Mid-term Exam #2 MATH 205, Fall 2014

Transcription

1 Mid-term Exam # MATH 05, Fall 04 Name: Instructions: Please answer as many of the following questions as possible Show all of your work and give complete explanations when requested Write your final answer clearly No calculators or cell phones are allowed This exam has 5 problems and 00 points Good luck! Problem Possible Points Points Earned TOTAL 00

2 (0 points) Consider the following two matrices and their reduced echelon forms: A , rref(a) , B , rref(b) (a) (4 points) Find rank A and rank B Solution: rank A 4andrankB (b) (8 points) Find a basis for Col A and state the dimension of Col A How does the subspace Col A compare to the subspace Col B? >< Solution: AbasisforColAis >:, , 6 4 5, 6 > 4 5 and >; 4 5 the dimension of Col A is 4 The columns of A and B both span all of R 4, so Col A ColB (c) (8 points) Find a basis for Nul B and state the dimension of Nul B Is there a relationship between the subspaces Nul B and Nul A? Give a one sentence explanation >< > Solution: AbasisforthenullspaceofB is 6 5 and the dimension of Nul B is There is no relationship between the null spaces of A 4 5 >: >; and B since Nul A is a subspace of R 4 while Nul B is a subspace of R 5 5

3 (0 points) This problem deals with P,thevectorspaceconsistingofallpolynomials with real coe cients of degree less than or equal to (a) (0 points) Determine whether each of the following sets is a basis for P Justify your answer in each case i A { +t, t+ t } ii B {, +t, t + t } Solution: i The set A is not a basis for P Thedimensionofthevectorsspaceis andtheseta has only vectors, therefore it cannot span P ii The set B is a basis for P TherearethreevectorsinBand dim P, so by the Basis Theorem it su ces to check if B is a linearly independent set The coordinate vectors of the set B withrespect to the standard basis E {,t,t } are 4 0 5, and 4 5 Puttingthesevectorsinto amatrixwehave A it is immediately obvious that the three columns are linearly independent Therefore B is a basis for P (b) (0 points) Choose a set from part (a) that is a basis for P Find the coordinates of p(t) 6 5t +t relative to that basis Solution: The only set that is a basis is B, soifind[6 5t +t ] B That is, I need to find c,c,c such that 5, 6 5t +t c () + c ( +t)+c ( t + t ) Mapping each polynomial to its coordinate vector relative to the standard basis E {,t,t },thisisequivalenttosolvingthematrixequation c c c To solve this, augment the matrix and row reduce, reduced echelon form Then c,c andc,so[6 5t +t ] B

4 0 (0 points) Let A 0 (a) (5 points) Write the characteristic equation of A and find all eigenvalues of A Solution: First, A I Then the characteristic equation of A is 0or( )( +)0 ThentheeigenvaluesofA are and (b) (0 points) Find a basis for each eigenspace Solution: : A + I, rref(a + I ) Then a basis for the eigenspace corresponding to : A I, rref(a I ) 0 0 is 0 0 Then a basis for the eigenspace corresponding to is (c) (0 points) Is A diagonalizable? If yes, find P, P, D such that A PDP Solution: The eigenvectors and are linearly independent and there are two of them, therefore A is diagonalizable Then P, P / / 0, D / / 0 (d) (5 points) Find A using part (c) Solution: A PD P,andD 0 0 Then D D, so 0 A PDP A by part (c) Therefore A Or we can 0 4

5 calculate A / / / / / / / / 5

6 4 (0 points) Let T : P! P be defined by T p(t) p 0 (t)+p(0) for all p(t) in P Here, p 0 (t) isthefirstderivativeofp(t) (a) (0 points) Show that T is a linear transformation Solution: Let p and q be vectors in P To begin, show that T (p + q)(t) T p(t) + T q(t) Wehave T (p + q)(t) p + q 0 (t)+ p + q (0) p 0 (t)+q 0 (t)+p(0) + q(0) p 0 (t)+p(0) + q 0 (t)+q(0) T p(t) + T q(t) Next let p be a vector in P and let c be a scalar Then Then T (cp)(t) ct p(t) T (cp)(t) cp 0 (t)+ cp (0) cp 0 (t)+cp(0) c p 0 (t)+p(0) ct p(t) (b) (5 points) Find a non-zero vector in the kernel of T Solution: I am looking for a polynomial p(t) ofdegreelessthanorequal to satisfying T p(t) 0,orp 0 (t) +p(0) 0 Let p(t) at + bt + c be an arbitrary element of P Then p 0 (t) at + b In order to satisfy p 0 (t)+p(0) 0, it must be that at + b + c 0 Thena 0andb c So a polynomial in the kernel of T will be p(t) t, for example (c) (5 points) Is T amappingontop (yes or no)? Give a one sentence explanation Solution: No, the mapping T is not onto P BythedefinitionofT,for every x in P,thepolynomialT(x) hasdegreeonelessthantheoriginal polynomial x Therefore no polynomial of degree is in the range of T Thus T does not map P onto P (d) (Bonus 5 points) Find a basis for the kernel of T and the dimension of the kernel of T Solution: From part (b), an element in the kernel of T has the form p(t) 0+bt b for b any real number Then a basis for the kernel of T is {t } and the dimension of the kernel of T is 6

7 5 (0 points) (a) (5 points) State the Rank Theorem Solution: The dimensions of the column space and the row space of an m n matrix A are equal This common dimension, the rank of A, also equals the number of pivot positions in A and satisfies the equation rank A + dim Nul A n (b) (5 points) Find a matrixwhosenullspaceisequaltoitscolumnspace, or explain why no such matrix exists Write your answer in complete sentences Solution: A matrixcannothaveitsnullspaceequaltoitscolumn space because of the rank theorem The matrix A has three columns This is an odd number so it is impossible for rank A and dim Nul A to be equal and sum to since both of those quantities are whole numbers