Show that for any matrix $A_{m \times n}$ , the row rank and column rank are equal
Looking for an intuitive explanation why the row rank is equal to the column rank for a matrix
Using THE before some countries I have been taught that it's grammatically proper to use the before countries like "Sudan and Yemen" but I was never given a reason (if there exists any, of course) when I asked why, except for "it is one of those ambiguous parts of English language that we &ndash; as non-native speakers &ndash; should just accept without trying to look for a reason". Now, I'd like to know if there is any answer to this "why" or I should keep on believing that there are no hard and fast rules when it comes to questions like these. Is it possible to tell a Valyrian steel sword from a very well crafted steel sword, by sight alone I am travelling to Athens, Greece, shortly, and have seen . Should I be worried? Should I cancel my trip?  Update  It's fine, taxis are currently on strike, but Syntagma Sqaure is fine: Can I load a .NET assembly at runtime and instantiate a type knowing only the name? Is there a package that supports the addition or subtraction of chemical equations, perhaps with automatic tabbing and a alignment.  Whilst it is obvious I could set up tables or use mathmode's align feature, it seems so much more efficient to have a package handle the tabbing automatically, since it is rather predictable in this application.  mhchem does not seem to have such a feature...      Another advantage is you could maintain the readability of your equations and you would not have the added clutter associated with tables, especially tabs &amp;, which diminish the readability of your .tex document.  And of course, the line should be automatically inserted before the last row. How can I use commercial at sign in Objective-C macro? Find a non-negative function on [0,1] such that $t\cdot m(\{x:f(x) \geq t\}) \to 0$ that is not Lebesgue Integrable Problem:  Find a non-negative function $f$ on $[0,1]$ such that $$\lim_{t\to\infty} t\cdot m(\{x : f(x) \geq t\}) = 0,$$ but $f$ is not integrable, where $m$ is Lebesgue measure.  My Attempt:  Let $f(x) = \frac{\chi_{(0,1]}}{\sqrt{x}}$. Then,  \begin{align*} \lim_{t\to\infty} t\cdot m(\{x \in [0,1]: f(x) \geq t\}) &amp;= \lim_{t\to\infty} t \cdot m(\{x\in (0,1]: 1/\sqrt{x} \geq t\})\\ &amp;=\lim_{t\to\infty} t \cdot m(\{x\in (0,1]: x \leq (1/t^2)\})\\ &amp;= \lim_{t\to\infty} t \cdot m((0, (1/t^2)))\\ &amp;= \lim_{t\to\infty} \frac{t}{t^2} = 0. \end{align*}  However, $f(x)$ is integrable over this interval. I have tried functions looking like $f(x) = 1/x^p$ but I cannot find any that will work here. There is a hint that says there is a monotonic function that fits this description.    Also, does anyone know of a list of non-Lebesgue integrable functions on $[0,1]$? I feel as though I could use this for many counterexamples if one were to exist. Thanks! How to download a file over HTTP? I have a small utility that I use to download an MP3 file from a website on a schedule and then builds/updates a podcast XML file which I've added to iTunes.  The text processing that creates/updates the XML file is written in Python. However, I use wget inside a Windows .bat file to download the actual MP3 file. I would prefer to have the entire utility written in Python.  I struggled to find a way to actually download the file in Python, thus why I resorted to using wget.  So, how do I download the file using Python? How can I mount the /system directory rewritable or read-only on my Android phone? This seems to be one of those tricky examples. I only know one proof which is quite complicated and follows by localizing $\mathbb{Z}[\sqrt[3]{2}]$ at different primes and then showing it's a DVR. Does anyone know any simple quick proof? plot(density(rexp(100))   Obviously all density to the left of zero represents bias.  I'm looking to summarize some data for non-statisticians, and I want to avoid questions about why non-negative data has density to the left of zero.  The plots are for randomization checking; I want to show the distributions of variables by treatment and control groups. The distributions are often exponential-ish.  Histograms are tricky for various reasons.    A quick google search gives me work by statisticians on non-negative kernels, e.g.:  .    But has any of it been implemented in R?  Of implemented methods, are any of them "best" in some way for descriptive statistics?  EDIT:  even if the from command can solve my current problem, it'd be nice to know whether anyone has implemented kernels based on literature on non-negative density estimation Update statement with inner join on Oracle Differentiability of $g:=f(\sqrt{x^2+y^2})$ for a $C^1$ function with $f'(0)=0$ NBHM $2008$ Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a continuously differentiable function such that $f'(0)=0$.  Define for all $x,y\in \mathbb{R}$,  $$g(x,y)=f(\sqrt{x^2+y^2})$$ Pick out the true statements :   $g$ is differentiable on $\mathbb{R}^2$ $g$ is differentiable on $\mathbb{R}^2$ if and only if $f(0)=0$ $g$ is differentiable on $\mathbb{R}^2- \{ (0,0)\}$   I do not have much idea of differentiation of multivariable calculus.  I just tried out some examples to get some idea..  I have to take some function whose derivative at $0$ has to be zero.  for simplicity I would take $f(x)=x^2$ Then I would get :     $g(x,y)=x^2+y^2$ which is differentiable on $\mathbb{R}^2$   This would suggest me that second option has more chances as $f(0)=0$ in this case.  I would now take $f(x)=x^2+1$ then :     $g(x,y)=x^2+y^2+1$ which is differentiable on $\mathbb{R}^2$ though $f(0)\neq 0$   So, I would now eliminate second option and third option also...  So, It is now got confirmed that i have to work out to "Prove" that first option is correct.  Now, I have to prove that $g(x,y)=f(\sqrt{x^2+y^2})$ is differentiable.  I have to look for partial derivatives first :  $$\frac{\partial g}{\partial x}= f'(\sqrt{x^2+y^2}).\frac{2x}{2\sqrt{x^2+y^2}}=f'(\sqrt{x^2+y^2}).\frac{x}{\sqrt{x^2+y^2}}$$  $$\frac{\partial g}{\partial y}= f'(\sqrt{x^2+y^2}).\frac{2y}{2\sqrt{x^2+y^2}}=f'(\sqrt{x^2+y^2}).\frac{y}{\sqrt{x^2+y^2}}$$  I have used here that $f$ is differentiable on $\mathbb{R}$  As $f$ is continuously differentiable, we have $f'$ is continuous and I know that :  $\frac{x}{\sqrt{x^2+y^2}}$ and $\frac{y}{\sqrt{x^2+y^2}}$ are also continuous.     As $f'$ and $\frac{x}{\sqrt{x^2+y^2}}$ are continuous So is the product $\frac{\partial g}{\partial x}=f'(\sqrt{x^2+y^2}).\frac{x}{\sqrt{x^2+y^2}}$      As $f'$ and $\frac{y}{\sqrt{x^2+y^2}}$ are continuous So is the product $\frac{\partial g}{\partial y}=f'(\sqrt{x^2+y^2}).\frac{y}{\sqrt{x^2+y^2}}$   So we have continuous partial derivatives for $g(x,y)$..  So, I am willing to conclude that:     $g(x,y)$ is  differentiable on $\mathbb{R}^2$   I would be thankful if some one can confirm if the answer correct.  I would be much more thankful if some one can let me know if there are any gaps in my argument.  This is just a proof verification Question..  Thank you. jQuery - Is it okay to use $('#ElementId') everytime? How long typically are paper reviews? Is there such a thing as “too long”? When you review papers submitted for publication, is there an “optimal” length for reviews? In my experience as an author and referee, I have seen a large range of review lengths (for reference, a paper in my field is typically between 3 and 8 printed pages):   Zero length: for some of the papers I have authored, the reviewer just clicked the “accept as is” or “reject” checkmark on the review sheet, without adding any comment (at least, not any comment visible to me). It's not so common, but it has happened. Short length: a lot of time, I received reviews who consisted of a single paragraph. Medium: one full page, maybe two pages.   Although I have never received any such long reviews, I have myself written on a few occasions reviews that exceed two pages, including once or twice a four or five-page review.  These were cases where the paper was good, but could be (in my view) much improved and some of the aspects/consequences had escaped the authors' consideration.  I am wondering how useful reviews of various length are to editors. I often consider that “unmotivated” reviews are useless, as they do not give any real insight about the paper to the editor. For example, if the editor gets two conflicting non motivated reviews, how is she to decide?  On the other hand, although I write some from time to time, I have never received long reviews, so… is this something frowned upon? There's something wrong with the vertical alignment on the :   (click to enlarge)  On the , which uses a similar list but without the Favorite column, the vertical alignment is correct.  (Tested in the latest Firefox, Safari and Chrome on macOS Catalina.) A non-trivial, non-negative, function bounded below by its derivative with $f(0)=0$? I did not know what to search to see if this existed elsewhere. But, I could not find it. Here's the question:  Does there exist a continuously differentiable function, $f: [0,1] \rightarrow [0, \infty)$, such that the following hold? (i) $f(0) = 0$ (ii) $f'(x) \le c f(x)$ for all $x$ ($c$ is a fixed constant) (iii) $f \not\equiv 0$  This was a fun problem someone asked me a long time ago. I have always been convinced there is no such function, but I cannot prove it. I have mainly tried using the mean value theorem (in multiple forms), and re-wording the problem in terms of the integral of a continuous function to no avail. I spent some time trying to use (ii), the definition of derivative, and squeeze theorem, but no luck. After much time trying to prove non-existence, I spent a little time trying to find such a function... also with no luck. This problems ability to avoid being solved has since taken away the original fun and replaced it with a twinge of annoyance. $M + N$ and $M \cap N$ are finitely generated $A$-modules implies $M$ and $N$ finitely generated using exact sequences Let $0 \to M_1 \to M \to M_2 \to 0$ be an exact sequence of $A$-modules.  i) Prove: If $M_1$ and $M_2$ are finitely generated, then $M$ is too.  ii) Let $M$ and $N$ be sub-modules of an $A$-module $L$. Assume that $M+N$ and $M\cap N$  are finitely generated $A$-modules. Prove that $M$ and $N$ are finitely generated $A$-modules.  If have proved the first part, but i don't see how to do the second part. I tried to use the first part, so to find an exact sequence with $M$ in the middle and finitely generated modules in place of $M_1$ and $M_2$, but I can't find a surjective map $M \to P$ (for some f.g. $A$-module) in the sequence:  $$0 \to M\cap N \to M \to P \to 0$$  any hints/help would be appreciated. Switch statement fall-through...should it be allowed? For as long as I can remember I have avoided using switch statement fall-through. Actually, I can't remember it ever entering my consciousness as a possible way to do things as it was drilled into my head early on that it was nothing more than a bug in the switch statement. However, today I ran across some code that uses it by design, which got me immediately wondering what everyone in the community thinks about switch statement fall-through. Is it something that a programming language should explicitly not allow (like C# does, though it supplies a workaround) or is it a feature of any language that is powerful enough to leave in the programmer's hands? Edit: I wasn't specific enough to what I meant by fall-through. I use this type a lot:     switch(m_loadAnimSubCt){         case 0:         case 1:             // Do something             break;         case 2:         case 3:         case 4:             // Do something             break;    }  However, I'm concerned about something like this.    switch(m_loadAnimSubCt){         case 0:         case 1:             // Do something, but fall through to the other cases             // after doing it.          case 2:         case 3:         case 4:             // Do something else.             break;    }  This way whenever the case is 0, 1 it will do everything in the switch statement. I've seen this by design and I just don't know if I agree that switch statements should be used this way. I think the first code example is very useful and safe. The second seems kind of dangerous. Is 'hair' singular or plural? When one says I was washing my hair, is it singular or plural? What is the singular for hair?