Let $X$ be a normed space and $Y$ be a finite dimensional subspace of $X$. Show that there is a projective $P\in B(X)$ such that $Im P=Y$. Hint: First Solve for $dimY=1$ then generalize the solution for any finite dimensioanl spaces. Here since $Y$is  finite dimensional subspace of a normed space so it is closed. However we are not given that $X$ is banach, so we cannot use any theorem about projective in banach spaces!. **The question stated in  is not quite the same, I think. Can someone, suggest a way to solve the question using the hint.
I am reading  that finite-dimensional subspaces of normed vector spaces have closed direct sum complements. This is the proof:      Let $\{e_1, ..., e_n\}$ be a basis for $\mathcal M$. Every $x \in   \mathcal M$ has then a unique representation $$x = \alpha_1(x)e_1 +...+  \alpha_n(x)e_n.$$ Each $\alpha_i$ is a continuous linear functional on    $\mathcal M$ (a linear map from  finite dimensional space is always    continuous) which extends to a member of $\mathcal X^*$,  by the    Hahn-Banach theorem ($\mathcal X^*$ is the dual of $\mathcal X$). Let    $\mathcal N$ be the intersection of the null spaces of these exten­sions. Then $\mathcal X = \mathcal M\oplus \mathcal N$.   I follow it until the last sentence. Clearly $\mathcal N$ is closed, as it is the intersection of closed sets (the inverse of $0$ under any continuous function, and in particular any linear functional on a vector space, is closed). Also it is easy to see that $\mathcal M \cap \mathcal N= \{0\}$, since if some $m\in\mathcal M$ is in every null space, it must have the zero coefficient for every basis vector, and so it must be zero.  My question: Why do we have $\mathcal X =\mathcal M + \mathcal N$? This seems to only hold if the extension given by Hahn-Banach is zero off of $\mathcal M$, but I see no reason why this should be the case.
Is there any way other than instanceof operator for object type comparison in java? I have a Git repo that I have deleted four files from using rm (not git rm), and my Git status looks like this:  #    deleted:    file1.txt #    deleted:    file2.txt #    deleted:    file3.txt #    deleted:    file4.txt   How do I remove these files from Git without having to manually go through and add each file like this:  git rm file1 file2 file3 file4   Ideally, I'm looking for something that works in the same way that git add . does, if that's possible. Deep learning model (LSTM) with temporal and non temporal attributes Convert Shapefiles to GeoJSON? I'm trying to convert Shapefiles from the  website into a GeoJSON file for use with Google Maps API. I'm using QGIS to facilitate but I can't get the projection/coordinates right. So far I've tried Google Mercator (EPSG:900913) and WGS 84 (EPSG:4326) with no luck. I'm using the OpenLayers to verify results and it keeps placing Texas near the  Gulf of Guinea (0 lat, 0 lng). I'm a new blender user, but have used other modelling programs before.   I've downloaded a  and unzipped it.   When I import the obj into blender, it is my understanding (from here: ) that the mtl should just import itself with it. However, when I import the obj, I just get a grey model with no texture. Is it actually importing the mtl but I'm not seeing it? Also for the record, I have changed no settings in blender.  Picture for reference: How to show that $L^p$ spaces are nested? Suppose $1&lt;p_1&lt;p_2&lt;\infty$, then show $L^{p_1}[a,b] \supset L^{p_2}[a,b]$.  I was able to show $\|f\|_{p_1} \le \|f\|_{p_2} (b-a)^{1/p_1 - 1/p_2}$ but I'm not sure how to proceed from here. DEFINE vs Variable in PHP Is “are” a borrowed word? I read somewhere that English is the only language to have borrowed a form of its to be verb from another language. I want to say, if memory serves, that it was are that was borrowed from an early Scandinavian language. Is that so?  I know Old English had sind for one of their pronouns inflections of be, the same as in German.  But did it switch to are naturally from some other gentler evolution, or was it borrowed? How do I translate strings in Twig templates? I am working on a (my first) Drupal 8 project witch is a multilingual site. Drupal 8 really delivers in terms of native support for content translation.  However, I have a few strings set directly in some if the theme .Twig template files printed out like ex.   {{ 'Read more'|trans }}   I am really stuck on where to handle the translation for strings like these within the Drupal 8 installation.  What steps do I need to take to be able to translate these strings into the language currently selected for on the site? Do I need any additional modules other than the ones included in Drupal 8 core?  Steps I have taken:   Activated the Content Translation &amp; Interface Translation modules Added Twig strings like {{ 'Read more'|trans }} Flushed cache and refreshed pages containing the strings Checked admin/config/regional/translate for the strings with no result User with low reputation shows as having modified thread How can this user , reputation 3, show in  as having "modified 7 hours ago", but on the modified thread, , not show at all?  This user also has two medals, which is somewhat puzzling. Who would get the longest road? Imagine the following setup. There are 3 (or more) players with an equally long road. What would happen if the longest road was broken? How is it decided who obtains the longest road from there on? Let the search find questions by ID $\{a_n\}$ be a sequence such that $ a_{n+1}^2-2a_na_{n+1}-a_n=0$, then $\sum_1^{\infty}\frac{a_n}{3^n}$ lies in... Let $\{a_n\}$ be a sequence of positive real numbers such that   $a_1 =1,\ \  a_{n+1}^2-2a_na_{n+1}-a_n=0, \ \ \forall n\geq 1$.   Then the sum of the series $\sum_1^{\infty}\frac{a_n}{3^n}$ lies in...  (A) $(1,2]$, (B) $(2,3]$, (C) $(3,4]$, (D)$(4,5]$.   Solution attempt:   Firstly, we figure out what $\frac{a_{n+1}}{a_n}$ is going to look like. We get, from the recursive formula, $\frac{a_{n+1}}{a_n}=1+\sqrt{1+\frac{1}{a_n^2}}$ (remembering the fact that $a_n&gt;0$, the other root is rejected).   We know that, if $\lim_{n \to \infty}\frac{a_{n+1}}{a_n}&gt;1$, then $\lim a_n \to \infty$.   Further, $(a_{n+1}-a_n)= \sqrt{a_n(a_n+1)}&gt;0$. (Again, the other root is rejected due to the same reason).   Hence, $(a_n)$ increases monotonically.   Therefore, the largest value of $\frac{a_{n+1}}{a_n}$ is approximately $1+\sqrt{1+\frac{1}{1}} \approx 2.15$  Now, the sum can be approximated as $\displaystyle\frac{\frac{1}{3}}{1-\frac{2.15}{3}} \approx 1.3$ (In actuality, $\mathbb{sum}&lt; 1.3$).  So, option $(A)$ is the correct answer.   Is the procedure correct?   I have been noticing a handful of this type of questions (based on approximations) lately, and the goal is to find out where the sum / the limit of the sequence might lie.   Is there any "definitive" approach that exploits the recursive formula and gives us the value, or does the approach varies from problem to problem? Install Numpy for Python3.4.0 on Ubuntu 14.04.01 LTS I want to filter a java.util.Collection based on a predicate. How to show that $\sum_{k=1}^n k(n+1-k)=\binom{n+2}3$? Showing $(a^{2}+2)(b^{2}+2)(c^{2}+2)\geq 9(ab+bc+ca)$ While the etiquette of dealing with duplicate answers , there hasn't been a great way to deal with these, "Uh, yeah, I said this, too" types of behavior than the almighty downvote.   It would be nice to have a "Duplicate Answer" option on the flag menu. Five flags auto-deletes the answer, as with any other flag. In order to prevent abuse, the flagged answer would have to have the last owner edit (including original post) be younger than the original answer's original posting by at least 2 minutes.  Also, this ability should be a 5,000+ rep privilege, since we currently have no fanfare of rep from 3,001 to 9,999, and a power like this has a bit more umph than closing a question.  Thoughts? How do I troubleshoot the "Error, press shutter release button again" message on my Nikon? The LCD on my Nikon D3000 is displaying the following error message:     Error. Press shutter release button again.   I have pressed the shutter release button but the camera still does not work. How to Query all opportunities ids from account I write a sample query to fetch account Ids from cases of a particular record type.Then i wrote another sample query parent to child relationship to fetch opportunity ids from accounts.  My sample code  List&lt;Case&gt; c=new List&lt;Case&gt;(); c=[Select Id,accountId from case where RecordTypeId = '01290000000sF3L']; List&lt;Id&gt; accountIds= new List&lt;Id&gt;(); List&lt;Account&gt; accounts=new List&lt;Account&gt;(); List&lt;Opportunity&gt; opps=new List&lt;Opportunity&gt;(); for(Case cl:c)  { accountIds.add(cl.accountId); } accounts=[Select Id,name,(select Id,name from opportunities) from account        where Id in:accountIds]; System.debug('Data in accounts '+accounts);   But my issue is how to get opportunity id from above Soql accounts means how to add those opportunity id's to the list.  for(Account ao:accounts) { //How to fetch Opportunity ids from above account list. }