysis?) 2) $d_1(x,x)=\frac{d(x,x)}{1+d(x,x)}=\frac{0}{1+0}=0$. 3) $d_1(x,y)=\frac{d(x,y)}{1+d(x,y)}=\frac{d(y,x)}{1+d(y,x)}=d_1(y,x).$ 4) $d_1(x,y)=\frac{d(x,y)}{1+d(x,y)}\le{\frac{d(x,z)+d(z,y)}{1+d(x,z)+d(z,y)}}=\frac{d(x,z)}{1+d(x,z)+d(z,y)}+\frac{d(z,y)}{1+d(x,z)+d(z,y)}\lt\frac{d(x,z)}{1+d(x,z)}+\frac{d(z,y)}{1+d(z,y)}=d_1(x,z)+d_1(z,y).$  However, #4 is strictly less, not less than or equal to, according to my analysis, so where did I go wrong? Why are effective pixels greater than the actual resolution? compares Canon EOS 550D and Canon EOS 500D cameras and mentions     18.7 million effective pixels   for 550D. However the best resolution possible using this camera is  5184 * 3456 = 17915904 ~ 17.9 million pixels   What are effective pixels, and why is that number greater than 17.9 million in this case? Prove that if $ f\mathrm{(}z\mathrm{)} $ is analytic at the infinity  ,then :  $ {\lim}_{{z}\mathrm{\rightarrow}\mathrm{\infty}}{f}\prime{\mathrm{(}}{z}{\mathrm{)}}\mathrm{{=}}{0} $ How do I URL encode a string Recently I have installed an older version of Ubuntu on my old machine. Whenever I try to install any software, I get an error saying it couldn't be found:  $ sudo apt-get install vlc Reading package lists... Done                Building dependency tree        Reading state information... Done     E: Couldn't find package vlc Black screen when creating movie I've rendered about 300 frames for a short clip. The frames look fine, but when I compile into a movie all I get is a black screen. I haven't been able to find anything to solve this problem. The individual frames are well lit, capture what I want them to, look great. Just creating the AVI or MPEG results in a clip that has nothing but black. Could humans choose to establish fibre digesting colonies in our guts? Why do some AC adapters and power supplies generate a whining noise, and what can I do about it? How to use aes-encryption in PostgreSQL? Are there fewer positive integers than all integers? In our 6th grade math class we got introduced to the concept of integers. With all the talk about positive and negative, it got me wondering. Is the amount of elements in $\mathbb{Z^+}$ less than the amount of elements in $\mathbb{Z}$?  Here is how I thought of it. If we have $\mathbb{Z^+}$ and add one element to the "back" of it ($\mathbb{Z}^{\geq-1}$) there are certainly more elements in that new set, so there must be more elements in $\mathbb{Z}$ than in $\mathbb{Z}^+$ but on the other side if we try to express the amount of elements in them "numerically" (In a loose sense of the word) they both have $\infty$ elements.    So what is the right answer? Is there even one?  (Please note I am a bit of a layman when it comes to mathematics)  Edit: About the possible duplicate, I am not looking for a bijection between the two sets, I am looking for if they even have the same number of elements and why a bijection shows that they do/don't