Let $A$ be a $n \times n$ matrix with complex elements, $L$ a $1 \times n$ matrix with complex elements and $C$ a $n \times 1$ matrix with complex elements. Show that $\det (A-CL)=\det(A)-LBC$, where $B=\textrm{adj}(A)$ is the adjugate matrix of $A$.
I would like a proof of the following result, given on .  For all square matrices $\mathbf{A}$ and column vectors $\mathbf{u}$ and $\mathbf{v}$ over some field $\mathbb{F}$, $$ \det(\mathbf{A}+\mathbf{uv}^\mathrm{T}) = \det(\mathbf{A}) + \mathbf{v}^\mathrm{T}\mathrm{adj}(\mathbf{A})\mathbf{u}, $$ where $\mathrm{adj}(\mathbf{A})$ is the adjugate matrix of $\mathbf{A}$.  Note that $\mathbf{A}$ may be singular. However, the proof given on wikipedia requires that $\mathbf{A}$ is nonsingular.
I've got a simple Leaflet map with a basemap and one map-layer coming from my own hosted GeoServer.  I made the layers show up nicely in a map.  Now I want to add popups on the point layer showing information stored in MapService.  I couldn't find too much information about that topic.   There's one link to a working example but the link seems to be broken. I found another example which I addopted. You can see it on . I'd like to display the content of the name field in the attribute table but nothing happens when the user clicks the map.  Any idea? Question: Are there infinitely many primes of the form $4n+3$ and $4n+1$?  My attempt: Suppose the contrary that there exist finitely many primes of the form $4n+3$, say $k+1$ of them: $3,p_1,p_2,....,p_k$  Consider $N = 4p_1p_2p_3...p_k+3$, $N$ cannot be a prime of this form. So suppose that $N$=$q_1...q_r$, where $q_i∈P$  Claim: At least one of the $q_i$'s is of the form $4n+3$:  Proof for my claim: $N$ is odd $\Rightarrow q_1,...,q_r$ are odd $\Rightarrow q_i \equiv 1\ (\text{mod }4)$ or $q_i ≡ 3\ (\text{mod }4)$  If all $q_1,...q_r$ are of the form $4n+1$, then $(4n+1)(4m+1)=16nm+4n+4m+1 = 4(\cdots) +1$  Therefore, $N=q_1...q_r = 4m+1$. But $N=4p_1..p_k+3$, i.e. $N≡3\ (\text{mod }4)$, $N$ is congruent to $1\ \text{mod }4$ which is a contradiction.  Therefore, at least one of $q_i \equiv 3\ (\text{mod }4)$. Suppose $q_j\equiv 3\ (\text{mod }4)$  $\Rightarrow$ $q_j=p_i$ for some $1\leq i \leq k$ or $q_j =3$  If $q_j=p_i≠3$ then $q_j$ | $N = 4p_1...p_k + 3 \Rightarrow q_j=3$ Contradiction!  If $q_j=3$ ($\neq p_i$, $1\leq i \leq k$) then $q_j | N = 4 p_1...p_k + 3 \Rightarrow q_j=p_t$ for some $1 \leq i \leq k$ Contradiction!  In fact, there must be also infinitely many primes of the form $4n+1$ (according to my search), but the above method does not work for its proof. I could not understand why it does not work. Could you please show me?  Regards My colleague is using an application that consumes a lot of memory which makes the system too slow. Is it possible to share memory with other PCs over the Internet?   The system has 8 GB of RAM, and the application consumes more than 6 GB. mysqli or PDO - what are the pros and cons? In our place we're split between using mysqli and PDO for stuff like prepared statements and transaction support. Some projects use one, some the other. There is little realistic likelihood of us ever moving to another RDBMS.  I prefer PDO for the single reason that it allows named parameters for prepared statements, and as far as I am aware mysqli does not.  Are there any other pros and cons to choosing one over the other as a standard as we consolidate our projects to use just one approach? Why does "ls -all" show time for some files but only year for others? If you issue ls -all command some files or directories have the year and some have the time? Why do some show the year while others show the time? Is the time representative of the time the file was created? What is the best way to increase my reputation and privileges? I'm not sure if this is a stupid question, but here goes anyway... I have only recently started learning PHP and Javascript and I'm reasonably proficient. One of the reasons I'm so proficient is because I LOVE Stack Overflow and find most of the solutions to my dilemmas here.   But I've found that I never get the chance to ask a question because it has normally been asked many times before and when I go through the unanswered questions to try and help someone else, they are generally WAY over my head.  So I'm not sure how to get more privileges. I can't even vote up answers that I like or that helped me. I can't comment to thank someone for their solution. It is most tragic and demoralising... How do I change the situation? A 398MB directory was only compressed to 393MB using 7Z and Normal ZIP compression. Is this normal? If so, why do people continue to use ZIP on Windows? Find the smallest disk in complex plane containing the eigenvalues of a matrix (NBHM 2012) Consider the given matrix  $$\left[\begin{matrix}1&amp; -2&amp; 3&amp; -2 \\1&amp; 1&amp; 0&amp; 3\\-1&amp; 1&amp; 1&amp; -1\\0&amp; -3&amp; 1&amp; 1&amp;\end{matrix}\right]$$  Find out the smallest disk like ($|z-1| &lt; r$ ) in the complex plane containing the eigenvalues of the given matrix.  One easiest method is to find out the eigenvalues from the characteristic equation of the matrix. I can do it so no problem with it.   I am looking for a method to solve the problem without the above method that can be evaluated in very short time. This type of method is possible, I believe.  Thank you for your help How to measure GPU usage? What is the mechanism behind the slowdown of light/photons in a transparent medium? So light travels slower in glass (for example) than in a vacuum. What causes light to slow down? Or: How does it slow down? If light passes through the medium, is it not essentially traveling in the "vacuum" between the atoms? Does a surjective ring homomorphism have to be surjective on the unit groups? Current sensing nA to mA I need to sense the current of a target  through various periods of its on/sleep/off state and then need to send this data back to another interface MCU to read out the value. They will be both on the same board, powered by 5&nbsp;V USB.  However these are the requirements I have:   5&nbsp;V USB PSU Must be able to sense currents in the range of nA to mA with high    accuracy/precision. (I figure 1&nbsp;nA to 500&nbsp;mA) Only measure the current of the target MCU and not the interface. Must output maximum 3.3&nbsp;V interface MCU   I have looked at available options and current sense amplifiers by Texas Instruments can't sense nA due to higher bias currents. So therefore I feel I would need a precision amplifier.  However I am stuck on how to proceed with this because, I need to somehow have a automatic dynamic current range and I am fairly new to current sensing and am not really sure on all the specifics. Name tikz pic as if it was a node I'm confused about tikz pics. I can freely define them without issue, and even add coordinates inside that will be available by name inheritance like (-A) as for example in \pic (S) at (0,0) {myPic}; \draw (S-A) -- ++(0,0);. This is all documented around p. 255 of the pgf manual.  However I don' understand why it is not possible to give a name to a pic like the one above if no coordinate is defined inside it. By default it would be great if it could behave like a normal node.  MWE:  \documentclass{minimal} \usepackage{tikz}  \tikzset{     myBlock/.pic = {         \node[draw,minimum width=1cm,minimum height=0.5cm] at (0,0) {};     },     myBlockNamed/.pic = {         \node[draw,minimum height=1cm,minimum width=0.5cm] (-m) at (0,0) {};     }, }  \begin{document} \begin{tikzpicture}     \pic () at (0,0) {myBock}; % can't give it a name     \pic (S) at (3,0) {myBlockNamed}; % can give it a name     \draw (S-A) -- ++(2,0); \end{tikzpicture} \end{document}   The underlying point is I'm working on a control blocs library (so that we can stop the chat below with @Alenanno). For example here is the definition of a math operator  \tikzset{     mySum background/.style = {         circle, minimum size=1.5em, fill=white,     },     mySum edge/.style = {         draw=black, circle, minimum size=1.5em, line width=.5pt,     },     pics/bOp/.style = {     code = {       \node [mySum background] {};       \foreach \t [count=\i] in {#1}{         \pgfmathsetmacro{\angle}{\i*90}         \node[anchor=center, font=\tiny] at (\angle:0.45em) {$\t$};       }       \node [mySum edge] {};     },   }, } Hello everyone interested. Here's a seemingly simple problem.  Suppose $G$ is a complete, simple graph with vertices $\{1,2,..,n\}$, $n\geq k\geq 3$.  How many cycles of length $k$ can we have on $G$?  The statement seems simple but I have trouble putting everything together.  So I say fix a vertex and start forming cycles with length k. But I get nowhere... What is the proper approach to this problem since counting is not easy? Is there a relevant theorem that I am unaware of?  Thanks for any help... Notation question: $\ll$ When I look at the definition given from the Mac OS X Dictionary (I have set American English as interface language, and the dictionary used is then the New Oxford American Dictionary), I read:     the /ði/ /ð(ə)/ /ði/ [called the definite article]   adjective   1. denoting one or more people or things already mentioned or assumed to be common knowledge: what's the matter? | call the doctor | the phone rang. Compare with a.   • used to refer to a person, place, or thing that is unique: the Queen | the Mona Lisa | the Nile.   • informal denoting a disease or affliction: I've got the flu.   • (with a unit of time) the present; the current: dish of the day | man of the moment.   • informal used instead of a possessive to refer to someone with whom the speaker or person addressed is associated: I'm meeting the boss | how's the family?   • used with a surname to refer to a family or married couple: the Johnsons were not wealthy.   • used before the surname of the chief of a Scottish or Irish clan: the O'Donoghue.   I have never thought of an article as adjective. Is normally an article defined as adjective? Command-line completion from command history So, I've looked at history and at Ctrl+R, but they are not what I thought I knew. Is there a way that I can type in the beginning of a command, and cycle through the matches in my history with some bash shortcut?  # mysq(some shortcut key)   Gives me:  # mysqldump  --add-drop-table -e -q -n -C -u  (some shortcut key) #  mysql -u ben.dauphinee -p Show that if $g \circ f$ is injective, then so is $f$. The Problem:  Let $X, Y, Z$ be sets and $f: X \to Y, g:Y \to Z$ be functions.  (a) Show that if $g \circ f$ is injective, then so is $f$.  (b) If $g \circ f$ is surjective, must $g$ be surjective?  Where I Am:  So, I really have trouble with these, for some reason. I can draw pictures and make sense of the problems, but writing down proofs is very difficult for me.   Basically, for (a), I ended up with some complicated statement involving an implication implying another implication and then tried to derive a contradiction. It just got so convoluted that I couldn't make sense of it anymore, and I know there's a quick, elegant way to show it.  For (b), I know that $g$ need not be surjective. Once again, though, proving it directly from definitions has given me a bit of a headache.  Any help here would be appreciated. Thanks in advance.  The Proofs!  Ok, I did it. Thanks for the help, everyone! Let me know if there's anything wrong with these proofs, or if they could bet any better.  (a) Suppose $f$ is not injective. Then $$ f(x_1)=f(x_2) \implies x_1 \ne x_2 \text{        }(*).$$ Let $f(x_1)=y_0=f(x_2)$ and let $g(y_0)=z_0$. Then $$ (g \circ f)(x_1) = (g \circ f)(x_2) = g(y_0) = z_0. $$ Since $g \circ f$ is injective, $$ (g \circ f)(x_1) = (g \circ f_2) \implies x_1 = x_2. $$ However, this contradicts $(*)$. Therefore, $f$ must be injective.  (b) Suppose $g$ is not surjective. Then  $$ \forall y \in Y, \exists z \in Z \text{ such that } g(y) \ne z \text{         }(**).$$  Since, $g \circ f$ is surjective,  $$ \forall z \in Z, \exists x \in X \text{ such that } g(f(x)) = z \text{       } (***). $$  Let $f(x) = y$. Then,  $$ g(f(x)) = g(y) = z. $$  Because of $(***)$, this is true for all $z \in Z$, which contradicts $(**)$. Therefore, $g$ must be surjective. How do I get an animated gif to work in WPF? Counter example for a result of intersection of subspaces I am struggling with this question from Halmos's text, please ignore the imperative language.  "Suppose that $L, M$  and $N$ are subspaces of a vector space. Show that the equation    $$L \cap (M + N) = (L \cap M) + (L \cap N)$$   is not necessarily true.  Since each of these subspaces has origin in them, clearly there intersections could not be empty. I wasn't able to formulate an example where this result did n't hold. Any help would be highly appreciated.