I'm having some trouble with a question in my linear algebra textbook:     Let $A$ and $B$ be positive definite matrices, and let $Q$ be a unitary matrix.   Prove that if $A=BQ$, then $A=B$.   I tried to play around with the given equality, but I'm not getting any results. Tips would be appreciated:)
Given $A$ and $B$ positive-definite matrices and $Q$ unitary matrix, prove that if $A = BQ$, then $A=B$. $Q$ is unitary, so $QQ^*=I$ If $A$ and $B$ are positive-definite, than $A=A^*$ and $B=B^*$. $A^*=(BQ)^*=Q^*B^*$ $A^2=AA=AA^*=(BQ)(Q^*B^*)=B(QQ^*)B^*=BIB^*=B^2$ $A^2=B^2$. I don't know how to use the fact that $A$ and $B$ are positive-definite to finish the proof.
Transforming the dummy values to be able to take logs I have a panel data model with double-log functional form. I have 4 variables, one of which is a dummy. What is the best way to transform the values of 0 for my dummy to be able to take natural logs (since you can't take logs from zero)? I have found 2 options so far, but I dont know which is the best (or perhaps there is a 3rd option).  1) Transform all the values of 0 into a very small value (0.00000000001) and then take the logs. I'm not sure this wouldn't still change the outcome of my model quite a bit. 2) Change the values of 0 into 1 and the dummy values of 1 into e (the base of natural logarithm). The log of this newly defined dummy then takes on the values of zero and one and the interpretation of my betas would remain the same. Elegant proofs that similar matrices have the same characteristic polynomial? I'm having some difficulty understanding how to interpret variable importance output from the Random Forest package. Mean decrease in accuracy is usually described as "the decrease in model accuracy from permuting the values in each feature".   Is this a statement about the feature as a whole or about specific values within the feature? In either case, is the Mean Decrease in Accuracy the number or proportion of observations that are incorrectly classified by removing the feature (or values from the feature) in question from the model?   Say we have the following model:  require(randomForest) data(iris) set.seed(1) dat &lt;- iris dat$Species &lt;- factor(ifelse(dat$Species=='virginica','virginica','other')) model.rf &lt;- randomForest(Species~., dat, ntree=25, importance=TRUE, nodesize=5) model.rf varImpPlot(model.rf)     Call:  randomForest(formula = Species ~ ., data = dat, ntree = 25,  proximity = TRUE, importance = TRUE, nodesize = 5)  Type of random forest: classification Number of trees: 25 No. of variables tried at each split: 2          OOB estimate of  error rate: 3.33% Confusion matrix:           other virginica class.error other        97         3        0.03 virginica     2        48        0.04     In this model, the OOB rate is rather low (around 5%). Yet, the Mean Decrease in Accuracy for the predictor (Petal.Length) with the highest value in this measure is only around 8.   Does this mean that removing Petal.Length from the model would only result in an additional misclassification of 8 or so observations on average?   How could the Mean Decrease in Accuracy for Petal.Length be so low, given that it's the highest in this measure, and thus the other variables have even lower values on this measure? If a function has a finite limit at infinity, does that imply its derivative goes to zero? I've been thinking about this problem: Let $f: (a, +\infty) \to \mathbb{R}$ be a differentiable function such that $\lim\limits_{x \to +\infty} f(x) = L &lt; \infty$. Then must it be the case that $\lim\limits_{x\to +\infty}f'(x) = 0$?  It looks like it's true, but I haven't managed to work out a proof. I came up with this, but it's pretty sketchy:  $$ \begin{align} \lim_{x \to +\infty} f'(x) &amp;= \lim_{x \to +\infty} \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \\ &amp;= \lim_{h \to 0} \lim_{x \to +\infty} \frac{f(x+h)-f(x)}{h} \\ &amp;= \lim_{h \to 0} \frac1{h} \lim_{x \to +\infty}[f(x+h)-f(x)] \\ &amp;= \lim_{h \to 0} \frac1{h}(L-L) \\ &amp;= \lim_{h \to 0} \frac{0}{h} \\ &amp;= 0 \end{align} $$  In particular, I don't think I can swap the order of the limits just like that. Is this correct, and if it isn't, how can we prove the statement? I know there is  already, but I think this is different in two aspects. First, that question assumes that $\lim\limits_{x \to +\infty}f'(x)$ exists, which I don't. Second, I also wanted to know if interchanging limits is a valid operation in this case. The no. of all possible matrices of order $3\times 3$ with each entry $0$ and $1$ and $ -1$. How does the built-in debugger decide what to highlight? It's a shame to type out an entire question, press post and then:     Sorry, you are only allowed to ask 50 questions in a 30 day period   Could you put some sort of notice on the page itself before submitting it to let you know the question won't be going through?  Or perhaps tell me how many questions I've asked in the last 30 days if it's >0?  Edit  This is not a complaint or statement about the limit, just a request that the error handling be done in a slightly better way to save people a lot of time. Let $p$ be a prime number. Let $R$ be the subset of all rational numbers $m / n$ such that $n \neq 0$ and $n$ is not divisible by $p$. Show that $R$ is a ring. My Proof-trying. $R1$. Associativity for under addition is clear. $R2$. $\dfrac {m} {n}+0=0+\dfrac {m} {n}$, so there exist the identitiy element for under addition. $R3$. $\dfrac {m} {n} + \dfrac {-m} {n}=\dfrac {-m} {n} +  \dfrac {m} {n}$, so there exist inverse element. $R4$. $\dfrac {m} {n} + \dfrac {m'} {n'}=\dfrac {m'} {n'}+\dfrac {m} {n}$, so commutativity. So $R$ is an additive abelian group. $R5$. $\dfrac {m} {n} ( \dfrac {r} {s} + \dfrac {t} {v})=\dfrac {mr} {ns} + \dfrac {mt} {nv}$ and $( \dfrac {r} {s} + \dfrac {t} {v})\dfrac {m} {n}=\dfrac {rm} {sn}+\dfrac {r} {s}+\dfrac {tm} {vn}$ $R6.$ $R$ is associativity for under multiplication, clear. $R7.$ $1 \dfrac {m} {n}=\dfrac {m} {n} 1 = \dfrac {m} {n}$, so there exist identity element for under multiplication. Can you check my answer? If false, can you write true answer? Pre-image of a measurable set A is always measurable? Let $f$ be a continuous function on a set $E$.  Is it always true that $f^{-1}(A)$ is always measurable if $A$ is measurable?  I say no.  We know that $\bar{\psi(x)}:=\frac{\phi(x)+x}{2}$ where $\phi$ is the Cantor (or Cantor-Lebesgue) function and $\bar{\psi}:[0,1] \rightarrow [0,1]$.  We know this function maps a measurable subset of $C$ (the Cantor set) into a non-measurable set $W$.  A few observations about $\bar{\psi(x)}$.  It is strictly increasing and continuous, so it has a continuous inverse $\bar{\psi(x)}^{-1}$.  Thus $\bar{\psi(x)}^{-1}(W)=c \subset C$; that is, $\bar{\psi(x)}^{-1}(W)$ is non-measurable, but c is.  Is this correct? In the review page, it shows stats of recent reviewers. But it is showing wrong stats. Because today (on Monday) it shows week count 53 (on Meta Stack Overflow) which is impossible as there is a limit of 20 reviews per day.    PS: This bug   and . But the bug has occurred again. In JavaScript, the "this" operator can refer to different things under different scenarios.   Typically in a method within a JavaScript "object", it refers to the current object.  But when used as a callback, it becomes a reference to the calling object.  I have found that this causes problems in code, because if you use a method within a JavaScript "object" as a callback function you can't tell whether "this" refers to the current "object" or whether "this" refers to the calling object.  Can someone clarify usage and best practices regarding how to get around this problem?     function TestObject() {             TestObject.prototype.firstMethod = function(){                       this.callback();                       YAHOO.util.Connect.asyncRequest(method, uri, callBack);              }              TestObject.prototype.callBack = function(o){               // do something with "this"               //when method is called directly, "this" resolves to the current object               //when invoked by the asyncRequest callback, "this" is not the current object               //what design patterns can make this consistent?               this.secondMethod();             }             TestObject.prototype.secondMethod = function() {              alert('test');             }         } If the matrix $A^3 = 0$, is $A$ singular? 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? Is there a term for grammatical mistakes as a result of trying too hard? Should I install programs from a source tarball (`.tar.gz`), from the Ubuntu Software Centre, or from elsewhere? is exactly what I'm looking for. I want my description environment to produce this:  Label is long   text  lab2            blalballalaa label3          possible second rule label four      hello world   Someone offered this:  \newenvironment{mydescription}[1]                                                   {\begin{list}{}%    {\renewcommand\makelabel[1]{##1:\hfill}%    \settowidth\labelwidth{\makelabel{#1}}%    \setlength\leftmargin{\labelwidth}    \addtolength\leftmargin{\labelsep}}}   {\end{list}}   Which works great (the parameter is the longest label), except I think the enumitem should be able to do this. Unfortunately, the documentation for it is extremely confusing to me. Could someone show me how to achieve the same result with enumitem? In Minecraft, there are a lot of cool sounds. I decided to find there location and move them to my desktop. I have Ubuntu, and if when I remove them from that folder will Minecraft be unable to play them or is it smart enough to figure out that they are gone and download them again. Before minecraft was installed, it had to download the files. I even moved my favorite song (creative 6) from the sounds. The sounds are not labeled and have seemingly random names like b0b12118a97ddd733a50e9382e91a21652568641 and how will I know what to call them that will be recognized without digging through the index if Minecraft fails to download them? Or do I not have to worry because it will download them again upon launch?  Note: Why does it not label them and call them random names? As far as I know, ping needs to create a  (which needs either root access or cap_net_raw capabilities). From my understanding the trend these last years has been to remove  binaries and replaced them with . However when I look at the ping binary on my Fedora 32, it doesn't look to have any: $ ls -la $(which ping) -rwxr-xr-x. 1 root root 82960 May 18 10:26 /usr/bin/ping $ sudo getcap -v $(which ping) /usr/bin/ping $     Does ping need to open raw socket on fedora? Or is there another way to give it the permission to open a raw socket? Using a foreign driver's license in California when staying with a J-1 visa I'm planning to attend a Californian university as a visiting scholar (not paid by the university) on a J-1 visa, for a total of four weeks; a colleague of mine will do the same, but for 12 weeks. I've been told (personally, ie. no citable sources) that, if you're staying on a J-1, you're effectively a California resident for the time you're there, and, therefore, need a Californian driver's license.  I've read up on this and found  that states:     If you are a visitor in California over 18 and have a valid driver   license from your home state or country, you may drive in this state   without getting a California driver license as long as your home state   license remains valid.      If you become a California resident, you must get a California driver   license within 10 days. Residency is established by voting in a   California election, paying resident tuition, filing for a homeowner's   property tax exemption, or any other privilege or benefit not   ordinarily extended to nonresidents.   I'm not planning to vote in any elections, nor will I own a home there, but I'm not sure what would constitute "privileges or benefits not ordinarily extended to nonresidents".  Do we need to get driver's licenses in California, or can we use our foreign licenses? How about day trips to Arizona or Nevada?  Note: this seems to be related to , but my question is specifically about staying there on a J-1. Detecting structured data in Google Webmaster Tools