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352, AMath Lecture #8 July 11, 2008 1 Recap of last lecture In the last lecture, we looked at the ideas of subspaces, span, and linear dependence/linear independence. let me recap the speci c example we were looking at. In the last lecture, we looked at the speci c example of the subspace v1 (1) S = v2 : 3v1 + v2 2v3 = 0 . v3 We also looked at three vectors to use while making sets that span S: 1 0 1 3 , f2 = 2 , f3 = 1 . f1 = 0 1 1 (2) In particular, we saw that the set {f1 , f2 , f3 } spans S and is linearly dependent, whereas the set {f1 , f2 } spans S and is linearly independent. 2 Basis and dimension Today, I ll start o by wrapping that up with the concepts of basis and dimension. To some extent, dimension is really the idea we ve been building towards, so we have a precise de nition for our intuitive idea of dimension. But all of these ideas that get us there span, linear (in)dependence, and basis are important parts of linear algebra in their own right. 2.1 Basis De nition. Let S be a subspace of the vector space V. The set F = {f1 , f2 , . . . , fn } is a basis of S if it has the following two properties: 1. F spans S. 2. F is linearly independent. 1 Let s go back to the example from last time. The set {f1 , f2 , f3 } spans S, but it isn t linearly independent. As a result, this set isn t a basis for S. On the other hand, if I just take {f1 , f2 }, this set both spans S and is linearly independent. So this set is a basis for S. 2.2 Dimension Now that we know what a basis is, the de nition of dimension is really simple. De nition. Let S be a subspace of the vector space V. The dimension of S, denoted dim(S), is the number of elements in a basis of S. So if someone gives me a subspace S, to nd its dimension, I just nd a basis and count the number of vectors. There s some work involved in showing that the de nition of dimension is well-de ned. That is, I need to be sure that if I nd one basis for S that gives me one number for the dimension, I m not going to nd another basis later on that gives me a di erent number. It all works out in the end, though, and we won t go into the details. For the example we ve been looking at, the set {f1 , f2 } is a basis for S. This set has 2 vectors, so dim(S) = 2. Conveniently, this agrees with our intuitive idea of dimension. Since S looks like a plane, it s dimension ought to be 2. This example is more of a sanity check than anything else the interesting thing about dimension is using it in complicated situations where we don t have intuitive ideas to work with. 3 Notation for general linear transformations It s common terminology that the domain of a function is its set of inputs, and the range of a function is its set of outputs. So for a general transformation T , the idea is T : domain range. (3) Keeping with this, I m going to use d and r as placeholders for the sizes of the spaces that T maps between. Maybe it s better to just write what I mean: T : Rd Rr . 2 (4) I m going to stick with this as best I can from here on out. As an added convenience, the r happens to agree with the number of rows in a matrix representation of T . Unfortunately d isn t the rst letter of columns, but at least it s pretty close to c. 4 4.1 Null space What inputs don t matter? Motivation and de nition Let s take a quick look back at orthogonal projections. Remember the idea behind orthogonal projections. Given a vector v R2 , I can also take a perpendicular vector v . Then for any x R2 , I can decompose x as x = v + v for some scalars and . The decomposition is pictured here: (5) u v v v v = Pv (u) The orthogonal projection Pv takes the part parallel to v itself, so Pv (x) = v. (In Lecture 4, we found a formula for , but that s not so important at the moment.) There s something a little peculiar here. Say I apply this projection to v . Well, v decomposes as v = 0v + 1v . Since Pv (v ) takes only the part parallel to v, I get Pv (v ) = 0v = 0. (6) 3 Now say I start with an arbitrary vector x. If I add any multiple of v to x, then I don t a ect the part of it that lies parallel to v. So Pv (x + v ) = Pv (x). The projection Pv doesn t care about the vector v at all. This feature isn t particular to projections. For a general linear transformation T : Rd Rr , suppose that T v = 0. Then for any vector x Rr , T (x + v) = T x + T v = T x. (8) (7) So adding vv to x doesn t change the output at all! This seems to say that if T maps v to 0, then as far as T is concerned, v might as well not exist. If I can gure all the vectors that don t matter, then perhaps I can focus my attention on the inputs that do matter. The rst step is to give the set of all these inputs a name. De nition. The null space of T , denoted Nullspace(T ) is the set of all vectors that map to 0. In set notation, Nullspace(T ) = {v Rd : T v = 0}. (9) 4.2 Null space is a subspace of Rd As it happens, Nullspace(T ) is always a subspace. So it s a mini-vector space inside of the domain. Theorem. If T : Rd Rr is linear, then Nullspace(T ) is a subspace of Rd . Proof. Let u and v be any two vectors in Nullspace(T ). I need to show that any linear combination of these vectors is in Nullspace(T ) as well. This is pretty straightforward: T ( u + v) = T (u) + T (v) = 0 + 0 = 0. (since T is linear) (10) (11) (12) So u + v is in Nullspace(T ). This proves that Nullspace(T ) is a subspace. 4 Now, you might say that this is all well and good, but kind of abstract. So, Nullspace(T is ) a subspace of Rd . Whoop-dee-doo. And while your sarcasm would be appreciated, it would be wrong. The fact that Nullspace(T ) is a subspace has one huge implication for linear systems. If a linear system has more than one solution, then it has in nitely many solutions. Here s the theorem, and the proof. Theorem. If the linear system Ax = b has more than one solution, then it has in nitely many solutions. Proof. Let x1 and x2 be two di erent solutions of Ax = b. Then A(x1 x2 ) = Ax1 Ax2 = b b = 0. (13) This means that x1 x2 is a nonzero vector in Nullspace(A). (It s nonzero since x1 and x2 are di erent.) Since Nullspace(A) is a subspace, this means it contains in nitely many elements. Now take any of these in nitely many elements, and call it v. Then A(x1 + v) = Ax1 + Av = Ax1 = b. This provides in nitely many solutions to Ax = b. (14) 4.3 Nullity The size of the null space is also important enough to get its own name. It s called the nullity. De nition. The nullity of T , denoted nullity(T ), is the dimension of Nullspace(T ). As an equation, nullity(T ) = dim(Nullspace(T )). (15) This gauges the amount of information that the transformation T destroys (which it does by mapping vectors to 0). 5 5 5.1 Range What outputs are possible? Motivation and de nition Here s another peculiar feature of linear transformations, again motivated by the projection Pv . At a glance, I d say that the outputs from Pv live in R2 . This is true, but it s not very precise. This projection always gives me an output that s parallel to v. In fact, span(v) fully describes the possible outputs of Pv . This subspace looks like a straight line, and it s much smaller than R2 . This begs the question: For an arbitrary linear transformation T : Rd r R , what vectors in Rr are actually possible as outputs? This is a big question for linear systems of equations, too. If my system is T x = b, then b has to be among the possible outputs if there s any hope of nding an exact solution. The set of all possible outputs from T is called the range. Here s how to de ne it. De nition. Let T : Rd Rr be linear. The range of T , denoted Range(T ) is the set of all possible outputs from T . In set notation, Range(T ) = {w Rm : w = T v for some v Rn }, where T : Rn Rm . (16) 5.2 Range is a subspace Same as the null space, the range of a linear transformation is a subspace. Only Range(T ) is a subspace of Rr , whereas Nullspace(T ) is a subspace of Rd . Theorem. Let T : Rn Rm be a linear transformation. Then Range(T ) is a subspace of Rm . Proof. I need to show that Range(T ) is closed under linear combinations. To do this, suppose that w and z are vectors in Range(T ). Then there are vectors u and v in Rn such that w = T u, z = T v. (17) Now if I take an arbitrary linear combination of w and z, w + z = T u + T v. 6 (18) Since T is linear, I can rearrange the right side: w + z = T ( u) + T ( v) = T ( u + v). (19) (20) This shows that w + z is the output of T when I apply it to u + v. So w + v Range(T ), and thus Range(T ) is a subspace. This doesn t have immediate implications that are as gratifying as what we did with null spaces. But it s still a big part of understanding linear transformations. The set of possible outputs always looks like a line, or a plane, or their higher dimensional analogues. It does, at least, tell me that I can precisely de ne the size of the range. 5.3 rank(T ) How big is Range(T )? Since Range(T ) is a subspace of Rm , I can say how big it is by nding its dimension. In fact, this gets a special name: rank. De nition. The rank of T , denoted rank(T ), is the dimension of Range(T ). More equation-like, rank(T ) = dim(Range(T )). (21) Actually nding rank(T ) is another matter, as it involves constructing a basis for Range(T ). We ll deal with that a little ways down the road. 5.4 How can I nd Range(T )? question. Say that I have the transfor t13 t1d t23 t2d . . . . . . . tr3 trd Now let s check out a more practical mation T : Rd Rr as a matrix, t11 t12 t21 t22 T = . . . . . . tr1 tr2 (22) (For instance, this is the matrix I d get with a r d linear system, T x = b.) How do I determine which vectors are actually in Range(T ). 7 Fortunately, that s pretty easy to gure out. Remember that the matrixvector multiplication T x takes the components of x, and uses them to make a linear combination of the columns of T : t11 t12 t13 t1d t21 t22 t23 t2d T x = . x1 + . x2 + . x3 + + . xd . (23) . . . . . . . . tr1 tr2 tr3 trd I can vary the values of x1 , x2 , . . . , xd as much as I want. So by picking di erent vectors x, I can make T x give me back any linear combination of the columns of T . That s pretty spi y because it tells me that Range(T ) = span(columns of T ). (24) 4 and think about the projection Pv . I know 1 that Range(Pv ) ought to be span(v), but I want to check out this span of the columns business. First, I need to nd the matrix representation for Pv . To nd this, I see what it does to the standard basis vectors: Example. Let s take v = Pv (e1 ) = Pv (e2 ) = 4 v e1 v = v, v v 17 (25) (26) v e2 1 v = v. v v 17 So the matrix representation for Pv is Pv = So I should have 4 1 Range(Pv ) = span( 17 v, 17 v). 4 v 17 1 v 17 = 16/17 4/17 . 4/17 1/17 (27) (28) But any linear combination of 4 v 17 and 1 v 17 is just a scalar multiple of v. So (29) Range(Pv ) = span(v), which is what I was hoping I d get! 8
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Path: Washington >> SPAN >> 352 Fall, 2008
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Path: Washington >> STAT >> 491 Fall, 2008
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Path: Washington >> STAT >> 498 Fall, 2008
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Path: Washington >> STAT >> 498 Fall, 2008
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Path: Washington >> STAT >> 524 Spring, 2008
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Path: Washington >> STAT >> 534 Fall, 2008
Description: #-#-#Cathee Kneeling #STAT 534 #Homework #1, Due 4/11/2006 #--#-# Pull in source files directory <- \"http:/www.stat.washington.edu/catheek/stat534/HW#1\" lib.filename <- \"home1-make-data-lib-3-30-05.txt\" lib.pathname <- paste(directory, lib.filename, ...
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Path: Washington >> STAT >> 534 Fall, 2008
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Path: Washington >> STAT >> 535 Fall, 2008
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Path: Washington >> STAT >> 535 Fall, 2008
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Path: Washington >> STAT >> 560 Fall, 2008
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Path: Washington >> STAT >> 560 Fall, 2008
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Path: Washington >> STAT >> 567 Fall, 2008
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Path: Washington >> STAT >> 567 Fall, 2008
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Path: Washington >> STAT >> 567 Fall, 2008
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Path: Washington >> STAT >> 567 Fall, 2008
Description: Linear Quantile Regression 11.5 11.5 Spline Quantile Regression 11.0 10.5 Logearnings 10.0 Logearnings observed quantiles quantile regression lines 9.5 9.5 10.0 10.5 11.0 observed quantiles quantile regression splines 9.0 9.0 70 75 80...
570-Day-2.pdf
Path: Washington >> STAT >> 570 Fall, 2008
Description: Stemming Tokenizing suffixes irrelevant Porter Stemmer, simple and efficient Website: http:/www.tartarus.org/...
573.pdf
Path: Washington >> STAT >> 573 Fall, 2008
Description: ...
576-07s.pdf
Path: Washington >> STAT >> 576 Fall, 2008
Description: BIOSTAT/STAT 576 Statistical Methods for Survival Data MW 9:00am - 10:20am, Spring 2007; HST T530 Instructor: Ying Qing Chen Tel: 206.667.7051 Email: yqchen@u.washington.edu Oce hour: Th 2:30pm - 3:30pm; H657 Course Web Site: http:/www.scharp.org/u...
576.pdf
Path: Washington >> STAT >> 576 Fall, 2008
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579-07au.pdf
Path: Washington >> STAT >> 579 Fall, 2008
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579-08su.pdf
Path: Washington >> STAT >> 579 Fall, 2008
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579-08sp.pdf
Path: Washington >> STAT >> 579 Fall, 2008
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581.day1.08.pdf
Path: Washington >> STAT >> 581 Fall, 2008
Description: STATISTICS 581: Advanced Theory of Statistical Inference Fall, 2008 Time: Place: Professor: Oce: Phone: e-mail: Oce Hours: Texts: 10:30 - 11:20 MWF (lecture) MEB 245 Jon A. Wellner B320 Padelford 206-543-6207 jaw@stat.washington.edu 1:30 - 3:30 MWF...
ref.08.pdf
Path: Washington >> STAT >> 581 Fall, 2008
Description: References for Statistics 581, Fall 2008 Analysis: Bartle, R. G., The Elements of Integration. Rudin, W., Principles of Mathematical Analysis. Royden, H. L., Real Analysis. Luenberger, D., Optimization by Vector Space Methods. Probability: Will...
ch2.figs-epsf.pdf
Path: Washington >> STAT >> 581 Fall, 2008
Description: Statistics 581, Chapter 2 Empirical Distribution Function and Empirical Process Figures Wellner; 10/24/2008 1 0.8 0.6 0.4 0.2 0.2 0.4 0.6 0.8 1 Figure 1: Uniform Empirical Distribution Function, n = 50. 1 0.75 0.5 0.25 0.2 -0.25 0.4 0.6...
exam1.06.pdf
Path: Washington >> STAT >> 581 Fall, 2008
Description: Statistics 581, Midterm Exam Wellner; 11/06/2006 This exam is to be taken without any books or notes. 1. (24 points) Dene any three of the following ve terms. (a) A uniformly integrable sequence of random variables. (b) Convergence in rth mean of a...
mt03.pdf
Path: Washington >> STAT >> 582 Fall, 2008
Description: Stat 582 W03 Midterm exam Please give as complete solutions as possible. More paper is available if needed. 1. Let (X,Y) be bivariate normal, mean zero, variance 1, correlation r. Find a minimal sufficient statistic for r. Is it complete? 2. Suppo...
final2.89.pdf
Path: Washington >> STAT >> 582 Fall, 2008
Description: STAT 582 FINAL EXAM 1. Let T (F) = m 2 = ( xdF(x)2 . Find, using the asymptotic theory for statistical functionals, the limiting F distribution of T (F n ) when m F 0. 2. Let (X 1 , Y 1 ), . . . , (X n , Y n ) be independent random variables with ...
