Can we conclude from $MM^T=I$ that $M^TM=I $ over $\mathbb F_2$ Let $M$ be nonsingular real matrix, dimension $n$. If we know that $$MM^T=I,$$ where $I$ is identical matrix. Is it true that $M^TM=I$ too. And it is all over $\mathbb F_2$ field, which is field that consists only of $0,1$.   My first thoughts was that it is not necessary true, but then:    If we multiply our main equation by $M$ on the right we get: $$MM^TM=M $$ If we multiply our main equation by $M^T$ on the left, we have: $$M^TMM^T=M^T $$   So, can we conclude from here, that $M^T=I?$ Or there can exist other matrix $A\neq I$ such that $AM^T=M^T$ and $MA=M?$
If $AB = I$ then $BA = I$ If $A$ and $B$ are square matrices such that $AB = I$, where $I$ is the identity matrix, show that $BA = I$.    I do not understand anything more than the following.   Elementary row operations. Linear dependence. Row reduced forms and their relations with the original matrix.   If the entries of the matrix are not from a mathematical structure which supports commutativity, what can we say about this problem?  P.S.: Please avoid using the transpose and/or inverse of a matrix.
Is $x^x=y$ solvable for $x$? Given that   $x^x = y$; and  given some value for $y$   is there a way to expressly solve that equation for $x$? How to merge cells vertically Is it possible to create a table like this one?    The usual structure of tables I use looks like this:  \begin{table}[H]     \begin{adjustbox}{width=\textwidth,center}     % \begin{adjustbox}{center}         \begin{tabular}{lll}             \hline             \multicolumn{1}{l}{\head{XXX}} &amp; \multicolumn{1}{l}{\head{XXX}} &amp; \multicolumn{1}{l}{\head{XXX}} \\             \hline             \ttfamily xxx &amp; \ttfamily xxx &amp; \ttfamily xxx \\             \hline         \end{tabular}     \end{adjustbox}     \vspace{ - 05 mm}     \caption{xxx}     \label{tab:xxx} \end{table} When Dirac found his equation for the electron $(-i\gamma^\mu\partial_\mu+m)\psi=0$ he famously discovered that it had negative energy solutions. In order to solve the problem of the stability of the ground state of the electron he invoked Pauli's exclusion principle and postulated that negative energy states well already filled by a "sea" of electrons. This allowed him to predict the positron, viewed as a hole in the sea.  This interpretation was ultimately discarded owing to its inapplicability to bosons and difficulties with explaining the invisibility of the infinite charge of the sea.  According to my understanding, the modern argument goes something like this. There is a discrete symmetry of the lagrangian called charge conjugation $\psi \rightarrow \psi^c$ which allows the negative energy solutions to be interpreted as positive energy solutions for a second mode of excitation of the electron field with opposite charge, called positrons. The decay of electrons to positrons is then suppressed by the $U(1)$ gauge symmetry of the lagrangian forcing conservation of electrical charge.  According to this interpretation What Dirac would have missed was the lagrangian formalism. Is this historically and physically correct? R2jags does not remove the burn in part sometimes? In a world without air, I understand they definitely would. However, with drag taken into account, I think they wouldn't. Since the drag force varies proportional to the square of speed (ignoring the change in coefficient of drag with speed), wouldn't the total impulse due to drag on a bullet fired from a gun from a specific height horizontal to ground be higher than the total impulse on an identical bullet dropped from the same height with no horizontal velocity?  Also, when bullet spin is taken into account, the bullet fired from a gun should resist change in orientation; therefore, it should maintain a small but non-zero angle of attack, and also have a lift force, correct? How to work around Eclipse Luna/GTK bug 339174 (change to not use oxygen) The bug is described here:  For me it fails with:  java: /build/buildd/oxygen-gtk3-1.0.2/src/animations/oxygencomboboxdata.cpp:87: void Oxygen::ComboBoxData::setButton(GtkWidget*): Assertion `!_button._widget' failed.   Problem is if I go to GTK+ Apperance in Control Module and change to anything other than oxygen Eclipse will look totally awful.  Regardless what theme I use (tried QtCurve and Raleigh) it has the same look with huge padding around all controls etc.  Is it possible to have a sensible look without oxygen GTK?  I have kubuntu-desktop installed over the stock LTS distro.   $dpkg -l libgtk* | grep -e '^i' | grep -e 'libgtk-*[0-9]'     ii  libgtk-3-0                             3.4.2-0ubuntu0.8                          GTK+ graphical user interface library     ii  libgtk-3-bin                           3.4.2-0ubuntu0.8                          programs for the GTK+ graphical user interface library     ii  libgtk-3-common                        3.4.2-0ubuntu0.8                          common files for the GTK+ graphical user interface library     ii  libgtk2.0-0                            2.24.10-0ubuntu6.1                        GTK+ graphical user interface library     ii  libgtk2.0-bin                          2.24.10-0ubuntu6.1                        programs for the GTK+ graphical user interface library     ii  libgtk2.0-common                       2.24.10-0ubuntu6.1                        common files for the GTK+ graphical user interface library Is there a way to exclude other particle systems from the effects of a Wind effect? I have a scene with a number of particle systems. I'd like to have a separate wind effect for my main scene and another for an extra: an emitter + wind-effect to blow leaves into my scene. I can't find any way to exclude, so here I am. At the very worst, I can just render the main scene and then render the wind effect and composite it, but I figured I'd check first. What is the agreed upon definition of a "positive definite matrix"? "Washroom", "restroom", "bathroom", "lavatory", "toilet" or "toilet room" I've always been confused by the terms washroom, restroom, bathroom, lavatory, toilet and toilet room. My impression is that Canadians would rather say washroom while Americans would probably say bathroom or Saint John's in the same situation.  I guess the difference here is not only in different kinds of English, but also in whether one is referring to a room in their house or in some public place.  Which do you usually use? Please specify the difference if you use more than two from those six with different meanings, and also where you are from (i.e. what type of English you speak). Linux Debian codenames I've got a few, quite silly, non-technical questions about giving  to  releases.  Each Debian release has its unique codename, which is (so far) a characters' name from  movies by .  Here is list of all assigned codenames so far:   release 1.1 is  (Buzz Lightyear) - the spaceman, release 1.2 is  - the tyrannosaurus, release 1.3.x is  (Bo Peep) - the girl who took care of the sheep, release 2.0 is  - the piggy bank, release 2.1 is  (Slinky Dog) - the toy dog, release 2.2 is  - Mr. Potato, release 3.0 is  -  the cowboy, release 3.1 is  - the sergeant of the Green Plastic Army Men, release 4.0 is  - the toy blackboard (Etch-a-Sketch), release 5.0 is  - the toy binoculars, release 6.0 is  - the name for the three-eyed aliens, release 7.0 is  - the name of the rubber toy penguin with a red bow tie, release 8.0 is  - the name of the yodelling cowgirl, release 9.0 is  - a purple rubbery octopus toy at , release 10.0 is  - 's pet dachshund (currently ).   List of upcoming major Debian releases' codenames after :   release 11.0 is  - 's horse (currently ), release 12.0 is  - intelligent worm toy with a built-in flashlight.   There are also:   special codename  () which is symbolic link to codename which is currently unstable,  which is symbolic link to codename which is currently stable,  which is symbolic link to codename which is currently testing.   The list of Toy Story  is quite  but at some time, there will be no more characters' names to assign.  My questions are:   What codenames will be assigned if we run out of characters' names? Who decides what is codename of next release (please don't answer ambiguously like: 'community')? How many releases' names are planned ahead?   BTW: Interesting quote from :     The  of using Toy Story names was  by  who   was, at the time, the Debian Project Leader and was working also at   , the company that produced the movies.     by  (: ). Laplace equation of composite function Suppose we have $f(x,y)$, which satisfies Laplaces equation: $\frac{\delta^2f}{\delta x^2}+\frac{\delta^2f}{\delta y^2}=0$  Does $f(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2})$ also satisfy Laplace equation? Matlab Recursion Loop When Apple was in the high $600 or so range, it split 6 times and made it more affordable to purchase more shares for less, even though the value was technically the same. I see companies like Google are extremely high per share and am wondering why or why not they would or wouldn't split their stock right now. Propagate all arguments in a bash shell script SOQL result ordering in the absence of an Order By clause In the absence of an  in a SOQL query, what order will Salesforce return results in?  Clearly if ordering is important then Order By should be specified, but I recently encountered a question about what the order would be if it was undefined.  My initial thoughts were Salesforce would return them in whatever order suited them. Much like .  For simple queries this would typically line up with the creation date descending and the Id ascending. For more complex queries it would depend on the joins being performed. The gravitational force on your body, called your weight, pushes you down onto the floor. $$W=mg$$ So, what is the weight equation through general relativity? Why do we say that irreducible representation of Poincare group represents the one-particle state? Only because   Rep is unitary, so saves positive-definite norm (for possibility density), Casimir operators of the group have eigenvalues $m^{2}$ and $m^2s(s + 1)$, so characterizes mass and spin, and It is the representation of the global group of relativistic symmetry,    yes? Im creating a community user from an Apex code. When our internal users (non-admins) run the code they are getting following error. (This was working fine when we test the feature in sandbox)     SObjectException: Field is not writeable: User.ProfileId   The code is       //Grab Community User Profile     Profile p = [Select Id, Name from Profile where Name = :COMMUNITY_PROFILE LIMIT 1];       User PortalUser = new User();          PortalUser.UserName = Con.Email+'-POC';         PortalUser.ContactId = Con.Id;         PortalUser.ProfileId = p.id;         PortalUser.Email = Con.Email;         PortalUser.EmailEncodingKey = 'UTF-8';         PortalUser.FirstName = Con.FirstName;         PortalUser.LastName = Con.LastName;         PortalUser.TimeZoneSidKey = 'America/Los_Angeles';         PortalUser.LocaleSidKey = 'en_US';         PortalUser.LanguageLocaleKey = 'en_US';          if (Con.LastName.length() &gt;= 8){             PortalUser.Alias = Con.LastName.substring(0, 7);             }else{             PortalUser.Alias = Con.LastName;         }             SObjToInsert.add(PortalUser); I bought a new MacBook Air and was able to connect to the internet at first but now it won't connect.  I have other devices as well as other computers on wifi but my Mac won't connect.  When I try and click on the wifi, it tells me that it can't be joined.  How can I troubleshoot? At most finitely many (Hamel) coordinate functionals are continuous - different proof