Let T be sqaure matrix and regarded as a linear operator on a finite dimensional vector space V such that $T^2 = 0$.     $\dim(\ker T \cap \ker T^t) =\dim \ker T+\dim\ker\ T^t−\dim(\ker T\dot+\ker T^t) $    ($\dot+ $ denotes direct sum)  Is the abvoe equation true? If so, why is it so?
$\dim(U_1+U_2) = \dim U_1 +\dim U_2 - \dim(U_1\cap U_2).$  I want to make sure that my intuition is correct. Suppose we have two planes $U_1,U_2$ though the origin in $\mathbb{R^3}$. Since the planes meet at the origin, they also intersect, which in this case is a one-dimensional line in $\mathbb{R^3}$. To obtain the dimension of $U_1$ and $U_2$, we add the dimensions of the planes (4), and the subtract the dimensions of the line (1), which results in (3).   *additional question(s):   Can we generalize this notion to $\mathbb{F^{n}}$?  Suppose we have an additional case where $U_1$ and $U_2$ are planes in $\mathbb{R^3}$, but $U_1 \subseteq U_2$. In this instance, $dim(U_1 + U_2) &lt; 3$, because the first two-dimensional plane is contained in the second and as a result, the dimensions of the subspaces when summed cannot exceed two. Since both subspaces $U_1,U_2$ are two dimensional and $U_1 \subseteq U_2$, then their intersection is also two-dimensional, concluding $dim(U_1+U_2)=2+2-2 = 2$.  Is this proper intuition?
Show (within ZF Set Theory without choice) that for any sets $A,B,C:$   If $|A|^{|B|} = |B|$, then $|A| = 1 = |B|.$   That is, if there is a bijection from $|A|^{|B|}$ to $|B|$, then  $|A| = 1 = |B|.$  Please help me.. Failed review audit - for me quite clearly off-topic, can it be removed from the audit set? $\DeclareMathOperator{\card}{card}$From Problem-Solvers Topology  Prove the following: CANTOR'S THEOREM If $A$ is a set, then $$\card A &lt; \card \mathcal{P}(A)$$ where $\card A$ stands for the cardinality of set $A$.  My Answer If $A = \emptyset$, then $\card A =0$ and $\card \mathcal{P}(A)=1$. If $A = \{a\}$, then $\card A=1$ and $\card \mathcal{P}(A)=2$. So suppose that $A$ has at least two elements. Define a function $f: A \rightarrow \mathcal{P}(A)$ such that $f(x) = \{x\}$ for all $x \in A$. Then $f$ is injective. But it cannot be surjective, because for any two distinct elements $a,b \in A$, there is no element in $A$ that is sent to the set $\{a,b\}$ in $\mathcal{P}(A)$. Therefore, there is no bijection, and $\card A &lt; \card\mathcal{P}(A)$. Do you think my answer is correct? Thanks in advance Whether the players killed a mage and got his spellbook(s) or they met with a nice wizard who let him or her learn a spell from his spellbook, can a player prepare a spell from a spellbook that isn't his or hers?  I'm asking because the PHB p. 114 says:     [...] Preparing a new list of wizard spells requires time spent studying your spellbook and memorizing the incantations and gestures you must make to cast the spell: at least 1 minute per spell level for each spell on your list.    emphasis mine   Further, the block on the left as a Copying a Spell into the Book ("the" Book as in "your" Book, since it appears in the block titled Your Spellbook), which explains how you can copy a spell from one book to another, assuming the book is from a stranger, it cost more and takes more time. It actually says:     [...] deciphering the unique system of notation used by the system of notation of the wizard who wrote it. [...]   It continues saying that you have to learn the various parts of the spell (verbose, material, gesture.)  This makes it sounds like a wizard would have a really hard time to prepare a spell from another's spellbook. Is that the case? I am trying to prove that all vector spaces can be written as the direct sum of two subspaces.  My attempt: If each vector in vector space V can be uniquely written as the sum of two other vectors in V, the two vectors can be categorized into subspaces and thus V can be written as the direct sum of these subspaces. I am not sure how to show that this is true for all cases though. Edit Sorry for the confusion,to clarify I meantGiven a vector space V with a subspace W, prove that there exists a subspace W' such that the direct sum of W and W' is V.I am not sure about my proof method at all. Thank you. Using a keyboard shortcut to paste text What is the structure in "as best you can"? I instinctively translate it "as best as you can", however this makes no sense.  What is the real structure behind this phrase?  I'll include an  to illustrate the historical presence of this phrase:    Whatever the structure is, I'm guessing that it has something to do with 19th century grammar. Line breaking with hyperref href with an underline in itemize I have a simple itemized list like the following MWE:  \documentclass[10pt,a4paper]{article} \usepackage[utf8]{inputenc} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{graphicx} \usepackage{hyperref} \usepackage{lmodern} \usepackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry} \usepackage{lipsum} \begin{document} \begin{itemize} \item \href{http://google.com}{[ID1]: \underline{\lipsum[1]}} \end{itemize} \end{document}   This unfortunately doesn't break the block of text as expected, as shown below:    How can I fix this item so it will line break properly? Is Dirichlet function Riemann integrable? Rich Wizards, Poor Muggles Let $D := \{z \in \mathbb{C}: |z| &lt; 1\}$ and $f\colon D \rightarrow \mathbb{C}$ be holomorphic. Suppose $\lvert f(z)\rvert \leq 1$ on $D$, show that $$\frac{|f(0)| - |z|}{1 + |f(0)||z|} \leq |f(z)| \leq \frac{|f(0)| + |z|}{1 - |f(0)||z|} \ \forall z \in D.$$  I have tried using Cauchy Integral Formula to $f$ and expanding $f$ at $0$, but I have no idea why $f(0)$ appears in the inequality in the former case. Both answers and hints are appreciated and thanks in advance! Here are two sentences:        Once he got the passport, he went to the Chinese consulate in Toronto and applied for a visa.   Once he got the passport, he went to the Chinese consulate in Toronto to apply for a visa.      Are the two sentences grammatical, and do they sound natural?  If they are both fine, what's the difference? Mapping ManyToMany with composite Primary key and Annotation: How to increase battery life on Ubuntu 20.04 and what power saving software should I install, if any? I recently installed Ubuntu 20.04 LTS on my Lenovo Ideapad Flex 5 and realized that battery life in Ubuntu is not as good as Windows. The battery drains faster in Ubuntu. I have seen the questions in Ask Ubuntu about this, but I haven't found anything that regards Ubuntu 20.04 LTS. I've done some research and found out that the battery drain is because most vendors optimize their drivers for Windows and with Linux, you have to hope that the kernel drivers do the job (which they usually do). After that, I researched how to increase my battery life and almost all sources I found recommended installing tlp or powertop or something similar to these. But I have a few questions.  Does tlp, powertop, and similar software actually affect battery life?  If the answer to 1 is yes (in a positive way), what would you recommend I install (such as whether I should  install tlp or powertop by itself or in a combination)?  Are there any other ways to increase battery life in Ubuntu 20.04?   Note: I am a beginner to Linux and Linux distros and do not want to mess with anything advanced that may negatively affect the system. I may not be able to immediately know how to tweak any of the aforementioned software, so please suggest any tweaks if they are needed to get results. And if it helps to know, here are my system specs: OS: Ubuntu 20.04.1 LTS x86_64 Host: 81X2 IdeaPad Flex 5 14ARE05 Kernel: 5.7.17-050717-generic CPU: AMD Ryzen 7 4700U with Radeon Graphics (8) @ 2.000GHz GPU: AMD ATI 04:00.0 Renoir (These were given by neofetch) P.S: If you need anything else, just ask. Thanks in advance! What exactly is a photon? Consider the question, .  The answers say, "an elementary particle" and not much else. They don't actually answer the question. Moreover, the question is flagged as a duplicate of,  &ndash; the answers there don't tell me what a photon is either. Nor do any of the answers to  mentioned in the comments. When I search on , I can't find anything useful. Questions such as,  look promising, but bear no fruit.  say things like, "the photon is an excitation of the photon field." That tells me  nothing. Nor does the , which says:     The photon is the quantum of the electromagnetic four-potential, and therefore the massless bosonic particle associated with the electromagnetic force, commonly also called the 'particle of light'...   I'd say that's less than helpful because it gives the impression that photons are forever popping into existence and flying back and forth exerting force. This same concept is in the  too - but it isn't true. As , "Virtual particles only exist in the mathematics of the model." So, who can tell me what a real photon is, or refer me to some kind of authoritative informative definition that is accepted and trusted by particle physicists? I say all this because I think it's of paramount importance. If we have no clear idea of what a photon actually is, we lack foundation. It's like what :     Photons seem to be one of the foundation ideas of quantum mechanics, so I am concerned that without a clear definition or set of concrete examples, the basis for understanding quantum experiments is a little fuzzy.   I second that, only more so. How can we understand pair production if we don't understand what the photon is? Or the electron? Or the electromagnetic field? Or everything else? It all starts with the photon.   I will give a 400-point bounty to the least-worst answer to the question. One answer will get the bounty, even if I don't like it. And the question is this:    What exactly is a photon? Is there a website with historical temperatures and weather data? I am interested in hiking this fall in France and camping along the way. In order to figure what gear I need to pack to camp (particularly for the sleeping bag), I would like to know what temperatures to expect.  I am interested in historical temperatures to figure the usual temperature and I might look at them for other seasons. If that matters, I am thinking about hiking in the Pays Basque (South West of France), the mountainous part.  Where (ideally online and for free) can I find such data? Are there resources that cover Europe, or even the whole world? What is the symbol on the supply voltage that looks like a shaded-in capacitor? I am looking at  (dual low-voltage power amplifier). On the fourth page, there is a sample circuit; right at the V+ input there is a symbol that looks like a capacitor. However, it looks like it is shaded in. What is that? How can I set an external display (monitor) as the primary one? Currently the MacBook's display works as primary, so every application I open gets open on the 13" and I have to drag it to the 24". I also don't have access to the dock on my 24".  Is is possible to turn off the MacBook screen and use only the one attached via DVI?  (I've just connected my 24" HP LP2475w via Mini DisplayPort to DVI to my MacBook Pro 13", which works great except for this one thing.) Does using links to other pages dilute the linking page's link juice/page rank? There are lot's of questions here about passing link juice to other pages, including .  I've read several of them over the months, but I always come away wondering the same question.  Does linking to another page, dilute or burn some of the link juice the linking page has already earned?  For example, if my current page has for example, a link juice of 10, and I link to another page, it passes a certain amount of juice (I'll use 90% as an example) of the linking page's juice, giving the page being linked to a .9 boost in it's own link juice.  However, does that mean the linking page's page's remaining juice is now less than 10?  While some of the questions would imply this to be the case, it seems counter intuitive as no one would ever want to link to any page outside of their own content for fear of losing page rank. Is there a way to forget node names between TikZ pictures? I have a document with several TikZ pictures in it.  I generally use simple names for the nodes such as:  \node (A) at .....  \node (B) at .....   and so forth.  Because of this I have name clashes between the various pictures.  That's generally not a problem.  However I find myself in a position where I may or may not have defined a certain node within a given picture.  So I use the following snippet to test whether the node has been defined:  \makeatletter \long\def\ifnodedefined#1#2#3{%%   \@ifundefined{pgf@sh@ns@#1}{#3}{#2}} \makeatother   Unfortunately this fails if I've already created a node by the same name in another picture.  Is there a way to scope the definitions of nodes or is there a way to undefine nodes once you no longer need them?  P.S.  I'm aware that I could embed each picture in a standalone file and import it using \includegraphics.  Unfortunately, this approach will not work for the document I'm creating.  MWE  \documentclass{article} \usepackage{tikz} \usetikzlibrary{calc}  \makeatletter \long\def\ifnodedefined#1#2#3{%%   \@ifundefined{pgf@sh@ns@#1}{#3}{#2}} \makeatother  \setlength{\parindent}{0pt} \pagestyle{empty} \begin{document}   There's no node \textbf{D} here, so I draw a triangle:\par \begin{tikzpicture}    \coordinate (A) at (0,0);   \coordinate (B) at (1,0);   \coordinate (C) at (1,1); %  \coordinate (D) at (0,1);    \ifnodedefined{D}     {  \draw (A) foreach \myn in {B,C,D} {  -- (\myn) } --cycle;  }     {  \draw (A) foreach \myn in {B,C}   {  -- (\myn) } --cycle;  }  \end{tikzpicture}     There is a node \textbf{D} here, so I draw a square:\par \begin{tikzpicture}    \coordinate (A) at (0,0);   \coordinate (B) at (1,0);   \coordinate (C) at (1,1);   \coordinate (D) at (0,1);    \ifnodedefined{D}     {  \draw (A) foreach \myn in {B,C,D} {  -- (\myn) } --cycle;  }     {  \draw (A) foreach \myn in {B,C}   {  -- (\myn) } --cycle;  }  \end{tikzpicture}      There isn't supposed to be a node \textbf{D} here.  I'm expecting a triangle, but get a square.\par \begin{tikzpicture}    \coordinate (A) at (0,0);   \coordinate (B) at (1,0);   \coordinate (C) at (1,1); %  \coordinate (D) at (0,1);    \ifnodedefined{D}     {  \draw (A) foreach \myn in {B,C,D} {  -- (\myn) } --cycle;  }     {  \draw (A) foreach \myn in {B,C}   {  -- (\myn) } --cycle;  }  \end{tikzpicture}  \end{document}   produces: