{VERSION 4 0 "SUN SPARC SOLARIS" "4.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 257 "" 1 14 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 259 "" 1 14 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 261 "" 1 8 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 262 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 263 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 264 "" 1 8 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 265 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 266 "" 1 18 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 267 "" 1 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 1 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 270 "" 1 8 0 0 0 0 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 271 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 272 "" 0 8 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 273 "" 1 14 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 274 "" 1 14 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 276 "" 1 8 0 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 277 "" 1 14 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 278 "" 1 14 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 279 "" 1 14 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 " " 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 } {PSTYLE "Maple Output" 0 11 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 3 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 11 12 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }1 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 256 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 257 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 258 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 11 259 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }1 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {PARA 256 "" 0 "" {TEXT -1 0 "" }{TEXT 256 0 "" }{TEXT 257 9 " MATH 2270" }}{PARA 257 "" 0 "" {TEXT -1 0 "" }{TEXT 258 0 "" }{TEXT 259 9 "PROJECT 3" }}{PARA 258 "" 0 "" {TEXT -1 12 "October 2000" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 382 "This pro ject is about the fundamental subspaces associated with matrices, find ing good bases for these subspaces, and about coordinates with respect to different bases. The project is relatively short (at least compa red to the last one), and you should be able to use its content to wor k the book problems you have also been assigned. This project is due \+ on Wednesday October 25." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 260 0 "" }}{PARA 0 "" 0 "" {TEXT -1 21 "First load our tools:" }}{PARA 0 " " 0 "" {TEXT -1 1 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "with (plots):with(linalg):" }}}{PARA 0 "" 0 "" {TEXT -1 156 "Here's the mat rix we will play with today. We will keep it moderately sized, but of course we could make it much bigger without causing Maple any problem s." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 107 "A:=matrix([[ 1,0,-1,2,3],\n [ 3,2,-2,1,-1] ,\n [1,2,0,-3,-7],\n [0,-2,-1,5,10]]);" }}}{PARA 0 "" 0 "" {TEXT -1 110 "We should be able to figure out dimensions of \+ the nullspace, row space and column space by looking at rref(A):" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "rref(A);" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 204 "1) What are the dimens ions of the three spaces, based on rref(A) and our general theory? Ex plain. Also, for the theorem that rank plus nullity equals n, what ar e the particular numbers in this example?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 90 "2) A basis for the rowspace is st aring at you in the rref(A) matrix. What is this basis?" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 162 "3) You can back-s olve for the homogeneous system using rref(A), to get a basis for null space(A). Do this by hand on a sheet of paper, and write your answer \+ here." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 142 "4) You can get a basis for the column space of A by taking a subset o f your original 5 columns. Which columns do you choose, and explain w hy!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 243 "5 ) Another way to get a (better) basis for the column space is to do c olumn operations, putting your matrix into reduced column echelon form . You can do this in Maple by transposing, computing the rref of the \+ transpose, and transposing back:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 103 "rcef:=B->transpose(rref(transpose(B)));\n #a procedure for computing reduced\n #column echelon form" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "rcef(A);" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}} {PARA 0 "" 0 "" {TEXT -1 251 "So, what is a nice basis for the column \+ space of A? Pick any one of the original columns of A and express it \+ as a linear combination of this nice basis, to illustrate how easy it \+ is to do. (Do this by hand, and type in your answer and explanation.) " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 310 "6) \+ Maple will compute bases for these spaces with single commands. Com pare the answers you've gotten above with Maple's answers: Since you \+ know bases are not unique, you can pretty well guess that Maple is usi ng methods close to the ones we used, since its answers should be quit e similar to some of yours:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "rowspace(A);# Gives a basis for rowspace of A" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 29 "nullspace(A);#nullspace basis" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "colspace(A);#nice column space basi s" }}}{PARA 259 "" 1 "" {TEXT -1 440 "7) As we remarked in class, the theorem that the nullspace is perpendicular to the rowspace is kind o f clear once you realize that x being in the nullspace means that Ax=0 so that each row of A dots with x to give zero. In particular, if we take a basis for the rowspace of A, and also one for the nullspace of A, then elements of the first basis should all be perpendicular to el ements of the second one. Let's verify that this is true:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "r1:=row(rref(A),1); #r1,r2 basis fo r rowspace(A)\nr2:=row(rref(A),2);" }}{PARA 11 "" 1 "" {TEXT -1 0 "" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 81 "S:=convert(nullspace(A),li st); #making a list\n #keeps track of order in a set." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "n1:=S[1];n2:=S[2]; n3:=S[3];#nullsp ace basis. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 235 "rowmat:=sta ckmatrix(r1,r2); #stack matrices\n #or vectors one on top of each \+ other. We could\n #have gotten the same matrix by using the delet e\n #row command, i.e. delrows(rref(A),3..4).\n #these rows ar e a rowspace basis" }}{PARA 11 "" 1 "" {TEXT -1 0 "" }}}{PARA 11 "" 1 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 127 "nullmat: =augment(n1,n2,n3); #augment matrices or\n #vectors from left to r ight. These columns\n #are the nullspace basis" }}{PARA 11 "" 1 " " {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 105 "evalm(ro wmat&*nullmat); #compute all six dot products\n #in one matrix mul tiplication. You should get:" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%'mat rixG6#7$7%\"\"!F(F(F'" }}}{PARA 0 "" 0 "" {TEXT -1 408 "8) Since the \+ nullspace is 3-dimensional, and the rowspace is 2 dimensional, and sin ce they are perpendicular to each other in R^5, it seems likely that t he collection \{r1,r2,,n1,n2,n3\} is a basis for R^5. Use Maple and y our theoretical knowledge to verify that this is true, by using some t est which checks whether 5 vectors in R^5 form a basis. You have to t hink of the commands here and from now on. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 405 "9) Let us call E=\{e1,e 2,e3,e4,e5\}, i.e. the set of standard basis vectors in R^5. Let us c all our new basis T=\{r1,r2,n1,n2,n3\}. So, any vector v in R^5 can \+ be uniquely expressed as a linear combintation of the vectors in the T -basis, v=c1*r1 + c2*r2 + c3*n1 + c4*n2 + c5*n3, and the list of coeff icients [c1,c2,c3,c4,c5] are what we have been calling the coordinate s of v with respect to the T-basis." }}{PARA 0 "" 0 "" {TEXT -1 165 " \+ Find the coordinates of the following vectors ,with respect to the T-basis, by solving the appropriate linear systems. (Well, you can s olve 9a by observation.)" }}{PARA 0 "" 0 "" {TEXT -1 17 "9a) (1,0,-1, 2,3)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 16 "9 b) (1,0,0,0,0)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 34 "10) Find the transition matrices " }{TEXT 262 1 "P" } {TEXT 261 5 "E <-T" }{TEXT -1 5 " and " }{TEXT 263 1 "P" }{TEXT 264 5 "T <-E" }{TEXT -1 87 ", which take T coordinates to E-coordinates and \+ visa versa. Verify that the product " }{TEXT 269 2 "[P" }{TEXT 270 5 "T <-E" }{TEXT 273 1 "]" }{TEXT 272 1 "[" }{TEXT 271 1 "v" }{TEXT -1 9 "] yields " }{TEXT 274 1 "[" }{TEXT 278 1 "v" }{TEXT 279 1 "]" } {TEXT 276 2 "T " }{TEXT -1 20 "for the two vectors " }{TEXT 277 1 "v" }{TEXT -1 8 " in 9ab." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 333 " Maple can also do Gram-Schmidt. Make sure you use b rackets, [.] ,to enclose your list of vectors, rather than set symbols , \{.\}, if you want Maple to Gram-Schmidt your vectors in the order y ou list them. Here's how the command works, as applied to the 3-dimen sional subspace of R^5 which is the nullspace(A), for the matrix A abo ve" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 81 "step1:=GramSchmidt([n1 ,n2,n3]); #completes step 1, i.e. gets\n#orthogonal vectors." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 149 "S:=GramSchmidt([n1,n2,n3], \+ normalized);\n#normalizes results of step 1 to get unit vectors.\n#the result is somewhat messier than the unnormalized one." }}{PARA 12 "" 1 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 224 "11) Let N=[n1,n2,n3] be the nullspace basis you found ea rlier in this project, and let S be the orthonormal basis of the same \+ nullspace, which you have just constructed above using GramSchmidt. F ind the transition matrices " }{TEXT 265 1 "P" }{TEXT 267 4 "N<-S" } {TEXT -1 5 " and " }{TEXT 266 1 "P" }{TEXT 268 5 "S<-N." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 403 "12) Use Maple to v erify that the vector [8,1,13,1,1] is a null vector for A. Find the c oordinates of this vector with respect to the S basis and the N basis. (If you're just looking at the two bases, the N-coordinates of a vec tor are actually the easiest to find, but you can use the dot product \+ to find the coordinates with respect to an orthonormal basis. (There i s a dot product command in Maple.)" }}}{MARK "49 0" 75 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }