Is it possible to customize the side panel, when you press N in the 3d view, in 2.92?
Too big addon list, what should I do? How expand names and keep it all?
A linear operator commuting with all such operators is a scalar multiple of the identity. The question is from Axler's "Linear Algebra Done Right", which I'm using for self-study.  We are given a linear operator $T$ over a finite dimensional vector space $V$. We have to show that $T$ is a scalar multiple of the identity iff $\forall S \in {\cal L}(V), TS = ST$. Here, ${\cal L}(V)$ denotes the set of all linear operators over $V$.  One direction is easy to prove. If $T$ is a scalar multiple of the identity, then there exists a scalar $a$ such that $Tv = av$, $\forall v \in V$. Hence, given an arbitrary vector $w$, $$TS(w) = T(Sw) = a(Sw) = S(aw) = S(Tw) = ST(w)$$ where the third equality is possible because $S$ is a linear operator. Then, it follows that $TS = ST$, as required.  I am, however, at a loss as to how to tackle the other direction. I thought that a proof by contradiction,  ultimately constructing a linear operator $S$ for which $TS \neq ST$, might be the way to go, but haven't made much progress.   Thanks in advance! How do they compute the Gold and Elixir that can be looted? How do they compute the Gold and Elixir that can be looted from a base? Is it 100% of the remaining? or the total (including the uncollected from mine)? scope of clause within complex sentence If I said, &quot;I will run, not if it is snowing, if I my shoes are comfortable.&quot; Does the &quot;if my shoes are comfortable&quot; modify the &quot;I will run&quot; clause? I agree the sentence would be clearer with the addition of &quot;and&quot; in place of the comma. However, without the &quot;and&quot; is it still clear that the &quot;if my shoes are comfortable&quot; clause modifies the &quot;I will run&quot; clause? CloudPagesURL() value not showing in HTML I am trying to pass the email value from a data extension to HTML. I am using the CloudPagesURL() in my email in attempt to grab the subscriber's email.  Here is my code syntax that I am attempting to use:  Email Code:  &lt;p&gt;&lt;a title="MyPage" href="%%=RedirectTo(CloudPagesURL(505,'Email Address',@EmailAddress))=%%" alias="ThisAlias" conversion="false"&gt;Click to unsubscribe page 505&lt;/a&gt;&lt;/p&gt;   CloudPage AMPScript Code:  %%[  VAR @emailAddress  SET @emailAddress = RequestParameter("emailaddr")  ]%%   Cloudpage HTML Code:  &lt;td width='155' valign='bottom'&gt;&lt;input type='text' name='Name1' value='%%=v(@emailAddress)=%%' id='Name1' /&gt;&lt;/td&gt;   Any reason as to why this doesn't work? My syntax might just be wrong or I might be using the CloudPagesURL() function incorrectly. Unable to boot into Windows after installing Ubuntu, how to fix? I installed Windows on my computer, followed by an installation of Ubuntu. However, now I'm unable to boot into my Windows install.  What can I do to fix this? References in externalized pgfplots I was writing on an document with a lot of pgfplots in it. Because of the memory limitation in LaTeX and in order to save some compiling time I tried out the externalize command. With most of the plots it perfectly worked out. But when I use the \ref command, namely \ref{&lt;legendname&gt;} the legend gets lost and instead there is a ??.  I think this question in a way is similar to  and . But no answer to these questions could solve my problem. Even the use of xr and zref packages showed no effect.  Am I just missing something?    MWE of the maindocument:  \documentclass{article} \usepackage{zref-user} \usepackage{xr-hyper} \usepackage{pgfplots} \usetikzlibrary{pgfplots.groupplots} \usepgfplotslibrary{external}  \tikzexternalize \externaldocument{part1}  \begin{document} \input{part1} \end{document}     With part1.tex looking like this:  \begin{tikzpicture}     \begin{groupplot}[       group style= {  columns=3,xlabels at=edge bottom,                 y descriptions at=edge left,                 horizontal sep=2cm,group name=plots},     width=0.4\textwidth,     xtick=data,     legend columns=3]     \nextgroupplot[xlabel=site 1,legend to name=grouplegend]     \addplot coordinates {(1,20)(2,23)(3,24.5)};     \addlegendentry {AaAaA,BbBbB,CcCcC}%     \nextgroupplot[xlabel=site 2]     \addplot coordinates {(1,14)(2,16)(3,17)};     \nextgroupplot[xlabel=site 3]     \addplot coordinates {(1,10)(2,11.5)(3,12.25)};     \end{groupplot}   % Legend     \node at (plots c2r1.north) [anchor=south] {\ref{grouplegend}}; \end{tikzpicture}     I'm writing with MikTeX 2.9 and the latest updates of the packages used. For compiling I used the command pdflatex -enable-write18 &lt;file name&gt;.  I hope the MWE is short enough ;) and some knows how to solve this problem. Ubuntu files worthless on Windows? I found the .bashrc file and I want to know the purpose/function of it. Also how and when is it used? How to show zero before decimal point in bc? echo "scale=3;1/8" | bc   shows .125 on the screen. How to show 0.125 if the output result is less than one? Proof using Prüfer encoding The question is as below:      Show that each vertex $v$ of a tree $T$ appears $deg(v)-1$ times in the Prüfer encoding of $T$, where $deg(v)$ is the degree of $v$ in $T$.   For this, I think there are 3 scenarios a leaf should be considered for: (1) when it is a leaf as part of a tree; (2) a leaf that has been discarded; (3) the remaining edge. I feel like this will result in a short proof which is only two (or so) statements long, but at the same time, feel it is too simple.   EDIT: I've had a look at this again and narrowed it down to 2 cases:   Case 1: The vertex $v$ in question has $deg(v)\ge2$. $v$ then loses its neighbours as the algorithm is applied repeatedly until $v$ becomes a leaf itseelf. At this point, $v$ is attached to another neighbour, leaving an edge. This means the algorithm can no longer be applied and we have obtained the final Prüfer Encoding for the tree $T$.  Case 2: The vertex $v$ has $deg(v)\ge 2$. It loses its neighbours as the algorithm is applied repeatedly but is deleted itself at some point in the process.   Have I gone about this wrong?  Thank you. Show that $e^{-\beta} = \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{e^{-u}}{\sqrt{u}} e^{-\beta^2 / 4u} du$. Show that $e^{-\beta} = \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{e^{-u}}{\sqrt{u}} e^{-\beta^2 / 4u} du$.  I'm not really sure of how I should proceed to show this, and it's pretty un-intuitive as well.  I've managed to manipulate the RHS into  $$\frac{e^{4\beta}}{\sqrt{\pi}} \int_0^\infty \frac{1}{\sqrt{u}} e^{-(\beta+2u)^2/4u} du$$ which now vaguely resembles a Gaussian. However, substituting $v = \beta + 2u$ in in order to to get something like $e^{-x^2}$ doesn't particularly help. I got $$ \frac{e^{4\beta}}{\sqrt{2\pi}} \int_0^\infty \frac{1}{\sqrt{v - \beta}} e^{-v^2/2(v-\beta)} dv. $$  I'm really unsure of how I should proceed; any help would be appreciated. Which QGIS installation for Windows comes with r.in.lidar? How to write file into another Everybody perhaps knows of Jon Favreau's movie  which is going to be released in 2016. But quite recently I learned (and to my utter surprise) that another Jungle book movie is coming in 2017 and it's called .   Now, this movie has big names too with Christian Bale as Baghera, Benedict Cumberbatch as Sher Khan, Cate Blanchett as Kaa and a young Indian boy in the live action role as Mowgli.   How often does this happen, with two big productions with big cast making a movie on the same subject? I want to ask why this is happening and if there are any really significant differences between the two (apart from the cast, of course). When users create a node, they should be redirected to a payment page; if payment is successful, the user can see the node.  Is there any way to achieve this? Group Membership Test for Permutations Prove there are no other solutions of functional equation $f(x+y) = \frac{f(x)+f(y)}{1-f(x)f(y)}$ I have the following functional equation. Find all continuous functions $f:(-1,1) \to \mathbb R$ such that $$ f(x+y)=\frac{f(x) + f(y)}{1 - f(x)f(y)} $$ The first obvious solution is $f(x) \equiv 0$. Another one I guessed, it is $f(x) = \pm \tan x$. I suspect, that there are no more solutions. The problem is that I don't know how to prove that.  Since we have a rational equation, I have no idea how to make any substitutions in order to get expression for $\tan x$ (as it is not a rational expression).  P.S. I can also show that it is true that (a) $f(0) = 0$ (take $x=y=0$), and (b) $f(-x) = -f(x)$ (take $y=-x$).  Update: there is a solution of this problem also here: How to keep glasses dry in the rain? Will my Chinese visa still be valid if I renew my passport? If so how does that work?  I have heard that you can travel with your new passport and just show the old passport with the valid visa when entering China. Has anyone done that? Recently?   I have also been told that you have to get a new visa, which seems somewhat logical but frustrating, especially when my Chinese visa is only a few months old.  My Chinese visa was issued in late Nov 2014 and I heard the rules changed between the US and China. My visa says it's good for 10 years and my US passport expires in 5. If $ a$, $ b$, $ x$, $ y$ are integers greater than $1$ such that $ a$ and $ b$ have no common factor except $1$ and $ x^a = y^b$ show that $ x = n^b$, $ y = n^a$ for some integer $ n$ greater than $1$.  I started this way: $x^a=y^b$ $=&gt;$ $x=y^{b/a}=(y^{1/a})^b$. Thus $x$ is an integer, and so is $y^{1/a}$, so $y=n^a$ for some integer $n$ greater than $1$. Then from the given relation, we get $x=n^b$.   I am not very convinced with my solution, as I feel I did some mistake. Can anyone point out where the mistake is? Any other solution is also welcome :)