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Solutions 8 - MAS212 Linear Algebra I Exercises 8(with...

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Unformatted text preview: MAS212 Linear Algebra I Exercises 8 (with solutions) © Francis J. Wright, 2006 Assessed submission deadline 5pm on Friday lst December 2006 in the blue box on the ground floor. Your submission must have your FULL NAME and STUDENT NUMBER PRINTED CLEARLY at the top and separate sheets must be stapled together. Marks out of 100 are shown in square brackets after each question number. 1. [30] Let Vbe a vector space. The identity map on V, IdV: V—> V, is defined byIdV(v) = v for all v E V. If Q, 3' are two ordered bases for V, determine the change-of—basis matrixP = (IdV; Q, 3') and its inverse P_1 when V, (8, Q” are as follows: (a) V= R3, Q9= 6’3 (the standard ordered basis on R3),Q%' = (1, 0, 0), (1,1, 0), (1,1,1); (b) V= R4, $=(—1,1,1, 0),(1,—1,0,1),(1,0,—1,1),(0,1,1,—1), (8' =64 (the standard ordered basis on 03“). V Solution (a) Construct the change-of—basis matrixP from Q to Q'. Solving the vector equation Id(1, 0, 0) = (1, 0, 0) =a (1, 0, 0) +b (1,1, 0) +c (1, 1, 1) forthe scalar coefficientsa, b, c gives Id(1, 0,0)=(1,0,0)=1(1,0 , 0)+ 0(1,1,0)+0(1,1,1),andsimi1ar1y Id(0,1,0)(0,1,0)—1(1,0, 0) +1 (1,1,0) +0(1,1,1),and Id(0,0,1)= (0,0,1)=0(1,0 0—)1(1,1,0)+1(1,1,1),hence 1 —1 0 1 1 1 P := 0 1 —1 . To findP_1 use matrix inversion to giveP_1= 0 1 1 . Check2PP_1= 0 0 1 0 0 1 1 0 0 0 1 0 . 0 0 1 Or construct the change-of—basis matrix P_1 from 93' to 9%. Solving the vector equation Id(1, 0, 0) = (1, 0, 0) =a (1, 0, 0) +b (0,1, 0) +0 (0, 0,1) forthe scalar coefficientsa, b, c gives Id(1,0,0)=(1,0,0)=1 (1,0,0) +0 (0,1,0) +0 (0,0,1) andsimilarly Id(1,1,0)=(1,1,0)=1 (1,0,0) +1 (0,1,0) +0 (0,0,1),and Id(1,1,1)=(1,1,1)=1 (1,0,0) +1 (0,1,0) +1 (0,0,1),hence 1 1 1 P_1= 0 1 1 . Note that because the target basisQis the standard basis, the columns of the 0 0 1 change-of-basis matrix P_1 are just the vectors in the source basis (8'. This must always be the case. (b) Construct the change-of—basis matrix P from Q3 to B'. Id(—1,1,1, 0)=(—1,1,1, 0) =—1 (1, 0, 0, 0) +1 (0,1, 0, 0) +1 (0, 0,1, 0) +0 (0, 0, 0,1), etc. Hence, because the target basis W is the standard basis, the columns of the change-of—basis matrix P are the elements of Q, i.e. —1 1 1 0 1 —1 0 1 _1 _1 P := 1 0 1 1 .To findP either use matrix inversion to giveP = 0 1 1 —1 -1 i i A 3 3 3 3 i _i A i 3 3 3 3 d _ rt() or rocee as in a a. L a _L L p p 3 3 3 3 2 L i _L 3 3 3 3 (One way to compute P_1 is first to compute the matrix of minors of P, using (say) diagonal expansion for each minor, to give 1 1 —1 2 l l 2 —l —1 2 1 1 2 —l l 1 from which the adjoint of P is 1 —l —l —2 —1 l —2 —l —1 —2 l —l —2 —l —l 1 Then Laplace expansion about (say) the first row gives the determinant of P as —3 and transposing 2 the adj oint and dividing by the determinant gives the inverse as above.) 1 2 3 0 2. [20] The linear map on: R4 —> R3 is represented by the matrixA = (on; 64, 63) = 0 —l 2 1 , 2 1 4 3 where 63, 64 are the standard ordered bases on R3, R4. Using information from your solution to Question 1 above where appropriate (or otherwise) write down or calculate the following matrices: (a) P= (Idm, 64, 8) andP_1, where 8= (—1, 1, 1, 0), (1,—1, 0,1), (1, 0,—1,1), (0,1,1,—1); (b) Q=(161m,63,@),[email protected]=(1,0,0), (1,1,0), (1,1,1); (c) A'= (a; 61$) =QAP‘1. V Solution —1 1 1 2 —1 1 1 0 ()Ph 'P_1'Q '1(b) P1 1—121 P_1 1—1 01 a ere 1s 1n uestlon , so = — , = 3 1 2 —1 1 1 0—1 1 2 l 1 —l 0 1 1—1 1 —1 0 (b) Qhere isPin Question 1(a), so Q= 0 1 —l 0 0 1 (c) HenceA' = QA P‘1 = —1 1 1 0 1—101230 3—3—15 1 —1 0 1 01—10—121 =_2_2_2_2_ 1 0 —1 1 0 0 1 2 1 4 3 3 4 l 2 0 1 1 —1 3. [50] Regard the matrixA in Question 2 above as a linear map at : R4 —> R3 given by y =A x where x, y are column vectors. (a) Find a basis (if one exists) for ker(A) and extend it (on the left) to an ordered basis for the domain (R4). (b) Choose a basis for im (A) from the image of the basis for the domain (excluding the basis for the kernel) and, if necessary, extend it (on the right) to a basis for the codomain (R3). (c) Obtain an invertible 4 x 4 matrix P and an invertible 3 x 3 matrix Q such that QA P has the I 0 general form r , where Ir is the r x r identity matrix, r = rank(A ), and the zero submatrices O 0 represented by 0 may have different shapes and some may be absent. V Solution —2 t - I - I t - u (a) SolvmgA x = 0 gives the followmg solution 0 , where tis an arb1trary real number, t which represents the kernel as a one-dimensional vector subspace. Hence, a convenient choice of —2 1 basis (by choosing t= 1) is 0 . We can then extend this basis on the left to give 1 1 0 0 —2 0 1 0 1 , , , as an ordered basis for the domain (13“). 0 0 l 0 0 0 0 l (b) Mapping only the extension of the basis for the domain (thereby avoiding the kernel) gives a 1 2 3 spanning set for the image as 0 , -1 , 2 . These 3 vectors are obviously linearly 2 1 4 independent and so must form a basis for the whole codomain (R3). (0) The change-of—basis matrix in the domain from the new basis to the standard basis is the 1 0 0 —2 0 l 0 1 matrix whose columns are the new basis vectors, i.e.P= 0 0 1 0 . Similarly, the change- 0 0 0 1 of—basis matrix in the codomain from the new basis to the standard basis is the matrix whose -1 _i l 1 2 3 4 8 8 _ 1 1 1 columns are the new basis vectors, i.e. Q1= 0 —1 2 . Inverting gives Q= 3 —1 —1 2 l 4 i i _i 4 8 8 l 0 0 0 and hence QAP= 0 1 0 0 . Optional remark: the rank ofA is 3. 0 0 1 0 ...
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