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352, AMath Lecture #9 July 14, 2008 1 Recap of last lecture Here are the main points from the last lecture. For a linear transformation T : Rd Rr , the nullspace of T is Nullspace(T ) = {v Rd : T v = 0}. (1) This is the set of all vectors that T maps to 0, and it s a subspace of Rd . It s dimension is called the nullity of T : nullity(T ) = dim(Nullspace(T )). (2) The range of T is the set of all possible outputs from T . In set notation, Range(T ) = {w Rr : w = T v for some v Rd }. This is a subspace of Rr , and its dimension is called the rank of T : rank(T ) = dim(Range(T )). We also saw that Range(T ) = span(columns of T ). (5) (4) (3) 2 How do I nd Nullspace(T )? I ve de ned the null space of a transformation, but I haven t actually shown how to nd it. Conveniently, this boils down to Gaussian elimination. And with a little bit of care along the way, it also steers me towards my next goal: nding a basis of the range. Take the transformation given by the matrix 4 1/2 3/2 T = 1 4/3 5/6 (6) 3 1 5/2 1 To nd Nullspace(T ), I need to know which vectors this maps to 0. So I ll take a generic vector v and set T v = 0. 4 1/2 3/2 0 1 v1 + 4/3 v2 + 5/6 v3 = 0 . (7) 3 1 5/2 0 I m about to turn this into a linear system of equations, but remember that we started out with this equation based on the columns it ll show up later. Looking at the three components individually, I get the 3 3 linear system 4v1 + v1 + 3v1 1 v 22 4 v 32 + v2 + 3 v 23 5 v 63 5 v 23 =0 =0 = 0. (8) And now I solve this using (painfully detailed) Gaussian elimination: 1 1 3 3 4v1 + 2 v2 + 2 v3 = 0 v1 + 8 v2 + 8 v3 = 0 scale 4 v1 + 4 v2 5 v3 = 0 v1 + 3 v2 5 v3 = 0 3 6 6 5 3v1 v2 + 2 v3 = 0. 3v1 v2 + 5 v3 = 0 2 v1 + v1 + 1 v 82 29 v 24 2 (9) + pivot 3 v 83 29 v 24 3 11 v 83 3 v 83 =0 =0 =0 =0 (11) =0 (10) 11 v2 + 8 1 v 82 + scale v2 11 v2 + 8 v3 = 0 11 v 83 1 3 v1 + 8 v2 + 8 v3 = 0 pivot v2 v3 = 0 0 = 0. (12) This system has one degree of freedom. From here I have a couple of options. The rst is to decide that v3 will be undetermined. (Conventionally, the rightmost variable possible is the 2 undetermined one.) In that case, the second equation gives me v2 = v3 . And then the rst equation gives me 3 1 v1 + v3 + v3 = 0 8 8 = 1 v1 = v3 . 2 (14) (13) Then my generic vector in the null space is 1 2 v3 v3 . v= v3 (15) For the second option, I ll nd vectors that span the null space. Since there s only one degree of freedom in the linear system, this means that Nullspace(T ) has dimension 1. So nding one vector will do the trick. This means I can pick speci c values. If I set v3 = 1, then I end up with v2 = 1 and v1 = 1/2. I end up with 1/2 (16) Nullspace(T ) = span 1 . 1 This also tells me that nullity(T ) = 1. (17) 3 How do I nd a basis for Range(T )? Range(T ) = span(columns of T ). (18) In the last lecture, we saw that Sticking with the same T from the previous section, this tells me that 4 1/2 3/2 Range(T ) = span 1 , 4/3 , 5/6 . (19) 3 1 5/2 However, it s completely possible that I have more information than I need here i.e., that the vectors on the right are linearly dependent. 3 To check for linear dependence, I set up the equation 4 1/2 3/2 0 1 1 + 4/3 2 + 5/6 3 = 0 , 3 1 5/2 0 (20) with the goal of determining 1 , 2 , and 3 . Now s the part where I go Whoaaaaaa. . . wait a second here. I already did that. Look back at (7). Sure it has v s instead of s, but otherwise it s the exact same thing. That means I can use v1 , v2 , and v3 that I found before as my values here. So for any v3 , I have 1/2 3/2 0 4 1 ( 1 v3 ) + 4/3 v3 + 5/6 v3 = 0 . (21) 2 1 5/2 0 3 To break this dependence, I need to throw out one of the vectors in this relationship. I ll get rid of the rst one (death to integers!), and I end up with 1/2 3/2 Range(T ) = span 4/3 , 5/6 , (22) 1 5/2 with the vectors on the right forming a basis of Range(T ). This also tells me that rank(T ) = 2. (23) 4 An example with nullity=2 Now I ll walk through the same process for another transformation, this one given by 1/2 8 1 U = 5/6 20 5/2 . (24) 1 24 3 To nd the null space, the rst thing I ll do is set U v = 0 for a generic vector v. This gives 1/3 me 8 1 0 5/6 v1 + 20 v2 + 5/2 v3 = 0 . (25) 1 24 3 0 4 Like last time, I turn this into a linear system, which I simplify using Gaussian elimination: 1 v1 24v2 + 3v3 = 0 v 8v2 + v3 = 0 31 5 v1 + 20v2 5 v3 = 0 6 2 v1 24v2 + 3v3 = 0 swap 5 5 v1 + 20v2 2 v3 = 0 6 1 v 31 (26) 8v2 + v3 = 0 v1 24v2 + 3v3 = 0 pivot 0=0 0=0 (27) This system has two degrees of freedom. One way to characterize the null space is by all vectors of the form 24v2 3v3 for all v2 , v3 . v2 v= (28) v3 Or, I can pick a couple of spanning vectors by taking speci c values for v2 and v3 : 24 1 v2 = 1, v3 = 0 = (29) 0 3 0 . v2 = 0, v3 = 1 = (30) 1 In this case, 24 3 Nullspace(U ) = span 1 , 0 , (31) 0 1 with the vectors on the right forming a basis. This also tells me that nullity(U ) = 2. (32) Now I ll move on to nd a basis for Range(U ). Just like the last time, each vector in the null space indicated a dependence between the columns of U: 1/2 8 1 0 5/6 (24) + 20 (1) + 5/2 (0) = 0 (33) 1 24 3 0 5 and 1/2 8 1 0 5/6 ( 3) + 20 (0) + 5/2 (1) = 0 1 24 3 0 (34) For each dependency, I have to throw out one of the columns of U . One way to do this is by throwing out the second column based on (33), and the third column based on (34). This leaves me with 1/2 Range(U ) = span 5/6 , (35) 1 where the right side is a basis for Range(U ). This also tells me that rank(U ) = 1. (36) You might notice, though, that things would be a little screwy if I made these choices di erently. Say I threw out the rst column, based on (33). Then it would seem that this breaks the second dependency in 34). But in fact, the second and third columns are dependent on one another as well. (Column 3 = ( 8) (Column 2).) How can I avoid this problem? Well, I ll de nitely have to throw out one column for every element in the basis of Nullspace(U ). The trick is to throw out the columns that correspond to the free variables in the linear system I got while nding the null space. In this case, the free variables were v2 and v3 , so throwing out the second and third columns will do the trick. 5 Rank-Nullity Theorem In the past two examples, we ve seen a connection between the null space and the range of a transformation. This translates into a really nice link between the rank and the nullity of a transformation, which is called the Rank-Nullity Theorem. I won t actually prove the theorem, but we ve already seen the reasoning behind it. For a transformation T : Rd Rr , I know that Range(T ) is spanned by the columns of T . This means that rank(T ) is potentially as large as the number of columns, which is d. 6 However, the value of nullity(T ) tells me how many dependencies there are among the columns of T . I have to throw out this many of the columns to get a basis for Range(T ). The number of columns left over tells me rank(T ). To summarize: (# of columns) (# of dependencies) = rank(T ). And this boils down to d nullity(T ) = rank(T ). Just for sake of reference, here s the theorem, as concisely as possible. Rank-Nullity Theorem. Let T : Rd Rr be a linear transformation. Then d nullity(T ) = rank(T ). (39) (38) (37) 6 Geometric interpretation You can make pictures of the null space an range of a transformation using the m- le nullrangeplot.m in the Lecture #9 section of the course page. To use this, enter the transformation T , and then pass it to the m- le. For example, >> T = [[4; 1; 3] [1/2; 4/3; -1] [3/2; -5/6; 5/2]]; Will input the matrix from the rst example. (I m entering it by columns, which is a little extra work, but I m trying to be consistent with the think in terms of the columns vibe.) Then >> nullrangeplot(T) will show you Rd and Rr side by side, with Nullspace(T ) in Rd and Range(T ) in Rr . For this example, here s what I get: 7 If you like, you can also use vector3d.m from the course page to plot some vectors. For instance, say I want to plot e2 in the left gure, and T e2 in the right picture. To plot e2 in the left picture, >> >> >> >> >> subplot(1,2,1) e2 = [0; 1; 0]; hold on vector3d(e2) light The command light isn t necessary, but it adds some shading, which makes the vectors look nicer. Then to plot T e2 in the right picture, >> >> >> >> subplot(1,2,2) hold on vector3d(T*e2) light You can also use nullrangeplot for 2 2, 2 3, and 3 2 matrices. Give it a try! 8
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Path: Washington >> STAT >> 534 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...
