Solutions to Homework Section 3.7 February 18th, 2005

Transcription

1 Math 54W Spring 5 Solutions to Homeork Section 37 Februar 8th, 5 List the ro vectors and the column vectors of the matrix The ro vectors are The column vectors are ( 5 5 The matrix ( (,,,, 4, (5,,,, ( ( ( 4 5 ( 4 is in ro echelon form Find a basis for its ro space, find a basis for its column space, and determine its rank Since A is alread in ro echelon form, its nonzero ros form a basis for RS(A b Theorem 37 Since all of the ros are nonzero, a basis for RS(A is (, 4, 3,, (,,, 7, (,, 4,, (,,,5 For the column space, e use Theorem 373, hich sas that the column vectors containing pivots form a basis for CS(A Since ever column has a pivot, a basis for CS(A is 4 3,, 4, We have This is ro equivalent to U = 7 A basis for RSU 7 consists of the vectors (3,, and (,, 7 Since RSU =RSA, these also constitute a basis for RSA The first to columns of U constitute a basis for CSU Thus, the first to columns of A, namel (3, 6,, and (, 3,,, constitute a basis for CSA Since all the bases here contain to elements, e see rk Note that V = Span{(, 4,,4,(4,,3,,(,6, 4, } = RSA, here A is ro equivalent to U = 5 3 Thus a basis for V = 6 4 RSA consists of the vectors (, 4,, and (,, 5,3

2 is ro equivalent to U = Thus 5 5, ith basis NS {( 7 (5t + s 4t s, (5t + s, t, s t, s R} 7 B = {( 7, 5,,, ( 5 7 7,,, } 7 This shos dimns Since U has to pivots, e see rk Sure enough + = 4 = n in this case In exercises -4, determine if b lies in the column space of A If it does, express b as a linear combination of the columns of A ( 4 6 ( 4, b = 6 The second column of A is 3 times the first, so ( ( x CS( Span{ } = { x R} 4 4x Since b cannot be expressed in the form 4 3, b = ( x 4x, it does not lie in the column space The vector b lies in the column space if and onl if b can be ritten as a linear combination b = = a + b + c 3 for scalars a, b and c So e have to solve the sstem of equations a + b c = a + b + c = a + 3b + c = A bit of ro reduction tells us nope, the sstem is inconsistent, so b is not in the column space Shortcut: For future reference, notice that the matrix associated to the sstem as just A itself, augmented b the vector b So if ou ant to save time, skip the first to steps and jump right into the ro reduction

3 39 Let Find bases for RS, NS, CS and LNS Find the rank of A and verif that dimrs+dim NS = n, dimcs + dimlns = m Since A is alread in ro echelon form, a basis for the ro space is given b the nonzero ros of A: (, 3,, 4, (,, 5, Since the ro space has to basis vectors, A has rank For the null space, set up a sstem of equations and rite everthing in terms of the free variables: x NS( {x = z R4 Ax = } This tells us that = { = { x z = { R4 5z + =, x + 3 z + 4 = } /5 /5 + z, R4, R} /5 /5 /5 /5, R} is a basis for NS(A, and e can no verif that dimrs + dim NS = + = 4 A basis for the column space consists of the columns of A hich have pivots:,, 5 Finall, a ro vector x = (x,, z lies in the left null space LNS(A if and onl if 3 4 (x,, z 5 = (,,,, 3

4 and this happens if and onl if x =, 3x =, x + 5 =, 4x + = In the solution to this sstem, z is a free variable and x = = Thus the left null space consists of all vectors of the form (,, z This is a one-dimensional space ith basis (,, We can no verif dimcs + dimlns = + = 3 43 True or false? [(a] If A is an n n matrix, then the ro space of A is equal to the column space of A False The matrix ( has ro space spanned b (, and column space spanned b ( These are not the same [(b] Even if A is square, the column space of A can never equal the null space of A False The matrix ( ( has CS( NS( { R} [(c]if A is an m n matrix and the columns of A are linearl independent, then Ax = b ma or ma not have a solution But if it has a solution, that solution is unique True Since the columns of A are linearl independent, the ro echelon form of A must have a pivot in ever column, so there are no free variables associated to the sstem Ax = b [(d]a 3 4 matrix never has linearl independent columns True Four vectors in R 3 can never be linearl independent [(e]a 4 3 matrix must have linearl independent columns False The zero matrix doesn t have linearl independent columns As ou can see, the zero matrix is ver useful for producing counterexamples! 44 We consider A as a collection of n columns in R m (a If these vectors are linearl independent, then the form a basis of CSA, in hich case rk dimcs n We must have n m since ou cannot have more than m linearl independent vectors in R m (b If these vectors span R m, e have b definition CS R m, and hence rk dimcs dimr m = m In this case, n m since one cannot have feer than m vectors spanning R m (c If these vectors form a basis of R m, then both (a and (b hold, in hich case n = m = rka, and e see that A is a square, invertible matrix In Ex 48 e suppose A is an nxn matrix and has a right inverse B such that AB = I [WARNING: We do not assume that A is invertible, as e are not told hether B I In fact, this is exactl hat e set out to prove!] 4

5 48 (a To sho that CS R n, it is enough to sho that given an v R n, e can find an x such that Ax = v (cf Ex 33 But notice that v = Iv = ABv = A(Bv Thus, setting x = Bv, e see that Ax = v, and e are done (b Since rk dimcs dimr n = n, e see A is invertible b 383d (c Since A is invertible, there exists a matrix A such that A AA = I Take the equation AB = I Multipling both sides on the left b A, e get B = A I = A, proving the claim 5