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.
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!
Word with the same meaning as "good moments" or "pleasant memories"? Is there a single word with the same meaning as "good moments" or "pleasant memories"? I'm looking for a word which describes good thing(s) that happened to someone. The only word I can think of is "joy".  Some examples might be:   Wayne has many pleasant memories of the days spent with the family. She could remember many good moments from the last weekend. How can see call history in iPhone one month back, i also tried with tried party software but that is not perfect , is the any solution from apple? $\{a_n\}$ 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? Does the fact that matrices $A, B$ are similar imply that $A+cI$ is similar to $B+cI$? Let $A, B$ be similar $n \times n$ matrices over (say) $\mathbb{R}.$  Is it the case that $A+cI$ is similar to $B+cI$?    Here is an argument (which seems too good to be true...)  Since $A, B$ are similar, they have the same Jordan form. So we can write $PAP^{-1}=QBQ^{-1}=J$ for $P, Q$ invertible matrices.   Hence we have $P(A+cI)P^{-1}=PAP^{-1}+cI= J+cI.$ Similarly, we find $Q(B+cI)Q^{-1}=J+cI.$  Hence $A+cI$ and $B+cI$ are both similar to $J+cI$ and hence similar to each other.   Is this correct? I have been reading for days about how to remove the Google Search Bar from the top of my Nexus 6 since I never use it. I have gotten mixed answers on exactly what file to delete and where it is.  I have heard quicksearchbar.apk but it's not in /system/app like in previous versions. Does anybody know where the file/files are I need to delete in order to get rid of the search bar, or else, how to remove it? Preferably not downloading another launcher, since I don't want to bloat my phone when I could reduce the size. Round a double to x significant figures Difference in output between SAS's proc genmod and R's glm I'm trying to replicate a model fitted in SAS in R but the fit I'm getting gives me slightly different coefficients and standard errors.  Data:  testdata &lt;- data.frame(matrix(c("f","Test", 1.75,   16, 0,  16, 0,  1,  1, "m",    "Test", 1.75,   15, 1,  16, 6.25,   1,  0, "f",    "Test", 2.75,   4,  12, 16, 75, 1,  1, "m",    "Test", 2.75,   9,  6,  15, 40, 1,  0, "f",    "WHO",  1.75,   15, 1,  16, 6.25,   0,  1, "m",    "WHO",  1.75,   14, 2,  16, 12.5,   0,  0, "f",    "WHO",  2.75,   2,  13, 15, 86.6667,    0,  1, "m",    "WHO",  2.75,   3,  13, 16, 81.25,  0,  0 ), ncol=9, byrow=TRUE)) names(testdata) &lt;- c("sex", "vaccine", "dose", "not_p", "para", "n", "pct",                       "vacnum", "sexno")   SAS:  proc genmod data=model_data; class sex; model para/n  = dose sex vacnum    /dist=bin    link=logit   type3; run;  Analysis Of Maximum Likelihood Parameter Estimates  Parameter   DF Estimate Std Error Wald 95% Conf Lim Wald Chi-Square Pr &gt; ChiSq  Intercept   1 -9.4020   1.6220    -12.5810  -6.2230  33.60           &lt;.0001  dose        1  3.9208   0.6460      2.6546   5.1870  36.83           &lt;.0001  sex f       1  0.5574   0.5184     -0.4587   1.5735   1.16           0.2823  sex m       0  0.0000   0.0000      0.0000   0.0000    .              .  vacnum      1 -1.3221   0.5483     -2.3967  -0.2475   5.81           0.0159  Scale       0  1.0000   0.0000      1.0000   1.0000        R:  testdata$sexno &lt;- as.factor(testdata$sexno)     a &lt;- contr.treatment(2, base = 1, contrasts = TRUE)  contrasts(testdata$sexno) &lt;- a  fitreduced &lt;- glm(para/n ~ dose + as.factor(sex) + vacnum,                    family=quasibinomial(link="logit"), data=testdata)  coef(summary(fitreduced))                    Estimate Std. Error   t value    Pr(&gt;|t|) (Intercept)     -9.4013750  1.7613982 -5.337450 0.005935450 dose             3.9173794  0.7001133  5.595351 0.005007179 as.factor(sex)1  0.5704671  0.5568436  1.024466 0.363525300 vacnum          -1.3336100  0.5887552 -2.265135 0.086189704   I believe I have the right contrasts to give me a type III SS but there is a small discrepency in values, have a missed something here? I have a problem with the interpretation of a test result in which the p-value is 8.6e-28. How should it be interpreted? What is the e for?  Specifically, I used White's Test for the homoscedasticity assumption of a linear regression:  White's general test statistic :  150.0535  Chi-sq( 9)  P-value =  8.6e-28 $|G|>2$ implies $G$ has non trivial automorphism Ubuntu 14.04 multi touch screen support Just installed 14.04. It seems to be converting all my screen touch gestures to mouse gestures. And it only supports one point touch. It is supposed to support ten touch points (tested with windows 8.1).  Am I missing something here? Do I have to install some kind of package? Or is it just not supported?  Test - being able to draw here: Matrix exponential for Jordan canonical form Let $X$ be a real $n \times n$ matrix, then there is a Jordan decomposition such that $X = D+N$ where $D$ is diagonalisable and $N$ is nilpotent.  Then, I was wondering whether the following is correct.  $e^{tX}(x) = \sum_{k=0}^{m} \frac{t^k N^k}{k!} \left(e^{t \lambda_1} \alpha_1 v_1+..+e^{t \lambda_n} \alpha_n v_n \right).$  Here $x = \sum_{i=1}^{n } \alpha_i v_i$ and $v_i$ are the eigenvectors of the diagonalisable matrix, $\lambda_i$ are the eigenvalues of $D$ and $m$ is the degree up to which $N^k$ is still non-zero.  Is this correct or am I doing something wrong? Cause I could not find a general equation for this matrix exponential, so I tried my best. (Thus, I am only asking for a verification or correction of this answer.)  If anything is unclear, please let me know. I'm having trouble finding something to clarify this. It's been mentioned in the Difficulty Class section of the DMG (238) that there should be a 50/50 chance that an average person can succeed at a DC 10 challenge, implying you have to beat it. However, I'd always thought that there's a 50/50 chance for an average person to hit another average person, but the chance is really 55/45, since you only have to equal their AC. Can you add new statements to Python's syntax? Can you add new statements (like print, raise, with) to Python's syntax?  Say, to allow..  mystatement "Something"   Or,  new_if True:     print "example"   Not so much if you should, but rather if it's possible (short of modifying the python interpreters code) How to change background-color on text links on hover but not image links Reference template variable within Jinja expression Is it possible to extend a tourist Schengen visa (12 more days)? My daughter applied at the Italian Consulate in Philadelphia, and they gave her two entries for 20 days (we thought it was for 20 days each entry). She was in Italy 12 days, then went to England for 1 week and flew to Venice to take a 10-day cruise. She couldn't make it because they said she has only 8 remaining days, but she still has a land tour already paid this week.  What can she do? Can she extend the Schengen visa for 12 more days? Right now, she is in Madrid, but next week she has to fly from Barcelona to Paris to start with the tour. What would happen if she decides to take the tour with the expired visa? She is a US permanent resident. Subscribe to tags I am interested in? How do I subscribe to the tags that I am interested in?  I like to get notified via email or so, when someone asks new questions or if someone responds to my questions. What is the difference between can and could in 'Can/could you please explain this to me?' Can you please explain this to me?          Could you please explain this to me?   I am unable to figure out which to use which situation. I did google, and some posts say they are both the same, even if the second one is more formal. Other posts say could is the past tense of can, but in the above example, I don't think could is used as past tense of can.  What is correct way to use these words? Why does "under the sea" mean "in the sea"? How do I technobabble without sounding like technobabble? On occasion, I've had reason to describe obviously fictional technical processes in-character. I'd like to be able to describe on these occasions without sounding like I'm saying utter gibberish. But it's not straightforward to do, particularly if I have to make stuff up on the spot.   How do I do it?