I figure it has something to do with minimal polynomials and the Jordan canonical form, I just can't piece it together. I would really appreciate a hint on how to start it.
$A$ is a $n\times n$ matrix such that $ A^m = 0 $ for some positive integer $m$.   Show that $A^n = 0$.   My attempt: For $n &gt; m$, it's obvious since matrix multiplication is associative.  For $n &lt; m$, $A^n\times A^{m-n} = 0$; not sure what to do next. Also I know that $\det A = 0$.
How do you handle uneven tiles while rendering a tile map? Here is what I want to do with my tile map that I am unsure of. As you can see the top walls are way larger then the bottom and side ones (this is also an issue with my corners which are odd shapes [more like an L then a square] as well as larger then the 40x40px that my current tiles are).    I have also attached the tile sheet I am using for the tiles to show you what the corners and tops look like compared to the rest of the tiles.  What I’m thinking is I might be able to just draw them using the same array I have if I use the tiles width and height instead of a set WxH? I dont know how well this will work, but were the blocks that will have different values will be blocks the player/NPCs can’t step on then it may not be an issue, again I’m not even sure if that is a viable solution.  My code in case you want to see how I'm currently handling it:  (pastebin) (edit:im not handling it, meant how I'm currently doing the map)  tl;dr - how can I take take my 2d looking tile map and change it to have different size tiles? or is there a better way to do what I am trying to do? Is there a name of a method that I should be looking for while googling? Is it possible to apply modifiers to multiple objects at once? Can I smoothly upgrade from one LTS to next LTS release? Question is about Ubuntu release process. Are migrations from n to n+1 releases expected?  Assuming we have server with installed Ubuntu 14.04 LTS. Will it be smoothly upgradable to Ubuntu 16.04 LTS via apt-get upgrade ?  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Thanks Moderncv package - cventry date width I am writing my CV using moderncv package, everythig is nice but I have a problem with \cventry command. I have something like this:   (it means 2010 - now)  I would want to have this in one line, rather than in two lines. I tried looking into the moderncv style files but with no success. What can I do to achieve the effect I want? Which is correct, "you and I" or "you and me"? Debug a java application without starting the JVM with debug arguments Normally to attach a debuger to a running jvm you would need start the jvm with arguments such as the following:  &gt; java -Xdebug -Xrunjdwp:transport=dt_socket,address=1000,server=y,suspend=n   Now if I want to debug a process that wasn't started in debug mode, what can I do?  This situation arrises when a production system (i.e. started without debug args) exhibits a 'random' (I use the term loosely) bug. So I can't restart the jvm with the appropriate arguments, because nobody knows how to reproduce the bug again. Is it impossible to attach to the JVM in this situation?  Just to clarify it is not possible to use tools like jdb to attach to already running JVMs unless they were started in debug mode  from the JVM man page     Another way to use jdb is by attaching it to a Java VM that is   already running. A VM that is to be   debugged with jdb must be started with   the following options: Club filter of $\kappa$ is $\kappa$-complete I'm trying to show that club filter of $\kappa$ is $\kappa$-complete for uncountable regular cardinal $\kappa$:  Let $\kappa$ be uncountable regular cardinal, let $C(\kappa)$ be the club filter generated by $\kappa$.  To show that $C(\kappa)$ is $\kappa$-complete it is enough to show that for every sequence $\langle \alpha_i\mid i&lt;\gamma\rangle$ with $\gamma&lt;\kappa$ and $\alpha_i$ club of $\kappa$ the set $\alpha=\bigcap_{i&lt;\gamma}\alpha_i$ is a club of $\kappa$.  Showing that $\alpha$ is close is easy:     If $\beta_n$ is a sequence in $\alpha$ then for every $i$ we that $\beta_n$ is a sequence in $\alpha_i$, because $\alpha_i$ is a club it is close hence $\lim \beta_n$ is in $\alpha_i$ for every $i$ so it is also in $\alpha$   But I have a problem with unboundedness, my guess will be for each $\lambda&lt;\kappa$ to find a sequence for each $\alpha_i$ such that all of the sequences converge to some $\mu&gt;\lambda$, that way $\beta&lt;\mu\in\alpha$. I think I'll have to use the fact $\kappa$ is regular, but I don't know how to proceed and prove my idea.  So, is my idea correct, and if yes how can I proceed? Suppose $A,B,C$ are sets. Prove that $A\mathbin\triangle B\subseteq C \iff A\cup C=B\cup C$. Suppose $A,B,C$ are sets.  Prove that $A\mathbin\triangle B\subseteq C \iff A\cup C=B\cup C$.  I proved that for $x\in A\setminus B$, $x\in A\cup C$ and for $x\in B\setminus A$, $x\in B\cup C$. How do I show that $A\cup C=B\cup C$, and that $A\mathbin\triangle B\subseteq C \iff A\cup C=B\cup C$? Current  enables one to upgrade to macOS Mojave from a stable release, but I want to try out a fresh installation instead of upgrading with previous data, so that I can assess it better.  I've downloaded an application named "macOS Mojave Beta" from the App Store, and running it would presumably result in an upgrade. Now I wonder what to do next. Would  and booting from it be a good idea, when I'm working with a Beta version? How can a distribution have infinite mean and variance? How to change Windows 7 date format? Is it possible to configure Windows 7 to use YYYY-MM-DD instead of DD/MM/YYYY as the date format? How? Where? 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Is there any way I can solve this problem? I'm currently stucked with this issue.  Plus, I don't know why my \task command does work. Why is the counter subtask not affected by tabularx's several trials?  \documentclass{scrartcl}  \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc}  \usepackage{etoolbox} \usepackage{booktabs} \usepackage{ltablex} \usepackage{xifthen}  \def\credits{} \newcounter{subtask} \setcounter{subtask}{1} \renewcommand{\thesubtask}{\alph{subtask})} \newcommand{\task}[1][]{%     \ifthenelse{\isempty{#1}}{%         \thesubtask%         \stepcounter{subtask}%     }{%         \thesubtask%         \gappto\credits{(#1)}%         \stepcounter{subtask}%     }% }  \begin{document} \begin{tabularx}{\linewidth}[l]{l@{\hspace*{1ex}}X}     \task[1]    &amp;   TEXT1\\     \task       &amp;   TEXT2\\     \task[3]    &amp;   TEXT3 \end{tabularx} \credits \end{document}       The solution given by David Carlisle below causes a table that is too wide. How can this be undone?  \documentclass{scrartcl}  \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc}  \usepackage[a6paper, showframe=true]{geometry}  \usepackage{etoolbox} \usepackage{booktabs} \usepackage{ltablex} \usepackage{xifthen}  \def\credits{} \newcounter{subtask} \setcounter{subtask}{1} \renewcommand{\thesubtask}{\alph{subtask})} \makeatletter \newcommand{\task}[1][]{%     \ifx\hfuzz\@tempdima% %   % tabularx trial do nothing     \else%         \ifthenelse{\isempty{#1}}{%             \thesubtask%             \stepcounter{subtask}%         }{%             \thesubtask%                 \gappto\credits{(#1 BE)\\}%             \stepcounter{subtask}%         }%     \fi } \makeatother   \begin{document}  \begin{tabularx}{\linewidth}[l]{l@{\hspace*{1ex}}X}     \task[1]    &amp;   Some longer text that exceedes the line width.\\     \task[4]    &amp;   TEXT2\\     \task[3]    &amp;   TEXT3 \end{tabularx} \credits  \end{document}