Prove that if A has rank r &gt; 0, then A has an r x r invertible matrix. I have tried looking at the properties of invertible matrices, but I can't prove those qualities using just the rank of matrix A.
Let $A$ be a $m \times n$ matrix with entries from some field $F$. Define the determinant rank of $A$ to be the largest possible size of a nonzero minor, i.e. the size of the largest invertible square submatrix of $A$. It is true that the determinant rank is equal to the rank of a matrix, which we define to be the dimension of the row/column space.  It's not difficult to see that $\text{rank} \geq \text{determinant rank}$. If some submatrix of $A$ is invertible, then its columns/rows are linearly indepedent, which implies that the corresponding rows/columns of $A$ are also linearly indepedent.  Is there a nice proof for the converse?
Change .jpg quality of terminal screenshots I have $a,b,c\in\mathbb{Q}$ not all zero. ($a^2+b^2+c^2\ne 0$), I want to show that the following determinant is then non-zero. I failed to arrive at an appropriate form of the polynomial. Help please. $$\left|\begin{bmatrix} a &amp; 2c &amp; 2b\\b &amp; a &amp; 2c\\ c &amp; b &amp; a\end{bmatrix}\right| = a^3+2 b^3-6 a b c+4 c^3$$    Second question, what is the easiest way to argue that $\{1,\sqrt[3]{2},(\sqrt[3]{2})^2\}$ is linearly independent in $\mathbb{Q}$?    Motivation: Prove that $\mathbb{Q}[\sqrt[3]{2}] = \{a+b\sqrt[3]{2}+c(\sqrt[3]{2})^2\;|\;a,b,c\in\mathbb{Q}\}$ forms a field.   Proof: Since $\mathbb{Q}[\sqrt[3]{2}] \subset \mathbb{R}$, we prove $\mathbb{Q}[\sqrt[3]{2}]$ is a subfield of $(\mathbb{R},+,\cdot)$    $\forall (a+b\sqrt[3]{2}+c(\sqrt[3]{2})^2)\in \mathbb{Q}[\sqrt[3]{2}]\backslash\{0\}.$ We want to find $(a+b\sqrt[3]{2}+c(\sqrt[3]{2})^2)$ such that    $(a+b\sqrt[3]{2}+c(\sqrt[3]{2})^2)(d+e\sqrt[3]{2}+f(\sqrt[3]{2})^2) =$ $ (ad+2ec+2bf)+(ae+bd+2cf)\sqrt[3]{2}+(af+cd+be)(\sqrt[3]{2})^2 = 1$    Since $\{1,\sqrt[3]{2},(\sqrt[3]{2})^2\}$ is linearly independent (?) over $\mathbb{Q}$, we show there is unique solution to: $$\begin{bmatrix} a &amp; 2c &amp; 2b\\b &amp; a &amp; 2c\\ c &amp; b &amp; a\end{bmatrix} \cdot \left[\begin{array}{l l} d\\e\\f \end{array}\right] = \left[\begin{array}{l l} 1\\0\\0 \end{array}\right] $$ Which is equivalent in showing the determinant is non-zero $$\left|\begin{bmatrix} a &amp; 2c &amp; 2b\\b &amp; a &amp; 2c\\ c &amp; b &amp; a\end{bmatrix}\right| = a^3+2 b^3-6 a b c+4 c^3=(?)$$    By subfield test, 1)2)3)4) is enough to say that $(\mathbb{Q}[\sqrt[3]{2}],+,\cdot)$ is a subfield of $(\mathbb{R},+,\cdot)$ therefore a field.  EDIT: If you have shorter way that prove the proposition without touching my 2 questions, that is even better. Probability of sampling with and without replacement In sampling without replacement the probability of any fixed element in the population to be included in a random sample of size $r$ is $\frac{r}{n}$. In sampling with replacement the corresponding probability is $\left[1- \left(\frac{1}{1-n}\right)^r\right]$.  Please help me show how this is proved. I was browsing flight price at Singapore Airlines website, and a one-way flight from SIN to SFO cost this much:    That's quite expensive. I tried browsing around and found that a return flight from SFO to SIN and back to SFO cost much cheaper:    It's really weird because the second ticket includes the exact same flight that cost 1900+, but it's almost half the price. The reason for this counter-intuitive situation is explained in the answers this question   If I am located in Singapore and my intention is to fly one-way to SFO, is it safe to book the flight in the second picture and only board SFO-SIN leg? Will they deny me because I miss the SIN-SFO flight? I have filed my 2012 tax return jointly with my wife, along with my wife's ITIN application, back in late March.  In April, I got a letter for IRS ITIN department saying that we should send my wife's original passport to them. So we sent my wife's passport on May 15th. USPS tracking number confirmed that IRS received it on May 20th.  However, last week (end of June) we have received another letter from IRS ITIN department saying that our ITIN application has been rejected, rejected because we have not sent the original identification document that they requested!  Okay, let's put our tax return and ITIN aside. Where is my wife's passport? Right now my wife does not have her personal identification, nor proof of legal status in U.S. Could IRS have lost it?  I tried to call the IRS several times. One time I waited for 1 hour and 20 minutes to talk to a person, and that person said he has only 10 minutes to talk before he has to leave. He said they'll investigate and someone(?) will call us back soon, but never called back. I wrote them a letter, but no response. I tried calling several times more, but machine is saying 'sorry too many calls right now, we can't accept your call'.  What can I do? Why cant I navigate to lightning component in communities? Im on a lightning community, i have a lightning component and i want to open another component with this code:  navigateToMyComponent : function(component, event, helper) {     console.log('1');     var evt = $A.get("e.force:navigateToComponent");     console.log('2');     evt.setParams({         componentDef : "c:CompetenciaComponent"     });     console.log('3');     evt.fire();     console.log('4'); }   this its the console log:   NOTHINGS HAPPEND AND I DONT KNOW WHY Do we have different set of badges in different SE sites. eg. I don't see any 'Pundit' badge in badges popup in Hinduism SE.  Also, if the above is true, then can we include the left badges in a SE site at a later stage. Why is it considered so dangerous to travel in South Africa? As, the above link shows, other countries have similar murder rates but they are not in the radar of international travelers.  For example, as seen in the list, Puerto Rico has similar statistics. But, personally, I never heard of Puerto Rico to be a dangerous place to travel.  Why is it considered so dangerous to travel in South Africa? Is that due to the incompetent law enforcement agencies? On this site I see hundreds if not a thousand questions being asked about issues with a Wireless LAN adapter not being detected/recognized by Ubuntu. Some mainstream adapters appear to be utterly broken in Linux support, while others just require some simple steps to get it to work.  The options to get it to work appear to be infinitely huge and I have no clue where to get started!  What is considered 'best practice'? What order of things to try is most efficient in terms of time, effort and imposes minimal risk to my Ubuntu installation? Please help me save time and guide me what to do, step by step. Can $(X_1,X_2) \cap (X_3,X_4)$ be generated with two elements from $k[X_1,X_2,X_3,X_4]$? Can $(X_1,X_2) \cap (X_3,X_4)$ be generated with two elements in the ring $R=k[X_1,X_2,X_3,X_4]$? Can it be generated with three elements? (Here $k$ is a field.)    Thanks for any help. I'm porting a module that currently works with Blocks/BlockForms to Paragraphs/NodeForms. I'm inserting an embedded form inside a Paragraph inside a Node Edit page using a hook_field_widget_entity_reference_paragraphs_form_alter. The general structure where the form gets inserted is: $form['field']['widget'][0]['settings'][form here]  After this the form displays as expected on the node edit page. The inserted form goes through a number of #process steps. The element in question is a select form that when the button underneath it is pressed, is supposed to retrieve some data via a custom submit handler and place it on the page with ajax:  The array for this element looks like this:  You can see the custom submit handler that is supposed to fire when 'Select Segment Set' is pressed After debugging, I've found that the entire NodeForm appears to be submitted to the backend, and the only handler that is found is from some other field that uses ajax. In short 'updateSegmentSet' is never discovered as a submit handler. When I debug the original module based on block/blockforms, I do see the correct submit handler being discovered. Now in the latter case, of course a blockForm is submitted to the back end with the correct plugin. Given that this is basically a form within a form within a form, is there something special I have to do to get the custom submit handler to run? I've read a few things online e.g. the 'settings' key and '#tree' = true. EDIT The Ajax property looks like this:       '#ajax' =&gt; [         'callback' =&gt; [$this, 'updateSegmentSetAjax'],         'wrapper' =&gt; $wrapper_id,       ],    public function updateSegmentSetAjax(array &amp;$form, FormStateInterface $form_state) {     $button = $form_state-&gt;getTriggeringElement();     // Go one level up in the form, to the widgets container.     return NestedArray::getValue($form, array_slice($button['#array_parents'], 0, -2));   }  As best I can tell, this is never called either. So I am using  package for code examples. But there is a problem. When I copy the code examples from the final PDF document, they are usually not the same - the indentation is gone, there are random spaces after parenthesis, etc. Does anyone know how to make sure that the code that I copy from the final document is the same as I wrote it in the LaTeX document?  Here is what I mean. Compile this document:  \documentclass[12pt,oneside]{memoir}  \usepackage{listings} \usepackage[T1]{fontenc} \usepackage{xcolor} \usepackage{textcomp}  \definecolor{codebg}{HTML}{EEEEEE} \definecolor{codeframe}{HTML}{CCCCCC}  \lstset{language=Awk} \lstset{backgroundcolor=\color{codebg}} \lstset{frame=single} \lstset{framesep=10pt} \lstset{rulecolor=\color{codeframe}} \lstset{upquote=true} \lstset{basicstyle=\ttfamily} \lstset{showstringspaces=false}  \begin{document}  This code example prints out all users on your system:  \begin{lstlisting}[language=c] #include &lt;stdio.h&gt; #include &lt;stdlib.h&gt; #include &lt;string.h&gt;  #define MAX_LINE_LEN 1024  int main() {     char line[MAX_LINE_LEN];     FILE *in = fopen("/etc/passwd", "r");     if (!in) exit(EXIT_FAILURE);      while (fgets(line, MAX_LINE_LEN, in) != NULL) {         char *sep = strchr(line, ':');         if (!sep) exit(EXIT_FAILURE);         *sep = '\0';         printf("%s\n", line);     }     fclose(in);     return EXIT_SUCCESS; } \end{lstlisting}  \end{document}   Now compile it, open it, select and copy the code example and paste it in a text editor. Here is the result I get:  # include &lt;stdio .h&gt; # include &lt;stdlib .h&gt; # include &lt;string .h&gt; # define MAX_LINE_LEN 1024 int main () { char line [ MAX_LINE_LEN ]; FILE *in = fopen ("/ etc / passwd ", "r"); if (! in) exit ( EXIT_FAILURE ); while ( fgets (line , MAX_LINE_LEN , in) != NULL ) { char * sep = strchr (line , ':'); if (! sep ) exit ( EXIT_FAILURE ); * sep = '\0 '; printf ("%s\n", line ); } fclose (in ); return EXIT_SUCCESS ; }   Jeez! What's this... The indentation is gone. There are spaces in included headers &lt;stdio .h&gt;, there are spaces after parenthesis [ MAX_LINE_LEN ].  Anyone knows how to fix this? I'd like the code examples in my book to be copy/paste-able into a text editor so people can try them out easily. Find a recurrence relation for the number of $n$-digit binary sequences with no pair of consecutive $1$s.  I know my base case:$$a_1 = 2$$ for either $1$ or $0$. Normally to construct the recurrent relation I would say fill in $n-1$ spaces then add a $1$ or $0$, which is simply:$$a_n = 2\cdot a_{n-1}$$However, I don't believe this is right, because this doesn't take into account that there can't be any consecutive $1$s.  Even when I use the base case of $2$ digits:$$a_2 = 3$$for either $01$, $10$, or $00$, I still get the recurrence relation of:$$a_n = 3\cdot a_{n-2}$$which also does not account for $01$ and $10$ occurring consecutively.   So how would I express this? Proof of this result related to Fibonacci numbers: $\begin{pmatrix}1&1\\1&0\end{pmatrix}^n=\begin{pmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{pmatrix}$? $$\begin{pmatrix}1&amp;1\\1&amp;0\end{pmatrix}^n=\begin{pmatrix}F_{n+1}&amp;F_n\\F_n&amp;F_{n-1}\end{pmatrix}$$  Somebody has any idea how to go about proving this result? I know a proof by mathematical induction, but that, in my opinion, will be a verification of this result only. I tried to find through the net, but in vain, if someone has some link or pointer to its proof, please provide, I'll appreciate that.  Thanks a lot! Semigroups and solutions of equation According to , POST is used to create a resource:  The POST method is used to request that the origin server accept the entity enclosed in the request as a new subordinate of the resource identified by the Request-URI in the Request-Line.  According to , PUT is used to create or replace a resource:  The PUT method requests that the enclosed entity be stored under the supplied Request-URI. If the Request-URI refers to an already existing resource, the enclosed entity SHOULD be considered as a modified version of the one residing on the origin server. If the Request-URI does not point to an existing resource, and that URI is capable of being defined as a new resource by the requesting user agent, the origin server can create the resource with that URI.  So which HTTP method should be used to create a resource? Or should both be supported? Using have gone to / been to / been in I understand the difference between these constructions: "has gone to", "has been to", "has been in":   "has gone to" &mdash; there or on his way to; "has been to" &mdash; someone has been there but he is on his way back; "has been in" &mdash; someone is still there;   But now I am puzzled by next statements:   Have you ever been to / in Greece? &mdash; Yes, it's a very beautiful country; When did you last go to Canada? &mdash; I've never been to Canada.   If I use "has you ever gone to Greece" construction in the first case or "I have never gone to Canada" in the second statement will it be uncorrect? Or just will sound odd? May I say "I have never been in Canada" in the second sentence. Or in these case all of the variants are possible, the problem is in how people usually say it (in what how it perfectly sounds)? Can I get tourist visa on arrival at Dubai airport? I was reading an example about describing homomorphisms and I'm having a bit of trouble with this one.  Show there is no group homomorphism $f : \mathbb{Z}_{10} \to \mathbb{Z}_{25}$ such that $f(1) = 3$. The first step of the proof is  $|1| = 10$ in $\mathbb{Z}_{10}$, so $|f(1)| \in\{ 1, 2, 5, 10\}$.  Does this mean that $|f(1)|$ can be $1$, $2$, $5$ or $10$? I see that $|1| = 10$ since $1$ is a generator of $\mathbb{Z}_{10}$, and that $|f(1)|$ seems to be a factor of $10$. If $f(1)$ is the image of $1$ under $f$, how would I find the order of $f(1)$?  Thanks for your help. What GUI applications are there to assist in generating graphics for TeX? I found  to be a useful tool to help generate code for graphs. I suppose there must be other tools to help draw diagrams such as Sets/Venn Diagrams or Electrical Circuits (Logic Gates). Which ones would you suggest?