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# Elementary Matrix and Linear Algebra MATH 340

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This 14 page Class Notes was uploaded by Zechariah Hilpert on Thursday September 17, 2015. The Class Notes belongs to MATH 340 at University of Wisconsin - Madison taught by Staff in Fall. Since its upload, it has received 19 views. For similar materials see /class/205301/math-340-university-of-wisconsin-madison in Mathematics (M) at University of Wisconsin - Madison.

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Date Created: 09/17/15

MATH 340 SEPTEMBER 2008 REVIEW MIDTERM ll SOLUTIONS Problem 1 a 1 2 2 0 1 0 1 2 2 0 1 0 0 1 1 2 1 0 7gt 0 1 1 2 1 0 71 0 0 4 1 0 0 0 0 0 1 0 The associated equation is 11 212 213 0 12 13 214 0 and hence 13 and 14 are free variablesl Set 13 t7 I4 s and solve for 11 12 to get 12 7t 7 237 11 727t 7 2s 7 2t 43 Therefore the general solution is given by 11 437 12 7t 7 237 13 t7 I4 8 t7 8 arbitraryl b Write the solution in a form of column vectors 11 4s 4 0 12 7 7t 7 2s 7 72 71 13 t S 0 H 1 I4 8 1 0 4 0 72 71 0 7 1 1 0 This is a basis of the solution space c There are 2 elements in the above basis and so the dimension of the solution space is 2 Problem 2 a A basis is 17 t7 t2 b Consider the de ning equation of the linear independence 51t7102t2 1 6 and see if this polynomial equation has only the trivial solution for cl 52 By turning the equation into the standard form7 0272 clt 701 02 0 we obtain 02 07 cl 07 1 02 0 Therefore cl 02 0 is the only solution and hence t 7 17 t2 1 is linearly independent c Just add the constant polynomial 1 to the set S t 7 1t2 171 is a basis extending 5 Problem 3 a 11u111v1 b We compute 12 72 22 3 and u1u2 lgtlt10 X 72 1 X 2 3 Therefore we ave 3 1 1117112 cost9 7 7 7 HulHHWH xi X 3 and s09 g c We start with the formula l l wow H l to obtain an orthogonal basis Then we normalize them to obtain an orthonormal basis i 7 315 0 if i 31 s Problem 4 a See Problem 6 for the similar problemi 9W633 b We can write and hence a basis is given by Then we express Problem 5 2 73 l 0 0 1 Q 2gt2 0Stiol 5 73 1 We reduce A l 72 l to a row echelon form 3 l 7 l 72 l 0 1 7 0 0 a To nd a basis of the row space of A we just take the nonzero rows of the row echelon form 4 1721gtlto17gtl To nd a basis of the column space of A we nd the pivot columns of the row echelon form In this example the pivot columns are the lst and the 2nd columnsi Then we take the corresponding column vectors of the original matrix A which are 5 73 172 31 b To nd a basis of the null space we solve the equation associated to the above rowechelon form 11721241 13 0 12713 01 Setting the free variable 13 t we solve 4t 12 77 11 Writing this into the vector form we obtain 2 1 I1 7 7 12 t 13 t 1 Therefore a basis of the null space is given by H lpb lh c rankA 2 nulityA 11 d Knowing that 5 73 172 3 1 is a basis of the column space of A we check if the equation 5 73 4 cl 1 CQ 72 1 3 1 1 has a nontrivial solution The associated augmented matrix is 5 73 l 4 1 72 l 1 3 1 l 1 Its row echelon form is 1 72 l 1 o 1 1 7 0 0 l 71 Therefore the associated equation is inconsistent and so the vector is not in the column spacer Problem 6 a We have only to prove 11 1f A1 and A2 are in W then A1 A2 is also in W1 21 If A lt2 is in W then A ltZ1 21gt is also in W for all A 1 1 We start with By the de nition of W which is the sum of its 1 1entry and 2 2 being zero the matrices A1 A2 satisfy a1 d1 0 a2 d2 01 We v a1a2 b1b2gt A1A2 ltC1C2 d14rd2 lts l7 lentry and 272entry are a1 a2 and d1 d2 respectively We write their sum a1a2d1d2 a1d1a2d2 000 Hence A1 A2 is in Wi For 27 we note that the corresponding entries of AA are Aa Ad respectively So their sum becomes AaAdad A00i Therefore AA is in W for all A This shows that W is a subspace of M2X2Ri From the equation a d 07 we have d 7ai Other than this7 the choice of a7 b7 0 are completely free So W is the same as the space consisting of the matrices of the form A 0quot V which can be rewritten as l 0 b 0 l 0 0 a 0 71 0 0 C 1 0 l 0 0 l 0 0 0 7l 7 0 0 7 l 0 Problem 7 We consider the de ning equation 013 z 12 022 21 512 034 7 312 6 So we obtain a basis for the linear independence Writing it into the standard form 301 202 403 01 202z 01 502 7 303z2 6 we obtain 301 202 403 0 01 202 0 01 502 7 303 0 By solVing this equation7 we nd that it has nontriVial solution eigi7 01 727 02 17 03 l and so the set is linearly dependenti Problem 8 a To nd the transition matrix from S to T we need to express the elements of S v17 v2 as linear combinations of T 101 w27 which are already given as v1 7101 41027 112 5101 7 3102 PltTesgt lt11 33gt Ils If you were asked to nd MT we would apply the coordinate change formula lt3 33gt ltgt lt1 Warning You should note that I swapped the roles of letters 5 and T di erently from the way how the textbook used This of course is not important since they are just dummy variables Any other letters as you want can replace the letters Problem 9 Therefore the matrix should be b By the de nition of Ms we have a 6 The zero polynomial b Look at the problem 5 a to see how to do it MATH 340 REVIEW FOR MIDTERM I FALL 2008 Name39 Score Problem 1 Problem 2 W Problem 3quotW Problem 4quotW Problem 5 W Total Instruction Show all work No work no credit even if you have a correct answer References and calculator are not allowed Problem 1 20 points A and B are 2 gtlt 2 matrices7 where 2 4 1 0 A13 Bi2 3 Compute AB and AT 2B Here AT is the transpose of A Problem 2 Consider the following system of linear equations 1 722 1 21 742 33 1 31 762 33 0 1 5 points Write down the associated augmented matrix 2 15 points Use Gaussian elimination to nd solutions of the above system of equa tions Problem 3 20 points Let 10 71 A 12 0 01 1 Is A singular or not If A is nonsingular7 nd its inverse A l Problem 4 10 points Compute the determinant 71 2 3 1 2 3 1 3 0 2 0 0 0 0 2 3 Problem 5 10 points If the determinants a1 a2 as b1 b2 b3 C1 02 Cs Evaluate a1 b1a1 C1a1 2 b1 11 12 13 b2 b3 d1 d2 d3 a2 as 52 a2 bs as 02 d2 Cs as Math 340 Review for Midterm2 1 25 pt Let the matrix A be 1 2 A 0 1 71 0 a 10 pt Solve equation Ax 0 b 10 pt Find a basis for the solution space of the above equation c 10 pt Find the nullity of A OHM Agtl0 2 25 pt Let P2 be the vector space of all polynomials of degree 3 2 a 5 pt Write a basis for P2 b 10 pt Given a set S t71t21 in P2 Verify that S is linearly independent c 10 pt Extend S to a basis for P2 3 20 pt In the standard inner product space R3 let S 101T 1722T a 5 pt Find the length of the vector 131 101T b 5 pt Find the angle 6 between the vectors 131 101T and 132 1 72 2T c 10 ptUse the Gram Schmidt process to obtain an orthonormal basis of the subspace W SpanS Span101T 1722T 4 20 pt Let M22 be the real vector space of all 2 gtlt 2 matrices a 10 pt ls the subset W in M22 consisting of matrices lb 2 J a subspace of M22 Why Here 11 are real numbers pt 1s a su space n a asis or en express 7 as a b10 lfW b db fWTh 23 linear combination of the vectors in the basis 5 Consider the matrix 5 73 1 A 1 72 1 3 1 71 a Find bases of the row space and column space of A respectively b Find a basis of the null space of A c Find the rank and the nullity of A 4 d Check if the vector 1 is in the column space of A 1 6 Consider the set W of all rnatrices Z 2 such that a d 0 a Prove that W is a subspace of M2X2R the set of 2 gtlt 2 matrices b Find a basis of W and its dimension of W 7 Check if the following set of polynomials is linearly independent 3zz2 22z5z2 473952 1 8 Let S 171172 and T 131162 be bases for a vector space V7 and suppose that 171 71171 41627 172 5161 7 31172 a Find the change of coordinate matrix from S to T b Find the coordinate vector flg for f 5171 3172 9 Answer the following questions You do not have to explain a Write down the zero vector in the vector space l of polynomials of degree less than equal to n b Let A be the matrix 1 i4 9 77 A 71 2 i4 1 5 i6 10 7 Its row echelon form is give by 1 0 71 5 R 0 72 5 i6 0 0 0 0 Find bases of the column space and the row space respectively Glossary You should know at least if not all the de nitions of all these terms and understand what they stand for Coef cient matrix7 Augmented matrix Row echelon matrix7 Reduced row echelon matrix7 Elementary row operations Back substitutions7 basic variables7 free variables Vector spaces7 zero vector7 addition7 scalar multiplication Subspace7 de nition of subspaces linear combination7 linear independence7 span7 basis7 dimension homogeneous equation coordinates7 linear transformation7 isomorphism7 change of basis7 transition ma trix j column space7 row space7 rank and nullity of a matrix k inner product de nition7 examples7 Cauchy Schwarz inequality7 triangle in equality7 l orthogonal7 orthonormal basis7 Gram Schmit process AA AAAAA A tron C lt17 CL 0 C7 D VVVVVVVVV MATH 340 SEPTEMBER 2008 REVIEW MIDTERMl SOLUTIONS Probleml AB7 721472 72043 7 71012 7 ll372 1033 7 75 9 T 7 721 2 0 7 01 A2B7lt43746709 Problem2 1 1 72 0 l 71 2 74 3 l 1 3 76 3 l 0 2 We apply the following elementary operations to the above system iR3 R3 7 3R1 iiR2 R2 7 2R1 iiiR3 R3 7 R2 iiR2 13R2l Then we get the system in RlRlElF 172071 0011 ooolo Therefore 12 is a free variable and 11 13 are basic variables and the associated equation is 11 7 212 7l 13 1 We set 12 t and then we obtain the solutions 12 tzg l 11 2t 71 t arbitrary In particular there are in nitely many solutions namely one for each value of 12 Problem 3 To see if A is singular or not we can calculate its determinant and decide We could also calculate the RlRlElF of A and if it is the identity then A is invertible Welll do it in the second way since in case A is invertible we can take the identity matrix and apply to it the elementary operations that we did to A to nd its RlRlElFl and this will give us A ll To this purpose consider the augmented system All 10 100 120010 01 001 and apply the following elementary operations iR2 R2 7 R1 iiR2 R2 7 R3 iiiR3 R3 7 R2 iiR1 R1 Rgl We get Therefore A is invertible with inverse Do not forget to Check your answer is correct by multiplying this to A and to get the identity Problem4 712 00 320 220 2320 02 12 02 32 3102 18 2i 8 1lt3l03l 2l3 3lgt 2lt2l0 3l 2l13lgt 13 03 772 a3 a a2 7 22 Problem 5 Since the determinant is linear for each row using the second row we have that al a2 a3 a1 a2 as a1 a2 as a1 171 a2 112 a2 bs a1 a2 a2 1 1 b2 b2 Clidl C27d2 Csids Clidl C27d2 Csids Clidl C27d2 Csids Since the rst matrix in the right hand side has two repeated rows its determinant is 01 On the other hand using again the linearity of the determinant in rows this time with the third row we have that a1 a2 as a1 a2 as a1 a2 a3 171 b2 b2 b1 b2 b2 7 b1 b2 b2 2 7 71 3 cl 7 d1 CQ 7 d2 Cg 7 d3 cl Cg d3 d1 d2 d3 hence a1 a2 a3 a1bi a2b2 a2bs 3 517d1 527d2 Csids Glossary You should know at least not all the de nitions of all these terms and understand what they stand for Coef cient matrix Augmented matrix H matrix column vector row vector matrix addition product linear combination Transpose of the matrix ABT BTAT Row echelon matrix Reduced row echelon matrix Elementary row operations P FWN Back substitutions basic variables free variables 7 De nition of invertible matrices inverse lnverse of elementary matrices Proofs of the identities AB71 B7114717 AT71 A71 T 81 Determinant Properties of determinants reduction to triangular matrices Determinant of triangular matrices 50 101 Cramer s rule adjoint A adj A

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