Let $T$ be a linear transformation on the real vector space $\mathbb R^n$ over $\mathbb R$ such that $T^2 =\mu T$ for some $\mu\in\mathbb R$ . Then which of the following is/are true?           $\|Tx\| = |\mu| \|x\|$ for all $x \in\mathbb {R^n}$     If $\|Tx\| =  \| x\| $for some non zero vector $x \in\mathbb R^n$, then $\mu=\pm1$     $T= \mu I$ where $I$ is the identity transformation on $\mathbb R^n$       If $\|Tx \|&gt;\|x\|$ for a non zero vector $x \in \mathbb R^n$, then $T$ is necessarily singular.         I am completely stuck on it. Can anybody help me please?
Let $T$ be a linear transformation $T:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ such that $T^{2}=\lambda T$ for some $\lambda\in\mathbb{R}.$ Then which of the following is/are true?  $1.\|T(x)\|=|\lambda|\|x\|$ for all $x\in\mathbb{R}^n.$  $2.$ If $\|T(x)\|=\|x\|$ for some non-zero $x\in\mathbb{R}^{n}$, then $\lambda=\pm 1.$  $3.T=\lambda I$ where $I$ is the identity transformation on $\mathbb{R}^{n}.$  $4.$ If $\|T(x)\|&gt;\|x\|$ for a nonzero vector $x\in\mathbb{R}^{n},$ then $T$ is singular.  According to me since $T^{2}=\lambda T$ we have $T$ satisfies the polynomial $x^{2}-\lambda x.$ So choices for its minimal polynomial are $x,x-\lambda,x(x-\lambda).$ So $3$rd option is clearly incorrect and $4$th option is not satisfied if $T=I$ and hence correct one. I am confused in first two options . Please help me. Thanks in advance.
A continuous function that attains neither its minimum nor its maximum at any open interval is monotone Let $f: \mathbb R\to \mathbb R$ be a continuous function such that $f$ attains neither its minimum nor its maximum at any open interval $I \subseteq  \mathbb R$ , then how to prove that $f$ is monotone ? UITableView cells strangely disappearing So...as what do we close too localized questions now? So we got  which works very well and is very nice and handy. But there is one question left, as what do we close ?  It actually helps , "too localized" was removed because it was misused most of the time:     5(A). "Too Localized" is no longer necessary since the specific   off-topic reasons now address its main use case. (new)      Too Localized was, by far, the most misused close reason in our   surveys, with both Community Managers and Moderators deeming over 50%   of randomly sampled TL closures to not have merited closure (including   on SO).      Today, TL is used helpfully on code dump questions on SO, but the new   OT reasons are where this is now correctly addressed. SO can use   "Large blocks of code with requests for debugging without meaningful   supporting info" as a specific OT reason.   But the new dialog leaves us with only four options:        Programming questions are off-topic on Meta Stack Overflow.   This question does not appear to seek input and discussion from the community. If you have encountered a problem on one of our sites, please describe it in detail.   Changes to the system or to the circumstances affecting the asker have rendered it obsolete. If you encounter a similar problem, please post a new question.   Other (add a comment explaining what is wrong)      Only the "other" reason does very well fit on "too localized" questions now, doe we close it with that? + or @ mark after running 'ls -al' In Mac OS X, run 'ls -al' gives me something like this.   drwxrwxrwx+  4 smcho  staff     136 May  5 09:18 Public drwxr-xr-x+  6 smcho  staff     204 Feb  1  2010 Sites drwxrwxrwx   9 smcho  staff     306 Feb  2  2010 backup drwxr-xr-x@ 36 smcho  staff    1224 Sep  4 22:51 bin    What's the + or @ at the end of the first column means? Is this unique to Mac, or common in UNIX?   ADDED  After Michael Mrozek's answer, I ran 'ls -ale' to get the following.   drwx------+ 66 smcho  staff    2244 Aug 30 13:40 Library  0: group:com.apple.sharepoint.group.3 allow search  1: group:everyone deny delete drwxr-xr-x   3 smcho  staff     102 Sep  4 15:01 Mail drwx------+ 13 smcho  staff     442 Aug 28 17:55 Movies  0: group:everyone deny delete drwx------+  6 smcho  staff     204 Jul  9 09:37 Music  0: group:everyone deny delete drwx------+ 11 smcho  staff     374 Aug 28 16:55 Pictures  0: group:everyone deny delete drwxr-xr-x   3 smcho  staff     102 Mar 18 15:43 Projects drwxrwxrwx+  4 smcho  staff     136 May  5 09:18 Public  0: group:everyone deny delete drwxr-xr-x+  6 smcho  staff     204 Feb  1  2010 Sites  0: group:everyone deny delete   What those appended messages mean? Why do I have them for some of the files? I don't remember doing anything particular for them. Is my C1/D visa valid for transit in the US, if I want to go on a vacation to Mexico? How to make a model cast shadows onto the world texture/background? In most of my models, I usually end up setting a world background using an HDRI image of some kind or another. This has always been really useful for lighting in general or as a background for my final render.  But what if I want my model to look like it is interacting or touching part of the background? For example, in the image below that I found, the modeled objects look like they're part of the scene, and even appear to cast shadows on the ground. How can I do this? Creating a Movie from a Series of Plots in R How to display special German characters "ÄÜÖß" in a map? I gave a scammer my telephone number, address, and name. How bad is it? Layers with different directional drop shadows in Photoshop I am wondering how to have 2 different directional drop shadows in the one file with Photoshop. e.g. 2 different type layers — one has a drop shadow at 90 degrees, the other at -45 degrees.  Currently it seems Photoshop only lets me have 1 universal direction for the whole file, which I find really pointless.   I know how to apply different drop shadows and merge down layers to achieve the same thing, but its tedious and not exactly streamlined for making future changes to that layer.  How is this achieved? Why can't I connect to Ubuntu One? I have purchase new laptop from Dell with ubuntu 12.04 preinstalled. while connecting to Ubuntu One following occurs and I am not able to sync with Ubuntu One.   WebClientErr u'Error downloading https://one.ubuntu.com/api/account/ - server replied:Not       Found' u'&lt;!DOCTYPE HTML PUBLIC "-//IETF//DTD HTML 2.0//EN"&gt;\n&lt;html&gt; &lt;head&gt;\n&lt;title&gt;404 Not Found&lt;/title&gt;\n&lt;/head&gt;&lt;body&gt;\n&lt;h1&gt; Not Found&lt;/h1&gt;\n&lt;p&gt;The requested URL /api/account/ was not found on this server. &lt;/p&gt;\n&lt;hr&gt;\n&lt;address&gt;Apache/2.2.22 (Ubuntu) Server at one.ubuntu.com Port 443 &lt;/address&gt;\n&lt;/body&gt;&lt;/html&gt;\n' I was browsing PhD position offers in Europe on different websites and I was quite surprised to find that several offers contained sentences such as     University of X value equality and diversity. We strongly encourage women and BAME applicants for this position.    where BAME stands for Black, Asian and Minority Ethnicities; or     The research group aims to increase the number of women in scientific positions. Female candidates are therefore encouraged to apply.   Is it considered normal in the academic environment to discriminate between  candidates according to their race and/or gender for "egalitarian" reasons?   How is the choice of a candidate based on their gender and/or race considered "egalitarian"? Is there a reason why "gn" in "reigning" is pronounced /n/ while in "regnant" it is pronounced [gn]? Both reigning and regnant are related to the same Latin noun, regnum.  Why is the ‹gn› spelling pronounced [n] in the first word but [gn] in the second? What are Covering Indexes and Covered Queries in SQL Server? Can you explain the concepts of, and relationship between, Covering Indexes and Covered Queries in Microsoft's SQL Server? How to determine callee function name in a script To make it short, doing something like:  -bash$ function tt {  echo $0; }  -bash$ tt   $0 will return -bash, but how to get the function name called, i.e. tt in this example instead? I'm taking Calculus II next semester so I don't have all the tools from it to back me up but something came up in a discussion this evening that asked me to prove that the following sequence converges. As I'm new to formal math I'm hoping someone can verify my proof, especially step $6$. I must prove that $$\lim_{n\to\infty}P_n=b$$ where $$P_{n+1} = P_n + \frac{b-P_n}{2}$$ $$P_0 = a$$ and $a,b \in \mathbb{R}$ such that $a&gt;b$ My proof goes as follows:   We must prove that $$\lim_{n\to\infty}P_n=b$$  That is, $\forall\epsilon&gt;0$ there exists a $\delta$ such that there exists an $n$ satisfying $$n&gt;\delta \iff b-P_n&lt;\epsilon$$  First, let us prove that $P_n$ is bounded by $b$ by contradiction. Let us choose an $i$ such that $P_i &lt; b$ and $P_{i+1}&gt;b$. Then we have $P_{i+1} = P_i + \Delta P$ where $\Delta P &gt; b-P_i$. but this is impossible because by the definition of the recurrence $\Delta P = \frac{b-P_i}{2} &lt; b-P_i$. Therefore, $P_i$ is bounded from above by $b$.  Additionally, it is trivial to show that $P_{i+1} &gt; P_i$.  With this in mind, choose an arbitrary $\epsilon$. We need to show that there exists an $n$ such that $P_n &gt; b-\epsilon$. Assume the contrary, that $P_n$ is bounded by $\epsilon$.  Using a similar argument to what which was used before, assume $P_n$ is bounded by $\epsilon$ and then there exists a $P_i$ such that $P_{i+1}\not &gt; P_i$. But this results in a contradiction with (4)  Therefore, $P_n$ is not bounded by $\epsilon$ and we have proved what we want to prove. I created a bitmap with transparent background but in the Windows Picture Viewer I see the white background and even if I reopen it in Photoshop CS3. how do I make sure the white colour in bitmap is really transparent or see if contain any transparent colour? Shortcut for toggling between different windows of same app? If $\dfrac{4x^2-1}{4x^2-y^2}$ is an integer, then it is $1$ The problem is the following:     If $x$ and $y$ are integers such that $\dfrac{4x^2-1}{4x^2-y^2}=k$ is   also an integer, does it implies that $k=1$?   This equation is equivalent to $ky^2+(1-k)4x^2=1$ or to $(k-1)4x^2-ky^2=-1$. The first equation is a pell equation (if $k$ is a perfect square) and the second is a pell type equation (if $k-1$ is a perfect square). I've tried setting several values of $k$ to get some solutions but i got nothing. I'm starting to think that $k$ must be $1$. The following identity is true for any given $x \in [-1,1]$: $$\arcsin(x) + \arccos(x) = \frac{\pi}{2}$$  But I don't know how to explain it. I understand that the derivative of the equation is a truth clause, but why would the following be true, intuitively?  $$\int^{x}_{C1}\frac{1\cdot dx}{\sqrt{1-x^{2}}} + \int^{x}_{C2}\frac{-1 \cdot dx}{\sqrt{1-x^{2}}} =\\  \arcsin(x) - \arcsin(C1) + \arccos(x) - \arccos(C2) = 0 \\  \text{while } \arcsin(C1) + \arccos(C2) = \frac{\pi}{2}$$  I can't find the right words to explain why this is true?    Edit #1 (25 Jan, 20:10 UTC): The following is a truth clause: $$ \begin{array}{ll} \frac{d}{dx}(\arcsin(x) + \arccos(x)) = \frac{d}{dx}\frac{\pi}{2} \\ \\ \frac{1}{\sqrt{1-x^{2}}} + \frac{-1}{\sqrt{1-x^{2}}} = 0 \end{array} $$  By integrating the last equation, using the limits $k$ (a constant) and $x$ (variable), I get the following:  $$ \begin{array}{ll} \int^x_k\frac{1}{\sqrt{1-x^{2}}}dx + \int^x_k\frac{-1}{\sqrt{1-x^{2}}}dx = \int^x_k0 \\ \\ \arcsin(x) - \arcsin(k) + \arccos(x) - \arccos(k) = m  \text{ (m is a constant)}\\ \\ \arcsin(x) + \arccos(x) = m + \arcsin(k) + \arccos(k) \\ \\ \text{Assuming that } A = m + \arcsin(k) + \arccos(k) = \frac{\pi}{2} \text{ ,for } x \in [-1,1] \end{array} $$ Using Calculus, why is that true for every $x \in [-1,1]$?    Edit #2:  A big mistake of mine was to think that $\int^x_k0 = m \text{ (m is const.)}$, but that isn't true for definite integrals.  Thus the equations from "Edit #1" should be as follows: $$ \begin{array}{ll} \int^x_k\frac{1}{\sqrt{1-x^{2}}}dx + \int^x_k\frac{-1}{\sqrt{1-x^{2}}}dx = \int^x_k0 \\ \\ \arcsin(x) - \arcsin(k) + \arccos(x) - \arccos(k) = 0\\ \\ \arcsin(x) + \arccos(x) = \arcsin(k) + \arccos(k) \\ \\ A = \arcsin(k) + \arccos(k) = \frac{\pi}{2} \text{ ,for } x \in [-1,1] \end{array} $$