Linear Algebra (Honors)
Linear Algebra (Honors) MATH 115AH
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This 4 page Class Notes was uploaded by Kaylin Wehner on Friday September 4, 2015. The Class Notes belongs to MATH 115AH at University of California - Los Angeles taught by C. Manolescu in Fall. Since its upload, it has received 98 views. For similar materials see /class/177807/math-115ah-university-of-california-los-angeles in Mathematics (M) at University of California - Los Angeles.
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Date Created: 09/04/15
Math 115 AH Practice Midterm 2 Linear Algebra You have 50 minutes No books and notes are allowed 1 TrueFalse Circle the correct answer No justi cations are needed in this exercise 1 point each In the questions below7 V is a vector space over a eld IF 1 A linear transformation T V a V is invertible if and only if T is injective T 2 Assume A B E Mnm are invertible Then AB is invertible and AB 1 A lB l F We actually have AB 1 B lA l 3 Assume T V a V is invertible7 and BC are ordered bases for V Then T 1 T71 We actually have T 1 TA F 4 If A is a matrix7 then rankA rankAt T 5 If A E M3X3R satis es At 7A then A is not invertible T Use determinants 6 Similar matrices always have the same eigenvalues T 7 If A is an eigenvalue of an operator T V a V then each vector in EA is an eigenvector for T F Zero is not an eigenvector7 but lies in EA 8 If W1 W2 W3 Q V are subspaces such that V W1W2W3 and W1 Wg Wg 0 then V W1 69 W2 69 W3 Consider W1 W2 Xy plane and W3 Z aXis in R3 9 If V is an inner product space and 1 E V is such that ltvwgt 0 for all w E V then 0 1 T 10 If V is an inner product space and S Q V is a subset7 then the orthogonal complement of S is a subspace of V T 2 10 points Determine whether the matrix A lt35 31 E M2X2R is diagonaliz able The characteristic polynomial is t2 7 tr At det A t2 9 which does not split over R it has no roots Hence A is not diagonalizable 3 10 points Let V EUR be the real vector space of real polynomials of degree at most one Consider the linear functionals fhfg E V given by up 1plttgtda f2p p 0 Show that f1 and f2 form a basis for V For a polynomial pt at b we have f110 02 t b and f210 a We prove that f1 and f2 are linearly independent Indeed7 suppose there exist 041 042 E R with a1f1 cvng 0 Then a 041E b 042a 0 for all ab E R Choosing a 0b 1 we get 041 0 Choosing a Lb 0 we get 042 0 Thus f1 and f2 are linearly independent Since V is two dimensional7 so is V Two linearly independent vectors in a two dimensional space must form a basis MATH 115A MIDTERM SOLUTION Monday7 February 9th 2009 Exercise 1 10 points Bases Let F be a eld Find bases for the following subspaces of F5 justify your answer W1 01702703704705 6 F5 3 a2 as i 04 0 W2 a17a27a37a47a5 E F5 a1 13 a4 and a2 a5 0 What are the dimensions of W1 and W2 For W1 it clear that the map ltlgt F5 7 F de ned by 19017 12703704705 a2 as i 04 is a linear transformation and Nltlgt W1 Applying the dimension theorem7 it follows that dim W1 4 and a basis for W1 is 0 0 0 1 7 71 0 OHHMO l 0 0 51 0 7 07 0 0 0 l A vector x 6 W2 is necessarily of the form x 1l117 7b So dim W2 2 and a basis for W2 is OOHO 1 0 6217 1 0 l H Exercise 2 10 points Lagrange Polynomials Let F be a eld and n E N be an integer Denote by PnF the vector space of all polynomials P with coef cients in F such that degP S n The indeterminate will be denoted by t Let zo an E F be n1 distinct scalars For k E 07 7n7 de ne the polynomial 1t 7 960 1t 7 mm 7 n1 1t 7 96a 96k 7 760 96k 7 mm 7 n1 9 7 ml 1 LIN 1 Compute Lkzj for all j k E 0 n Then show that the subset L0 L1 Ln is a basis for A straightforward computation shows that Lkzj jk ie Lkxj1ifj k and 0 ifj 51 k Let now a0a1an be 71 1 scalars and assume that aOLO alLl anLn 0 the zero polynomial This implies in particular that for any j 0 0010 11111 anLnj 01le 07quot Consequently a0 a1 an 0 and the subset B L0L1Ln is linearly independent Since dim PnF n 1 it follows that B is indeed a basis for Let y0y1 yn E F be any scalars Use the previous question to show that there exists one and only one polynomial P E PnF such that Pzk yk for all k E 0 n Assume that such a polynomial P exists We may decompose P using the basis 6 Then there are scalars a0a1 an such that P Z aij j0 Necessarily yj Px ajLJxj 17 So P is necessarily unique and P does exist It suf ces to set P 2 ijj j0 D Exercise 3 10 points Applying the dimension theorem Let V be a nite dimensional vector space and let T V gt V be a linear transformation We assume that nullityT nullityT2 1 Using the assumption together with the dimension theorem prove that NT NT2 and RT RT2 The inclusion NT C NT2 is obvious Since dimNT dim NT2 it follows that NT NT2 Applying the dimen sion theorem twice for T and T2 we get dimV nullityT rankT nullityT2 rankT2
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