Suppose $U$ is k-dimensional vector space on complex field, and $S\in L(U)$. Assume $S$ has at least $m$  distinct nonzero eigenvalues. Show that $\def\null{\operatorname{null}} \null (S^{k-m})= \null (S^{k-m+1})$  Here is my approach:  Suppose $ S \in L(U) $ and k is non-negative integer and $S$ has at least $m$ distinct non-zero eigenvalues. If $S^{k-m} v = 0 $ for all $v \in U$, then $S^{k-m+1} v = S (S^{k-m})=S(0)=0$. Thus, $ \null (S^{k-m}) = \null (S^{k-m+1})$.  Please give me feedback on my answer. If it is wrong, direct me to right track.  Thanks in advance!
I am given that $V$ is n-dimensional vector space over $\mathbb{C}$ and $T \in L(V)$. And $T$ has least $m$ distinct nonzero eigenvalues. How do I show that $\text{null}(T^{n-m}) = \text{null}(T^{n-m+1})$?  I'm really not sure how to start here so just the first couple steps would be very helpful. Thanks!!
Tricks and tips for finding information in man pages Does anyone have any tricks and tips for finding information in man pages? is there any uncertainty on the free particle with a definite momentum $\vec p$? The probability amplitude for a free particle with momentum $\ p$ and energy $E$ is the complex wave function: $$\psi_{(\vec x , t)}=e^{i(\vec k\cdot \vec x -\omega t)}$$  is there any uncertainty on the free particle with a definite momentum $\vec p$!? How to use a special character as a normal one in Unix shells? Many questions like 'How to type the double-quote char (")?' are being asked, and we don't want to clutter our community with the same answer (Type it as \" if not enclosed in 's, " if enclosed in 's.) So, the question is here.   You can't type special chars into a terminal like normal ones, e.g. this command will fail:  echo Updates (11)   So, how to type in these chars in the terminal as if they were normal ones?  !#$^&amp;*?[](){}&lt;&gt;~;'"\|&lt;space&gt;&lt;tab&gt;&lt;newline&gt; With all the new sites up and running it gets more and more difficult for users to really know the scope of a site and thus a lot of questions a migrated to places where they are (probably even more) off-topic, see e.g. . So I request the opposite of :  Allow migration if you have enough rep on both the source and the target site (the target reputation required could be a bit lower, say 5001)  That way questions could only be migrated by users familiar with the scope of both sites who can really estimate where the question is better. Maybe those who don't have the rep on the target site could still be able to cast a flagging-only-but-not-counting-as-vote vote...  1) Courtesy of Fantasy piece involving different worlds and a crippled protagonist This is a novel that I think I remember reading somewhere in the 1990s courtesy of my brother (who does not remember the book). Only a few details are coming to mind at the moment. It's a male protagonist. Something was not quite right with his legs, which led to him being nicknamed Frog or Leapfrog due to his usual method of locomotion, which consisted of pushing off from his legs to his hands and vice versa. Something happens and he winds up moving between a number of worlds (I want to say that he fell into a well or a hole in a field). There was a part which stuck with me where he's in a bottomless hole and finds he keeps falling faster and faster without end with the rushing wind becoming actively painful (I don't remember if he actually reaches a terminal velocity or if some aspect of the world prevented it while still allowing the rushing air to affect him physically).  Unfortunately, that's all I can remember. I don't remember whether he eventually gains companions on the voyage or how it ended. The most vivid memory is that of the nonstop falling. How do I open the Dahl chest in Sanctuary? Could anyone provide an example of a nonempty subset $U$ of $R^2$ such that $U$ is closed under scalar multiplication, but $U$ is not a subspace of $R^2$ To my understanding, this is the default configuration of /etc/crontab:  # m h dom mon dow user  command 17 *    * * *   root    cd / &amp;&amp; run-parts --report /etc/cron.hourly 25 6    * * *   root    test -x /usr/sbin/anacron || ( cd / &amp;&amp; run-parts --report /etc/cron.daily ) 47 6    * * 7   root    test -x /usr/sbin/anacron || ( cd / &amp;&amp; run-parts --report /etc/cron.weekly ) 52 6    1 * *   root    test -x /usr/sbin/anacron || ( cd / &amp;&amp; run-parts --report /etc/cron.monthly )   Why were these specific times chosen for the default entries?  At first I assumed that they are somewhat randomized on installation to even out the system load, but that doesn't seem to be correct. How to calculate $\lim_{x\to 1^+} \log (x)^{\log(x)}$? How to calculate $\lim_{x\to 1^+} \log (x)^{\log(x)}$ ? i know that its "1", but why? How can i calculate this?  Thank you very very much =) Crash adding comment The SE iOS app crashes when adding comments to certain posts.  It was discovered to be if the "Add an Answer" button was visible, not just responding to a deleted comment.  Edited to match answer. Previous steps were not the only use case, and this error keeps getting re-reported in different ways. Unable to activate Windows 8 It is stated as a problem in Spivak's Calculus and I can't wrap my head around it. If banks offer a fixed rate lower than the variable rate, is that an indication interest rates may head down? This is one of the major banks in Australia:   At present time, the variable rate is 7.80% but the fixed rate is 6.99% for one year.  And it seems most banks in Australia, if you fix the rate (at the moment), you will get a lower rate.  Is this an indication that in the eyes of the banks, that they believe the interest rates are going to down? I just ask because it seems most people in the media think the rates are going up and they've risen about 6 times in a row now.  Could it be that the banks are secretly preparing for rates to drop? Giving a single user write permissions I want to grant a single user write permissions to a directory owned by root over command line. With chmod, it seems that I can only change the permissions for owner, group and other, not for a specific user. Poll on best VNC / remote desktop software for assisting others on Windows/Mac machines from Ubuntu?    I've heard good things about TeamViewer and Fog Creek Copilot, but I'm wondering if the included GNOME Vinaigre VNC client is good enough for this.  To specify, I'm looking for best option based on:   SIMPLEST ease-of-use for client to download/use on their end. See #1. Works cross-platform I am able to control client's mouse and/or keyboard from remote machine. Use /testfor to make a custom command Basically, I'm trying to recreate the Faction plugins /f map in my single-player without mods as my PC doesn't run them very well. (I know it sounds strange.) I am trying to use the /testfor to test if I have written '/f map' in chat, and if I have, it would show me a custom generated map. I don't know how to word this better xD Essentially what I want is for it to test if I have written a command in chat, and if I have, display something. Upgrading from a modern smartphone camera without going all in on a DSLR kit I'm looking for a word that applies to all the things a person can wear, e.g. clothes, shoes, accessories, etc. It musn't be too generic, e.g. product, item, etc. Understanding the proof for showing that there are infinitely many $n\in\mathbb{Z}^+$ such that $n!+1$ is divisible by at least two distinct primes There is another post pertaining to this question; however, the other post does not address some specifics that concern me in the proof. Proof: Let $n=p-1, p \geq 7$. Then $p|(p-1)!+1$. We will show that $(p-1)!+1=p^n$ has no solutions. Suppose not. Suppose $(p-1)! + 1 = p^n$ has a solution for $p\geq 7$. Then $(p-1)! =p^n-1  = (p-1)(p^{n-1}+\cdots+p+1)$. So $(p-2)!\equiv n (\mod p-1)$. Now suppose that $(p-1)|n$. Then $p-1\leq n$. So $(p-1)!+1&lt;p^{p-1}\leq p^n$. Hence, there are no solutions. It remains to show that $(p-1)|(p-2)!$. But $p-1=2m$ means $2m|(p-2)!$. QED  Questions:  My understanding is that at the start of the proof, Wilson's theorem is used to show that at least one prime divides $n!+1$.  Why do we want to show that $(p-1)!+1=p^n$ has no solutions? The way I see it is that $p^n$ is composite, and we are basically saying that $p^n\nmid (p-1)!+1$, so no composite number divides $n! +1$. But how and where in the proof do we actually show that another prime $q \neq p$ divides $n!+1$?    Thanks in advance. example of discontinuous function having direction derivative Is there a function (non piece-wise unlike below) which is discontinuous but has directional derivative at particular point? I have a manual that says the function has directional derivative at $(0,0)$ but is not continuous at $(0,0)$. $$f(x,y) = \begin{cases} \frac{xy^2}{x^2+y^4} &amp; \text{ if } x \neq 0\\  0 &amp; \text{ if } x= 0 \end{cases}$$  Can anyone give me few examples which is not defined piece wise as above?