$a_k = \mathrm{rank}(A^{k+1}) - \mathrm{rank}(A^k)$ is increasing
$\operatorname{rank}(T^3) + \operatorname{rank}(T) \geq 2\operatorname{rank}(T^2)$ for finite-dimensional $V$
These two operators commute...but their eigenvectors aren't all the same. Why? The Hamiltonian   $$H =  \left[ \begin{array}{cccc} a &amp; 0 &amp; 0 &amp; -b \\ 0 &amp; 0 &amp; -b &amp; 0\\ 0 &amp; -b &amp; 0 &amp; 0\\ -b &amp; 0 &amp; 0 &amp; -a \end{array} \right] $$  commutes with the qubit exchange operator  $$ P = \left[ \begin{array}{cccc} 1 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 1 &amp; 0\\ 0 &amp; 1 &amp; 0 &amp; 0\\ 0 &amp; 0 &amp; 0 &amp; 1 \end{array} \right] $$  So I would expect the two to have the same eigenvectors. The eigenvectors of $P$ are easily seen to be $(1,0,0,0)^T; (0,0,0,1)^T; (0,1,1,0)^T ; (0,1,-1,0)^T$. The latter two are also eigenvectors of $H$, but the first two are not. Why? I thought commuting operators shared the same eigenbasis? regular expressions / pattern matching on ubuntu command line What is the $\sqrt{i^4}$?   $i^4$ = $(i^2)^2$ So is $\sqrt{i^4}$ = $\sqrt{(i^2)^2}$ = $i^2$ = $-1$?  Or is $\sqrt{i^4}$ = $\sqrt{1}$ = $1$?  When I plug it into my TI-89 Titanium, I get $1$.  Edit: I now see why $i^4 \neq -1$. $$\sqrt{x^2} = \left|{x}\right| $$ So$$\sqrt{(i^2)^2} = \left|i^2\right| = \left|-1\right| = 1$$ Is there any event that fires when WPF animation ends? How do we escape the cycle of bad college eating? Mouseover text: And yet I never stop thinking, 'sure, these ingredients cost more than a restaurant meal, but think how many meals I'll get out of them! Especially since each one will have leftovers!'    Randall Munroe really hit the nail on the head with this comic and especially the mouseover text. There's this sort of cycle that I seem to follow when I try cooking, and I need to get my way out of it. It goes like this:   Decide I should cook something. Thumb through my recipe books, find one or two recipes I like. Write down the ingredients for the recipes in a list. Go to the grocery store, stumble around for a while and eventually find exactly those ingredients. Go back home; at some point in time soon after, cook each recipe. Whew, that was a lot of work! Spend the next week or two eating junk again, until I decide I should cook something. (repeat)   Very similar to the comic; sometimes I too let the leftovers go bad (usually not). And every time I rack up a huge bill at the supermarket and think to myself it's justified because of the leftovers or the amount of food. Not to mention, that's a lot of work for just a couple of meals, and a long list of ingredients.  I love cooking fresh meals but that's just so much overhead and so time consuming! So I always resort back to Easy Mac, Ramen, frozen dinners and frozen pizza. Yuck.  How can I escape this cycle and cut down on the overhead? How can I focus on more than 2 recipes at a time without buying giant lists of ingredients and spending tons of money? Is it really worth the trouble or should I just chock it up to being a college student and wonder about the years ahead of me where I'll have much more time to spend cooking one meal at a time? future or present with when Homology Groups of $S^n$ Virtualbox Windows 10 Guest installation is freezing up in 14.04 I've installed VirtualBox using the Software Centre and I'm running a system on Ubuntu 14.04 with 8gb RAM, 64-bit dual core processor.  I want to run Windows 10, I've tried both 64 and 32 bit versions, in virtualbox. However, every time I start up windows 10 it gets to the point where the dots go round in a circle, then it freezes up and crashes back to the virtualbox manager. The only information it gives me is that the virtual machine was aborted.  I've changed the settings around a lot and nothing is working.  Any help would be appreciated.  Oh, and I've run Windows 10 technical preview on this machine before, a while ago  Thanks for any help in advance. $R^2$ of 1 but RMSE > 0 Let $a\ge1$ and define the sequence $(x_n)$ recursively by: $$x_1 = \sqrt{a}$$ $$x_{n+1}= \sqrt{a+x_n}$$  Here's what I did:  Plugging in some values makes it seem as if the sequence is increasing. I first solve the equation $t=\sqrt{a+t}$, equivalently $t^2 -t -a =0$, to get an idea what's going on. The roots are $$t_{1,2}= \frac{1 \pm \sqrt{1+4a} }{2}.$$ Therefore, between the roots, that is for $$t\in \left( \frac{1 - \sqrt{1+4a} }{2}, \frac{1 + \sqrt{1+4a} }{2}\right)$$  we have $t^2-t-a \le 0 \iff t \le \sqrt{a+t}$. Here's my attempt at showing inductively that $$\forall n\in \mathbb{N}, \ x_n \le \frac{1 + \sqrt{1+4a} }{2}$$  For $n=1$, the statement is true (simple arithmetic). Suppose it's true for $n$. Consider  $$x_{n+1}= \sqrt{a+x_n}\le \sqrt{a+\frac{1 + \sqrt{1+4a} }{2}}=\frac{1 + \sqrt{1+4a} }{2},$$  because $\frac{1 + \sqrt{1+4a} }{2}$ satisfies $x=\sqrt{x+a}$.  So this establishes the sequence $(x_n)$ is bounded. Now, by induction again, I try to show it's monotonically increasing. The base case $x_1\le x_2$ is obvious. Assume (strong induction) $x_{n-1}\le x_n$ for all $n&gt;1$. But since we have $$x_1\le x_n \le \frac{1 + \sqrt{1+4a} }{2}, $$   then $x_n$ lies in the appropriate interval, so $x_n \le \sqrt{a+x_n}=x_{n+1}.$  Is this proof correct? Can the condition on $a$ be relaxed to $a&gt;0$? Maybe even small negative $a$? And could someone provide me with another proof of convergence of $x_n$? I absolutely hate this one! I've had a promising Skype interview with a great professor from a top-level university in US. Our research interests are closely related and he encouraged me to apply BUT he says that all decisions are finally made by admission committee there.  The question is "Is a professor in such a high-level university able to affect admission committee if he really wants a student?" or "Does admission committee decision-making process become easier if a professor really wants an applicant?"  The main point is the difference between my transcript (which is a real catastrophe!) and my research background which is pretty good so the professor's decision might be different from admission committee. How can I take a picture with my webcam in Windows 7? I know it existed as sort of a folder in Windows Explorer in at least Windows XP, but I just can't seem to find it in Windows 7...  How can I take a picture with my webcam in Windows 7 without using any 3rd party tools? Intuitive explanation of variance and moment in Probability While I understand the intuition behind expectation, I don't really understand the meaning of variance and moment.   What is a good way to think of those two terms? Remove ajax call from regular links with jQuery Mobile Extend or remove Data Explorer timeout I know that we don't have automatic login for the Data Explorer, but what's with the automatic timeout/logout?  There have been times that I am logged into the site, maybe I walk away for a little bit, and I come back to being logged out.    Is there a reason for logging us out of Data Explorer?  If so, is it possible to either remove the timeout or extend the period of time before timeout? How do I make my program watch for file modification in C++? I have just updated Windows 7 to Windows 10.  My touchpad scolling (right-side) is not working anymore.    I have no idea how to fix this in Windows 10.  I have hit the start button and typed "synaptic", but have found nothing useful.  I used to have an icon on the taskbar that showed where I was touching the touchpad.  I used to be able to right-click that icon to change settings, but Windows 10 has not brought that icon over.  Also, the asus website has no synaptic touchpad drivers available for Windows 10.  I have tried the enaltech drivers, but upon the installation atempt I am told that my hardware is not correct. Windows & Ubuntu dual boot.No menu to choose OS? I just installed Ubuntu alongside Windows 7. Now I cant boot Windows 7 because it automaticaly boots Ubuntu without displaying boot menu. How can I get menu with options to choose which Os to boot? Can anyone help? Why Do Jedi Use One Lightsaber in Combat? While I've seen Ahsoko Tano, in The Clone Wars, fight with two lightsabers (one in each hand), this seems to be extremely rare for Jedi.  We've seen Sith do this frequently, like Asajj Ventress, with two lightsabers.  Darth Maul used a dual lightsaber against Qui-Gon Jinn and Obi-Wan Kenobi and then there's General Grievous, who uses multiple lightsabers with his multiple limbs.  But most Jedi use just one blade, and their lightsaber hilts tend to be simple, just a straight cylinder without much adornment, while Sith sometimes use curved grips or other more exotic stylings.  Is there a reason Jedi stick with the simplest lightsaber designs and why the vast majority (at least in the movies and in The Clone Wars series) use only a single lightsaber? LuaLaTeX, TeXLive 2016, standalone: undefined control sequence