Prove that for the $\mathbb R^{n\times n}$ vector space doesn't exist a basis containing  $n^2$ matrices, such that AB=BA for every two matrices $A$, $B$ in the base. $(n&gt;1)$  I think it is true because I did not get any example which makes it false. Thanks in advance.
I know that matrix multiplication in general is not commutative. So, in general:  $A, B \in \mathbb{R}^{n \times n}: A \cdot B \neq B \cdot A$  But for some matrices, this equations holds, e.g. A = Identity or A = Null-matrix $\forall B \in \mathbb{R}^{n \times n}$.  I think I remember that a group of special matrices (was it $O(n)$, the ?) exist, for which matrix multiplication is commutative.  For which matrices $A, B \in \mathbb{R}^{n \times n}$ is $A\cdot B = B \cdot A$?
Does a MCMC fulfilling detailed balance yields a stationary distribution? I guess I understand the equation of the detailed balance condition, which states that for transition probability $q$ and stationary distribution $\pi$, a Markov Chain satisfies detailed balance if $$q(x|y)\pi(y)=q(y|x)\pi(x),$$  this makes more sense to me if I restate it as:   $$\frac{q(x|y)}{q(y|x)}= \frac{\pi(x)}{\pi(y)}. $$  Basically, the probability of transition from state $x$ to state $y$ should be proportional to the ratio of their probability densities. Let's say that you want to output or concat strings. Which of the following styles do you prefer?   var p = new { FirstName = "Bill", LastName = "Gates" }; Console.WriteLine("{0} {1}", p.FirstName, p.LastName); Console.WriteLine(p.FirstName + " " + p.LastName);   Do you rather use format or do you simply concat strings? What is your favorite? Is one of these hurting your eyes?  Do you have any rational arguments to use one and not the other?  I'd go for the second one. How to solve $x^3=-1$? How to solve $x^3=-1$? I got following:  $x^3=-1$   $x=(-1)^{\frac{1}{3}}$  $x=\frac{(-1)^{\frac{1}{2}}}{(-1)^{\frac{1}{6}}}=\frac{i}{(-1)^{\frac{1}{6}}}$... The following is a well-known result in functional analysis:     If the vector space $X$ is finite dimensional, all norms are equivalent.     Here is the standard proof in one textbook. First, pick a norm for $X$, say  $$\|x\|_1=\sum_{i=1}^n|\alpha_i|$$ where $x=\sum_{i=1}^n\alpha_ix_i$, and $(x_i)_{i=1}^n$ is a basis for $X$. Then show that every norm for $X$ is equivalent to $\|\cdot\|_1$, i.e.,  $$c\|x\|\leq\|x\|_1\leq C\|x\|.$$ For the first inequality, one can easily get $c$ by triangle inequality for the norm. For the second inequality, instead of constructing $C$, the  is applied to construct a contradiction.   The strategies for proving these two inequalities are so different. Here is my question,      Can one prove this theorem without Bolzano-Weierstrass theorem?   UPDATE:     Is the converse of the theorem true? In other words, if all norms for a vector space $X$ are equivalent, then can one conclude that $X$ is of finite dimension? Did the Nazgûl retain their Rings of Power? In The Fellowship of the Rings, Frodo sees the true appearance of the Nazgûl:      He was able to see beneath their black wrappings. There were five tall figures: two standing on the lip of the dell, three advancing. In their white faces burned keen and merciless eyes; under their mantles were long grey robes; upon their grey hairs were helms of silver; in their haggard hands were swords of steel. Their eyes fell on him and pierced him, as they rushed towards him.   The Fellowship of the Ring - J.R.R. Tolkien - chapter 11, A Knife In the Dark   There's no mention of the Nazgûl wearing the rings of power that Sauron gave them, although their hands are referenced. Did the Nazgûl retain their rings of power once they became wraiths? Or did something else happen to the rings once the Nazgûl emerged?&#185;  &#185; As an FYI, I am currently re-reading The Lord of the Rings trilogy and am just pages from being done with The Fellowship of the Ring. I'm very familiar with the Peter Jackson movies. So that's where I'm coming from as far as canon knowledge goes. "Absolutely continuous random variable" vs. "Continuous random variable"? In the book Limit Theorems of Probability Theory by Valentin V. Petrov, I saw a distinction between the definitions of a distribution being &quot;continuous&quot; and &quot;absolutely continuous&quot;, which is stated as follows:  &quot;...The distribution of the random variable $X$ is said to be continuous if $P\left(X \in B\right)=0$ for any finite or countable set $B$ of points of the real line. It is said to be absolutely continuous if $P\left(X \in B\right)=0$ for all Borel sets $B$ of Lebesgue measure zero...&quot;  The concept I am familiar with is:  &quot;If a random variable has a continuous Cumulative Distribution Function, then it is absolutely continuous.&quot;  My questions are: are the two descriptions about &quot;absolute continuity&quot; in (1) and (2) talking about the same thing? If yes, how can I translate one explanation into the other one? If $0 &lt; \alpha &lt; 1$, then       $$ (x + y)^{\alpha} &lt; x^{\alpha} + y^{\alpha} $$      for $x$, $y$ positive.   Is this inequality true in general?  I tried using Young's Inequality: For $z,t &gt; 0$, and for $n$, $m$ such that $n+m=1$, then   $$ z^n + t^m \leq nz + mt $$  So, using this we have   $$ (x+y)^{\alpha} \cdot 1^{1 - \alpha} \leq \alpha(x+y) + (1-\alpha) = \alpha x + \alpha y + (1-\alpha)$$  which is not as tight as I want. Need help in the longtable with the figure How to change keyboard layout via command line cmd.exe on Windows XP/7? Is is possible to change the keyboard layout via command line on Windows XP/7? Blender resizing my image in half I have a problem. I'm trying to render an image and the resolution is 500x280. After I render the image and save it, for some reason the image gets resized to 250x140. I'm unsure why it happends. Normal text is invisible when using beamer with notes and XeLaTeX I'm using beamer with xelatex to prepare a talk.  If I switch on notes on right I'm experiencing strange behaviour:  When using csqoutes in the title, all normal text afterwards becomes invisible (or white). The text selectable in the output pdf. Note that the text is also visible in the small preview on the note page.  This does not happen when using pdflatex or lualatex.  MWE:  \documentclass{beamer} \usepackage[autostyle=true,german=quotes]{csquotes}  \usepackage{pgfpages} \setbeameroption{show notes} \setbeameroption{show notes on second screen=right}  \begin{document}    \begin{frame}         \frametitle{\enquote{Enqouted Title}}     \end{frame}      \begin{frame}         \frametitle{This is the second slide}         Lorem ipsum dolor sit amet, consectetuer adipiscing elit.     \end{frame} \end{document}   Update:  csqoutes is not the culprit. Seems to be a general problem when using xelatex and beamer with notes enabled.  New MWE:  \documentclass{beamer}  \usepackage{pgfpages} \setbeameroption{show notes} \setbeameroption{show notes on second screen=right}  \begin{document}     \begin{frame}         \frametitle{First Slide Title}         Here be text     \end{frame}      \begin{frame}         \frametitle{This is the second slide}         Lorem ipsum dolor sit amet, consectetuer adipiscing elit.     \end{frame} \end{document} This is NOT an installation problem question. I've had both op sys working fine side-by-side for a couple of weeks now. I'm using Ubunutu 12.04 64-bit.  This morning when I tried to switch from Win8 to Ubuntu, it just rebooted into Win8. No grub screen whatsoever. This is the 3rd machine I've had dual-booting &amp; have never seen this, but then Win8 is a 1st for me &amp; has thrown some pretty awesome blocks in the path to get to here. Atrocious op sys...  I'm assuming the ASUS updater did this yesterday. (ASUS R503U)  So, do I just put the Ubuntu disk in &amp; reinstall grub? Who paid for Harry's Nimbus 2000? The Nimbus 2000 was the fastest broom at the time of the Harry's first year at Hogwarts.  Professor McGonagall wanted to win the Quidditch Cup so she spoke with Dumbledore and got an exception from the rule that no first-years could have a broom for Harry.  My question is who paid for the broom?  It couldn't have been a cheap broom.  They wouldn't have taken gold out of Harry's Gringotts account without permission.  It wouldn't be right for the school to pay for it and I doubt McGonagall paid for it out of her own pocket.  So who paid for the broom? How to read a csv file and use the values as x and y points in blender? Prove the following inequality:   $$\dfrac{e^x + e^{-x}}2 \le e^{\frac{x^2}{2}}$$   This should be solved using Taylor series.  I tried expanding the left to the 5th degree and the right site to the 3rd degree, but it didn't help.  Any tips? Center of $D_6$ is $\mathbb{Z}_2$ The center of $D_6$ is isomorphic to $\mathbb{Z}_2$.   I have that $$D_6=\left&lt; a,b \mid a^6=b^2=e,\, ba=a^{-1}b\right&gt;$$ $$\Rightarrow D_6=\{e,a,a^2,a^3,a^4,a^5,b,ab,a^2b,a^3b,a^4b,a^5b\}.$$ My method for trying to do this has been just checking elements that could be candidates. I've widdled it down to that the only elements that commute with all of $D_6$ must be $\{e,a^3\}$ but I got there by finding a pair of elements that didn't commute for all other elements and I still haven't even shown that $a^3$ commutes with everything. For example, I have been trying to show now that $$a^3b=ba^3$$ and haven't gotten too far yet but if I had to answer a question like this on the exam, I feel it would be difficult, is there any kind of trick or hints other than brute force using the relations to get that $a^3$ commutes with everything?  For the solution once I have that the center of $D_6$ is what I think then as there is only one group of order $2$ up to isomorphism, it must be isomorphic to $\mathbb{Z}_2$.  Ideally a way that doesn't appeal to $D_6$ as symmetries of the hexagon if that seems possible. Suppose $z\neq -1$ is a complex number of norm 1m $(|z| =1)$. I know that $z= n + mi$, what is the most straightforward way of solving this problem? I was also given the following sketch for a possible approach. Process Unicode files with BOM correctly with POSIX tools Trying to use grep today, I ran into the  of the  in a Unicode file (UTF-8, in this case). Specifically, I was trying to find a file beginning with XYZ with the pattern grep '^XYZ', but of course grep treated the BOM as three separate characters and did not match the first line of the file if the first line started with XYZ. I even tried to update the regular expression to ignore spaces ('^[[:space:]]*XYZ'), but to no avail.  Other questions have dealt with  or , but I want to know if POSIX tools have a general option to handle Unicode files correctly. If grep handled the Unicode file correctly, it would consider the file contents to start after the BOM and match XYZ on the first line just like any other line. How do I reverse a list using recursion in Python? I want to have a function that will return the reverse of a list that it is given -- using recursion. How can I do that? How can I expand a partition into non adjacent free space using GParted?