If we have two positive compact , self adjoint operators; $A$, $B$. Is the product $AB$ a positive operator?
I want to show that for $S,T\in B(H)$ bounded operators on Hilbertspace with $S\geq 0,T\geq 0$ and $ST=TS$,  we have $S+T\geq 0$, and $ST\geq 0$.  $T\geq 0$ means $(Tx,x)\geq 0$.    To me it seems that $((S+T)x,x) = (Sx,x)+(Tx,x)\geq 0$. But for $ST$ i like some help.   There is a theorem in Rudin (12.32) that says that for every $T\in B(H)$ we have an equivalence:   $T$ is positive $\Leftrightarrow$ $T$ is self adjoint, i.e. $T=T^*$ and $\sigma(T)\subset [0,\infty)$.   Since $S, T$ are self adjoint we know that $ST=TS = S^*T^* = T^*S^*$ ($ST$ is selfadjoint since $S,T$ commute). Also I read somewhere that if $S,T $ commute then $\sigma(ST)\subset \sigma(S)\sigma(T)$, but I am not sure if this is true. Clearly I cant prove it, but if it were true by this ''Theorem'' it would follow that $ST \geq 0$.   Is there a more straightforward way?
Will getting a Software Testing Certification benefit me in my career? I have 2 years of experience in manual testing, and I am currently learning automation testing in my spare time. I was thinking that will getting a foundation/entry testing certification really help me in promotions and increase my salary or just increase my knowledge and nothing more? Even if I wish to try to get certified and pass the exams, is it too late or can I try after a few more years?  Kindly suggest! $(this) inside of AJAX success not working Methods to compute $\sum_{k=1}^nk^p$ without Faulhaber's formula As far as every question I've seen concerning "what is $\sum_{k=1}^nk^p$" is always answered with "Faulhaber's formula" and that is just about the only answer.  In an attempt to make more interesting answers, I ask that this question concern the problem of "Methods to compute $\sum_{k=1}^nk^p$ without Faulhaber's formula for fixed $p\in\mathbb N$".  I've even checked this  without finding what I want.  Rule #1: Any method to compute the sum in question for arbitrary $p$ is good, either recursively or in some manner that is not in itself a closed form solution.  Even algorithms will suffice.  Rule #2: I don't want answers confined to "only some values of $p$".  (A good challenge I have on the side is a generalized geometric proof, as that I have not yet seen)  Exception: If your answer does not generalize to arbitrary $p$, but it still generalizes to an infinite amount of special $p$'s, that is acceptable.  Preferably, the method is to be easily applied, unique, and interesting.  To start us off, I have given my answer below and I hope you all enjoy. Doubts on the axiom of union? I'm reading Guerino Mazzolla's Comprehensive Mathematics for Computer Scientists 1:     Axiom 3 (Axiom of Union) If $a$ is a set, then there is a set      $$\{x| \text{ there exists an element } b \in a \text{ such that } x\in b \}$$      This set is denoted by $\cup a$ and is called the union of $a$.    This axiom seems to take elements from inside of a set of sets, for example in the set: $a=\{\{u_1,u_2 \},\{u_3,u_4 \} \}$, there are two elements $b\in a$, then the mentioned set created by the axiom is $\{u_1,u_2,u_3,u_4 \}$.           The axiom seems to imply that the set $a$ must have a set inside of it for it to work, what if it  have an element that is not a set? (I'm not sure if this is possible, but I remember of having read something about the difference of sets and elements).         There's also an extension of my proposal, In the same book (and also in others) I've seen that the natural numbers are defined as:   $$\begin{eqnarray*}   {0}&amp;=&amp;{\emptyset} \\    {1}&amp;=&amp;{\{\emptyset\}} \\    {2}&amp;=&amp;{\{ \{\emptyset\}\}} \\    {3}&amp;=&amp;{\{ \{ \{\emptyset\}\}\}} \\   {\vdots}&amp;=&amp;{\vdots} \end{eqnarray*}$$  Then the singleton $\{ 3\}$ could be seen as $\{ \{ \{ \{\emptyset\}\}\}\}$, which could mean that   $$\cup \{ 3 \}=\cup\{ \{ \{ \{\emptyset\}\}\}\}= \{ \{ \{\emptyset\}\}\}=\{2 \}$$          Then the union could also be seen as the antecessor of a number?            Using the framework I provided, I see some ambiguity, look:   $$\{ \{ \{\emptyset\}\}\}=\{2 \}$$          I decided to interpretate it this way, but from the notation it's not clear in which pair of brackets I should stop, I guess that $\{ \{ \{\emptyset\}\}\}=\{2 \}$ but it could also be the case that $\{ \{ \{\emptyset\}\}\}=3$. I could want a singleton with a $2$ inside of it and a $3$ outside a set, no? I had Windows 8 installed on my system, which I screwed up, and now I run Ubuntu 12.04. While installing, Ubuntu used all the 461 GB of HDD space (one partition).   Now I want to make a new partition of maybe 100 GB or so, in order to install Windows 7 alongside Ubuntu (with a dual-boot). But I can't even get a partition done. I can't change the partition in GParted either - the option is disabled. How has the age of the Universe been derived from the observations made by the Planck mission? Review counts in top-bar and /review don't match So far, I thought that the number shown in the orange box in the top-bar was the number of items available for review. However, when I acually visit /review, the numbers don't match:    Note that top-bar shows 15 items (?) but the queues only add up to 8.  Since I can't think of a reason for this being by design (which, as experience shows, is not a good indicator for "this is not an SE feature) I'm tentatively labelling this a (UI) bug. How to make a circle with grid on the outside ring I made something like this:    I did this with the Ellipse Tool and the Line Tool.  Now the circle is accurate, but the lines inside are very inaccurate. Is there another tool that does the same thing but with a lot more accuracy? Solve the recurrence relation:$ T(n) = \sqrt{n} T \left(\sqrt n \right) + n$ $$T(n) = \sqrt{n} T \left(\sqrt n \right) + n$$  Master method does not apply here. Recursion tree goes a long way. Iteration method would be preferable.  The answer is $Θ (n \log \log n)$.  Can anyone arrive at the solution. Matryoshka words Possible Duplicate:             What is correct? The jury was divided or The jury were divided? I am told that the latter is accurate because all of the jury do not have the same opinion in the particular case. Then, how does the subject verb agreement fit in 'The jury is still out' ? I mean, 'The jury is out' would allude to a situation were the opinion on a matter is (are?) still divided. Prove the following $A \setminus (B\setminus C) = (A\setminus B)\cap (A \cup C)$ I want to prove the following so I decide to rewrite the RHS  $$A \setminus (B\setminus C) = (A\setminus B)\cap (A \cup C)$$ $$ A\setminus(B\setminus C) = A\setminus (B \cap C^c) = A\cap (B\cap C^c)^c = A\cap(B^c\cup C) = (A\cap B^c)\cup(A\cap C) $$ there is way that $(A\cap B^c)\cup(A\cap C)\rightarrow (A\setminus B)\cup(A\cap C) = (A\setminus B)\cap (A \cup C)$?  thanks. EDIT  LHS: $$x\in A \vee x\in A\cap C \rightarrow x\in A \wedge x\in C$$ RHS: $$x\in A \wedge x\in A\cup C \rightarrow x\in A \vee x\in C$$  how it can help me? Okay so I remember almost nothing of this show. The only thing I know is that I used to watch this show on Cartoon Network and it had a hero who used a sword to fight monsters and a girl who used bow and arrows. Also, the monsters kind of looked like computer generated and there was a book and stones which were somewhat related to the story? The Internet is awash with criticisms (or is that praise) of how Star Wars Episode VII: The Force Awakens borrows, steals, homages, references, archetype-uses or parallels from Star Wars Episode IV: A New Hope.  What are those similarities? Java FileReader encoding issue I tried to use java.io.FileReader to read some text files and convert them into a string, but I found the result is wrongly encoded and not readable at all.  Here's my environment:   Windows 2003, OS encoding: CP1252 Java 5.0   My files are UTF-8 encoded or CP1252 encoded, and some of them (UTF-8 encoded files)  may contain Chinese (non-Latin) characters.  I use the following code to do my work:     private static String readFileAsString(String filePath)     throws java.io.IOException{         StringBuffer fileData = new StringBuffer(1000);         FileReader reader = new FileReader(filePath);         //System.out.println(reader.getEncoding());         BufferedReader reader = new BufferedReader(reader);         char[] buf = new char[1024];         int numRead=0;         while((numRead=reader.read(buf)) != -1){             String readData = String.valueOf(buf, 0, numRead);             fileData.append(readData);             buf = new char[1024];         }         reader.close();         return fileData.toString();     }   The above code doesn't work. I found the FileReader's encoding is CP1252 even if the text is UTF-8 encoded. But the JavaDoc of java.io.FileReader says that:     The constructors of this class assume   that the default character encoding   and the default byte-buffer size are   appropriate.   Does this mean that I am not required to set character encoding by myself if I am using FileReader? But I did get wrongly encoded data currently, what's the correct way to deal with my situtaion? Thanks. Find the number of ways in distributing $8$ distinct balls into $6$ distinct boxes such that there is at least $1$ ball in each box.  We are well acquainted with the traditional Inclusion-Exclusion principle in solving this and it will come up as $191520$ ways.  I tried in this way:  When we distribute $8$ balls into $6$ boxes we have two possible cases  Case $1.$ There are exactly $2$ boxes with $2$ balls each  Case $2.$ There is exactly $1$ Box with $3$ balls in it.  For Case $1$, the distribution is done by choosing $2$ boxes from $6$ boxes and $2$ balls from $8$ balls and those two balls can be arranged in $2!$ ways. But I require another two balls from remaining $6$ balls to be placed in each of these two boxes which can be done in $\binom{6}{2} \times 2!$ ways. Finally remaining $4$ balls can be distributed in remaining $4$ boxes in $4!$ ways. So total number of ways for Case $1.$ is  $$\binom{6}{2} \times \binom{8}{2} \times 2! \times \binom{6}{2} \times 2! \times 4!=604800$$  For Case $2$, the number of ways is  $$\binom{6}{1} \times \binom{8}{3} \times 5!=40320$$  I am pretty sure the value obtained for Case $2.$ is correct. but I could not find where I went wrong in Case $1.$ Allow retracting a moderator flag if the post has been deleted and I don't have enough reputation to see deleted posts Currently, custom flags for moderator attention are not automatically invalidated (marked helpful) when a post gets deleted. Ordinarily, one can retract a moderator flag by going to the post, clicking &quot;flag&quot;, and then &quot;Retract Flag&quot;. This works even if the post has been deleted, provided they can see the post (they have 10,000+ rep, it's their own post, or they have answered it). However, it's not possible to self-remove a flag on a deleted post if they can't see the deleted post, because they can't get to the &quot;flag&quot; button. Is this intended? If not, can this ability please be implemented? As Aurora0001 commented, there is no server-side check, the button is just not visible. Can this please be implemented on the website, or is the fact that it's possible server-side a bug that will soon be fixed by removing it? Test Class for a Custom Controller Inaccurate shading on object Boot partition full, can't remove kernels