Prove a set is still linearly independent when transformed using an injective linear map Given a transformation T:V-W that is injective, if {v1,v2,...,vn} is a linearly independent subset of V, I need to prove that the set {T(v1), T(v2),...,T(vn)} will be a linearly independent set of W. To my understanding no values in {v1,v2,...,vn} are equal since they are independent and because there is a one to one mapping from V to W the set {T(v1), T(v2),...,T(vn)} should also have no equal values. This somewhat proves that {T(v1), T(v2),...,T(vn)} is independent. I am struggling to understand how to prove that the set {T(v1), T(v2),...,T(vn)} doesn't contain values that are linear combinations of other values in the set.
Showing that a one-to-one linear transformation maps a linearly independent set onto a linearly independent set Problem 4.3.32 in Linear Algebra, Lay:  Let $V$ and $W$ be vector spaces, let $T:V\to W$ be a linear transformation, and let $\{\mathbf{v}_1, \dots , \mathbf{v}_p \}$ be a subset of $V$. Suppose that $T$ is a one-to-one transformation [...]. Show that if the set of images $\{T(\mathbf{v}_1), \dots ,T(\mathbf{v}_p)\}$ is linearly dependent, then $\{\mathbf{v}_1, \dots , \mathbf{v}_p \}$ is linearly dependent. This fact shows that a one-to-one linear transformation maps a linearly independent set onto a linearly independent set.  All I can think of doing: If $\{T(\mathbf{v}_1), \dots ,T(\mathbf{v}_p)\}$ is linearly dependent, then there exist weights $c_1, \dots, c_p$, not all zero, so that $$c_1 T(\mathbf{v}_1) + \cdots + c_p T(\mathbf{v}_p) = \mathbf{0}$$ Since $T$ is linear: $$c_1 T(\mathbf{v}_1) + \cdots + c_p T(\mathbf{v}_p) = T(c_1\mathbf{v}_1) + \cdots + T(c_p \mathbf{v}_p)=T(c_1 \mathbf{v}_1 +\cdots + c_p \mathbf{v}_p)=\mathbf{0}$$ So the set $\{\mathbf{v}_1, \dots , \mathbf{v}_p \}$ must be linearly dependent. Am I missing something? And why must $T$ be on-to-one for this to be true? An example showing that it is not true if $T$ is not one-to-one would be great!
A method for the OP to accept comments? Question: What if the author of a question can accept a comment?   The problem:  Time and time again, answers are posted as a comment.  Usually, the person who commented has the best of intention (e.g. it was a trivial answer [to the answerer], and they weren't sure if the OP would immediately edit the question after the comment).  But this leaves the frustrating situation of an unanswered question with a perfectly good answer.  This problem accumulates over the years.     Candidate solution:  The OP can just "accept" the comment, after which a duplicate of the comment is made an accepted CW answer, with a preface indicating that it was originally a comment, e.g. This answer was originally a comment posted by [[blah]] that has been accepted by the author of the question.   Comments on this scheme:   The author of the comment has indicated they're not interested in gaining/losing reputation through their comment (thus CW answer). The original comment is still there. If desirable, the "author" of the accepted question could be a "bot", so the answer is not attributed to a genuine account.  (I suppose it would already be technically possible for Joe Blow to create a script for a bot that creates an answer from a comment in the proposed way.)     A similar question was asked before  (over 3 years ago), and in numerous other questions which have been deemed "duplicates" (,  and ; see also ), so there seems to be demand for such a feature.  The answers to these questions seem (to me, at least) quite weak:   "ask the commenter to add the comment as an answer"   The OP already did this by asking the question in the first place.  And what if the author of the comment doesn't respond (or takes a long time to respond)?  This also adds more chit-chat irrelevant to the question.   "add an answer of your own, referring to the comment, and accept it."   A hacky workaround.  How could this possibly be better than accepting a comment?  (Yet, this is the most up-voted response.)  Moreover, this adds to the user's list of "answers" (thus making the user's page harder to look through).   "When i leave a comment, i do so because i want to leave a comment."   Great; now what about everyone else?  In the immortal words of the FAQ: "Other people can edit my posts?!"  And it's not even an edit, it's copying the post from one box to another (a more hacky version of which is regarded as the "accepted" method). Prove that between any two roots of $f$ there exists at least one root of $g$ $$f(x)= 1 - e^x\sin(x)$$   $$g(x)= 1 + e^x\cos(x)$$   Prove that between any two roots of $f$ there exists at least one root of $g$.   I know Rolle's Theorem and the Intermediate Value Theorem (I think) need to be used. Can someone show a step by step proof for this? Proving that 1- and 2-d simple symmetric random walks return to the origin with probability 1 How does one prove that a simple (steps of length $1$ in directions parallel to the axes) symmetric (each possible direction is equally likely) random walk in $1$ or $2$ dimensions returns to the origin with probability $1$?  Edit: note that while returning to the origin is guaranteed $(p = 1)$ in $1$ and $2$ dimensions, it is not guaranteed in higher dimensions; this means that something in a correct justification for the $1$- or $2$-d case must fail to extend to $3$-d (or fail when the probability for each direction drops from $\frac14$ to $\frac16$). Proof of -(-v)=v in a vector space The exercise is: Prove that $-(-v)=v$ for every $v \in V$  Proof Suppose  $v \in V$ and $V$ is a vector space. Then $-(-v) \in V$ as result of the scalar multiplication property and $-(-v)=-(-1\cdot v)=-1 \cdot(-1\cdot v) =(-1 \cdot-1)\cdot v = 1 \cdot v = v $ The desired result $-(-v)=v$ holds.  The solution manual gives the proof:  Proof   I just wanted to make sure that the way I did the proof isn't missing anything?Thanks Shapefile: "table1.shp"   Field name: "Name" Field name: "Result"   Flat file: "referenceTable"   Field name: "STREET" Field name: "AKA"   I need to compare the value "Name" in "table1.shp" to the value "STREET" in "referenceTable"  If the values are the same then put the "AKA" value into the "Result" in "table1.shp"     Here is my code.  I got stuck on accessing the elements from the array  import arcpy, os, sys, string from array import *   table1 = "G:\\xStreetNew\\table1.shp" table1_column = ['Name', 'Result'] referenceTable = "G:\\xStreetNew\\Test.gdb\\referenceTable" referenceTable_column = ['STREET', 'AKA']  referenceTable_List = [] with arcpy.da.SearchCursor(referenceTable, referenceTable_column) as rCur: for r_row in rCur:     reference_row = r_row[0].split(",")     referenceTable_List.insert(0,reference_row) ##print sorted(referenceTable_List)  Table1_Update_Cursor = arcpy.da.UpdateCursor(table1, "Name") for row in Table1_Update_Cursor:   if row[0] == None:   row[0] = row[0] else:   row[0] = row[0].replace(";",",")   Table1_Update_Cursor.updateRow(row)  table1_List = [] with arcpy.da.SearchCursor(table1, table1_column) as tCur: for t_row in tCur:     row = t_row[0].split(",")     table1_List.insert(0,row) ##print sorted(table1_List)  Table1_row = len(table1_List) Table1_column = len([table1_List]) ref_row = len(referenceTable_List) for i in range(0,Table1_row):  #Table Row   for ii in range(0,Table1_column):   #Table Column     print table1_List[i][ii]   Below was my old post.  Sorry for the confusion.  I have a shapefile named "table1.shp".     I would like to get all values from fielname "Name" from this table and put them into a list. Able to output each element from the list     So far, I'm able to create a list but not too sure how to get to value "AVENUE 33" or "CHAPMAN ST"  This is my code:  import arcpy, os, sys, string from array import *   table1 = "G:\\xStreetNew\\table1.shp" referenceTable = "G:\\xStreetNew\\Test.gdb\\referenceTable" cursor = arcpy.da.UpdateCursor(table1, "Name") for row in cursor: if row[0] == None:     row[0] = row[0] else:     row[0] = row[0].replace(";",",") cursor.updateRow(row) table1_column = ['Name'] table1_curObj = arcpy.da.SearchCursor(table1, table1_column) table1_List = [] for row in table1_curObj:    table1_row = row[0].split(",")    table1_List.insert(0,table1_row)    print table1_List[0] SF Story about human adapting to alien environment This is a story that was read on BBC radio about thirty years ago, and this is about as right as I can remember it: The human pilot of a spacecraft is doing some reconnaissance on an alien planet. He crashes but survives. For some reason, no one will know where to look for him. He sees something in the distance that looks like a building of some sort. When he gets there, it seems to be a home. Very much like a human home. There's furniture, but it's strange and uncomfortable . There's food, but it's inedible. There's a faucet but what comes out of it seems more like acid. Everything looks as though it should be right, but it's not. He leaves the house, wanders and after a few days is desperate for water. But there isn't any water anywhere. He's close to death, crawling on the ground and then something happens. There are cracks in the soil and there appears to be water bubbling up from perhaps some underground stream. The water revives him. He goes back to the house and tries the food again. It's delicious and he eats enough to gain back his strength. Then he decides to try the water again. It's pure and wonderful. He takes a shower and the last line of the story is how refreshing the water is on his skin and something about him lifting his tail. Using strtok with a std::string I have a string that I would like to tokenize. But the C strtok() function requires my string to be a char*. How can I do this simply?  I tried:  token = strtok(str.c_str(), " ");    which fails because it turns it into a const char*, not a char* Per line text gradient (ala beeline reader) Making a WinForms TextBox behave like your browser's address bar When a C# WinForms textbox receives focus, I want it to behave like your browser's address bar.  To see what I mean, click in your web browser's address bar. You'll notice the following behavior:    Clicking in the textbox should select all the text if the textbox wasn't previously focused. Mouse down and drag in the textbox should select only the text I've highlighted with the mouse. If the textbox is already focused, clicking does not select all text. Focusing the textbox programmatically or via keyboard tabbing should select all text.   I want to do exactly this in WinForms.  FASTEST GUN ALERT: please read the following before answering! Thanks guys. :-)     Calling .SelectAll() during   the .Enter or .GotFocus events won't   work because if the user clicked the   textbox, the caret will be placed   where he clicked, thus deselecting all   text.      Calling .SelectAll() during the .Click event won't work because the user won't be able to select any text with the mouse; the .SelectAll() call will keep overwriting the user's text selection.      Calling BeginInvoke((Action)textbox.SelectAll) on focus/enter event enter doesn't work because it breaks rule #2 above, it will keep overriding the user's selection on focus. Misleading chat usercard on SE 2.0 chat Is it acceptable to change the vid column of a term programmatically in the database? PrintWriter and PrintStream never throw IOExceptions Some days ago I realized that  (as well as ) never throw an IOException when writing, flushing or closing.  Instead it sets an internal flag (trouble=true) when an error occurs. It's not possible to get the exact exception, but only if there was some exception ().  My question is: why would one want to have such behavior? Isn't that bad API design? Install a .NET windows service without InstallUtil.exe Deep Format of a Harddrive on a Linux System I've got an original Asus Eee PC Netbook that I'd like to sell.  When I've sold old computers in the past, I've typically done a deep/secure/zeroing out style format.  How can I do this on a Linux Netbook?  I'm looking for   The command that will let me do this Advice on how I can do this on a device with (seemingly?) non-removable media   Thanks in advance! Minecraft Survival Mode I am playing Minecraft on survival mode. I can break blocks like sand, dirt, and gravel and it will give me a block. When I try to break stone or iron or anything else, it didn't give me a block. What can I do so that I can get a block? I am on single player and my internet is fine. I tried deleting the app and then getting the app again but that didn't work. This is happening with all my worlds in survival mode. I accidentally disabled my admin account.  After I login to another account I found I can not get admin rights, because all admin accounts on my computer are disabled.  I clicked on 'run as administrator' but only see a smart card choice (all the admin accounts are disabled so no choice provided)    I don't want to re-install my OS, help!    Details:  I have account A on my computer.    I got a new computer so I want to give the old one to my mother.    Creating a new user account B for her and disable account A.    Logout and restart.    Successfully login B, now I found I could not get admin rights because no account choice is provided.    More informations about my machine:  OS is Windows 10 and my admin account is a Microsoft Account.  I have an arch-linux installed on my computer and dual-boot it using GRUB, so maybe I can not use safe mode. Why does the experience-to-next-level not change between some levels? At every level from 1st to 11th, the experience needed to reach the next level (from the current level) increases every level. However, this pattern gets broken for most of the later levels:   Going from 10th to 11th takes 21,000xp (= 85k - 64k), but 11th to 12th takes only 15,000xp! (= 100k - 85k). Both 13th and 14th level only take extra 20k experience each, i.e. the amount needed doesn't increase between those two levels. Levels 16 and 17 similarly only need 30k experience each, and 18 and 19 only need 40k experience each.   Why does the last half of the experience chart have some of these levels maintain the same experience costs, and even briefly decrease the experience cost for 12th level? Why does a proof of $\exists f: X\to Y$ injection $\iff \exists g: Y \to X$ surjection requires the axiom of choice? Why does a proof of $\exists f: X\to Y$ injection $\iff \exists g: Y \to X$ surjection requires the axiom of choice?  This question is answered here:   I'm not interested in a proof itself of this theorem, I want to know why this requires the Axiom of Choice. How does Windows 7 licensing work for running the OS as Virtual Machines? I plan on buying Windows 7 Ultimate Edition to run on an old Pentium 4. I also want to run Windows 7 virtual machines inside Windows 7. My question is two fold:   Do I need a license for each Virtual Machine? Does the answer change if I use a third party virtualization platform (e.g. , , , etc) as opposed to ? How do users navigate to Meta?