18.06 Problem Set 4 Due Wednesday, Oct. 11, 2006 at 4:00 p.m. in 2-106
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1 8. Problem Set Due Wenesy, Ot., t : p.m. in - Problem Mony / Consier the eight vetors 5, 5, 5,..., () List ll of the one-element, linerly epenent sets forme from these. (b) Wht re the two-element, linerly epenent sets? () Fin three-element set spnning subspe of imension three. Cn you fin three-element set spnning subspe of imension two? One? Zero? () Whih four-element sets re linerly epenent? Explin why. Solution {[ () Only the zero vetor,, is linerly epenent by itself. (b) Only the sets ontining 5. 5 n some other vetor.? We n reple one of these by {[ () How bout 5 to get imension. There re only two vetors on eh line (= -D subspe), so we n t hve three-element set {[ spnning only -imensionl subspe unless we llow uplites; if we o, then something like works. For zero imensions ( point!), the only wy is three opies of the zero vetor () All four-element sets re linerly epenent in three-imensionl spe. 5. Problem Mony / Consier the mtrix A = () Whih vetors (b) Whih vetors () For b 7 Solution 5= b 7 b () All liner ombintions of b. 5 will mke the olumns of A linerly epenent? 5 will mke the olumns of A bsis for 5, ompute bsis for the four subspes. 7 5 n 7 5. x y z w : y + w =? (b) All vetors whih re not liner ombintions of these, n whih stisfy b+ =. For exmple,
2 b = () Sine. For the set of ll possible exmples, A + = 5 + B + C : A., we know the first two olumns spn the olumn spec(a) n [ T is in the nullspe N(A). Sine the first two olumns re linerly inepenent, we onlue the first two olumns re bsis for C(A), so A hs rnk n [ T is bsis for the one-imensionl N(A). You n o the sme for A T : the fourth row is the negtive of the first, so the seon n thir row sum to three times the first, so 7 5 is in N(A T ), n 7 5 is in N(A T ). These re linerly inepenent, so the left nullspen(a T ) hs t lest two imensions. But the rnk of A is so tht s everyboy! Those two linerly inepenent elements form bsis for the left nullspe. Now o the row spe C(A T ): Any two bsis elements will o, sy the first n seon rows, or the first n thir rows {(but not the seon n fourth rows, beuse they re linerly epenent!) So one bsis woul be [ T T}. Of ourse, you n lso o elimintion. Problem Mony / Do Problem #5 from setion.5 in your book. Solution () These re linerly inepenent: the only liner ombintion x (,,)+x (,,)+x (,,) = (,,) is x = x = x =. (You n hek this by writing this liner ombintion s the system of equtions Ax = n then solving by elimintion to get x =.) (b) These re linerly epenent: (,,) + (,, ) + (,,) = (,,). Problem Mony / Do Problem # from setion.5 in your book. Solution {[ () The pivots re in the first two olumns, so one possible bsis for C(A) is. (Notie they re ifferent!) C(U) is {[ (b) Both A n U hve the sme nullspe N(A) = N(U), with bsis {[. () Both A n U hve the sme row spe C(A T ) = C(U T ), with bsis {[ Problem 5 Mony / n for.
3 Do Problem #7 from setion.5 in your book. Solution 5 I use the bsis {,x,x,x } for the first prt, n { x,x,x } for the seon prt. Problem Mony / Do Problem #9 from setion. in your book. Solution () Elimintion gives (b) Elimintion gives [ b b b b b b b b b b b b [ b b b b b b b b b b b b + b b, so N(A T ) hs bsis, so N(A T ) hs bsis {[.,. Problem 7 Mony / Consier the mtrix A = [ [. () Wht is the rnk of A? Wht re the imensions of the four subspes? (b) Give bsis for eh of the four subspes. () Now, for eh of the four subspes, fin the set of equtions tht ll vetors in the subspe must stisfy. (For exmple, if Ax = b for some x, wht re the onitions on the omponents b i of b?) () Give the omplete solution to A T y = Solution 7. () A hs rnk, sine the left ftor (ll it L) is invertible, n the right ftor (R) hs rnk. So the subspes hve imensions: r = for the row spe n olumn spe; n r = for the nullspe, n m r = for the left nullspe. (b) It s esy to give bses for the four subspes of R. Look t wht tht sys bout the four subspes of A: The olumn spe: A s olumns re just L times the olumns of R. Sine bsis for C(R) is {[ { {[, the orresponing bsis for C(A) must be L[,L[ = The nullspe: Ax = if n only if Rx =, so N(A) = N(R) =,. The row spe: Sine N(A) = N(R), the row spes (their orthogonl omplements!) must lso be equl: C(A T ) = C(R T { ) = [ T T}. The left nullspe: You n o this by the metho of Worke Exmple.A, the metho you use {[ bove in Problem #9, or by orthogonlity to the olumn spe. The bsis for N(R T ) is { [ so the bsis for N(A T ) is (L ) T =. {[.,
4 () Here re equtions efining eh of the four subspes. (Note tht there re other possible sets of equtions, whih n be got by tking liner ombintions of the given equtions.) The olumn spe: b + b + b =, by elimintion on [ A b. The row spe: + =, + = ; I use orthogonlity to the nullspe, but you oul use elimintion on [ A T. The nullspe: The equtions efining Ax = re just the rows of A: x +x +x =,x +x =. The left nullspe: equtions re the olumns (pivot olumns suffie): y +y +y =,y y =. () One we hve prtiulr solution (try y p = ), the omplete solution is just y = y p + N(A T ). (I got this prtiulr solution by bk-substitution: A T y = b beomes R T L T y = b. So let L T y = z n solve R T z = b (solution: z = ), n then fin y.) Problem 8 Mony / [ Using Mtlb, tke some rnom -by- mtries (try using the rn(m,n) funtion) n look t their four subspes. (A onvenient wy to lulte the subspes is the fourbse.m tehing oe; type in type fourbse t the Mtlb prompt for informtion on how to use it. ) Wht re the imensions of the four subspes for typil -by- mtrix? Cn you explin why? (Hint: wht re the os pivot is extly zero?) Now try -by-5 mtries. Wht re the imensions of the four subspes now? Now guess wht imensions the four subspes of rnom m-by-n mtrix will most likely hve. Solution 8 Here s one mtrix: >> [R,N,C,L = fourbse(rn(,5)) R = N = C = If you nee to ownlo the file fourbse.m from the Web site, on t forget to put it in the urrent iretory where Mtlb n fin it. [
5 L = Empty mtrix: -by- Unless we re very unluky, we lwys get similr results: the row n olumn spe hve imension, the nullspe hs imension, n the left nullspe hs imension (ontining only the zero vetor). (In other wors, lmost ll rnom mtries hve full rnk.) The hne tht the number I pik for ny igonl entry A(i:i) is extly zero fter elimintion is negligible, so we n hoose the igonl entries s our pivots. Problem 9 Friy / () Fin n iniene mtrix A for the grph bove. (b) Fin one solution to Ax = n two linerly inepenent solutions to A T y =. () Wht onitions on the omponents of b o we nee for Ax = b to hve solution? Tell whih of Kirhhoff s lws this illustrtes. Wht re the urrents? Wht re the voltges? () Compute A T A. You get positive numbers on the igonl these numbers ount the number of eh noe hs. When re the off-igonl entries, n when re they zero? (e) Wht is N(A T A)? Why oes A T Ax = f hve solution only when f + f + f + f =? Solution 9 () A = (b) A = n A T = A T =, for exmple. (Sine A hs rnk, these re bsis!) () b is in the olumn spe C(A), whih is orthogonl to the left nullspe N(A T ). Just hek bsis vetors: b T =, b T =, or in other wors b + b + b 5 =, b + b + b 5 =. This is Kirhhoff s voltge lw: the voltge hnge b i on eh ege i sums to zero roun eh loop. Here, the omponents of x re the voltges t eh noe, n the omponents of b re the voltge Suppose you ve lrey pike the erlier entries of A, n then you pik A(i:i) t rnom let s sy there s less thn hne A(i:i) will be ny prtiulr. (In symbols, Pr(A(i:i) = ).) Now o elimintion: this subtrts some multiple of eh of the previous rows from row i, to mke A(i:) through A(i:i-) zero. You hven t even looke t entry A(i:i) yet, so whtever you subtrt from A(i:i) only epens on the erlier entries, not on A(i:i). So A(i:i) isn t pivot if n only if A(i:i) equls tht prtiulr. An tht mens Pr(A(i:i)isn t pivot) = Pr(A(i:i) = ). 5
6 ifferenes long eh ege. Voltge ifferene is proportionl to urrent flow (remember V = ir from 8.?) so the omponents of b represent urrents through eh ege. () A T A =. The igonl entries ount the number of eges eh noe hs. The off-igonl entries s ij re - if noes i n j re onnete by n ege, n if not. (e) N(A T A) = N(A) is gin the set of multiples of 7 5. Sine A T A is symmetri, the left nullspe is the sme! When A T A = f hs solution, f is in the olumn spe C(A T A), whih is orthogonl to the left nullspe: so f T 7 5=.
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