University:
California State University, Northridge
Course:
MATH 262 | Introduction to Linear Algebra
Academic year:

2023
Views:

420

Pages:

3
Author:

anon

100NAME:M262 Linear Algebra : Eigenstuff Quiz.Please show all your work, unless explicitly instructed not to do so.1. Consider the following Linear Transformation (you may assume that it is a linear transformation).T:R3->30In each of the following refer to the information above. Justify your answers.(a) (10 points) Let 21 = be the standard basis for R3. Find [T], (that is the matrix of T with respect tothe standard basis for R3).(T]2= 100(b) (10 points) Find RREF of [T]a.RREF: [(c) (10 points) Use part (b) to compute the rank and nullity of [T]a?rank of [T]o = 2rank + nullity = dim R32 nullity = 3 I nullity of (i)==+(d) (10 points) Find a basis for the kernel of T.()=basis for the bernel of T={[i]} 2. Still usingT:R3->3(a) (10 points) Confirm that the characteristic polynomial of T isA==(1-A) -(b) (10 points) One eigenvalue is 1. What is its algebraic multiplicity? What is the other eigenvalue of T?det(A-2I)= 0 11=1, multiplicity-=-(1-2)2(2)= 0A=1,2=02=0, multiplicity(c) (10 points) Find a basis for the eigenspace associated to A=1.ker (A-AI), A=1ker (A-I)0x2=03=0x1 anythingBasis for E,= {[|]] Still usingT:R3->3X1-X2I2I30(d) (10 points) Is T is diagonalizable? If SO find a basis B for R3 consisting of eigenvectors for T:R3-> R3 andgive[T]q. If not state why no such basis exists.E1= span {C:1} ,Eo=span {Ci]}TEigenbajis = E[8I,C9J}T is not diagonalizable since we onlyhave two linearly independent ciger vectors(e) (10 points) Finddet[T], = (dot[[]2)= (1.1.0)"=0(f) (10 points) Is T invertible?T is not invertible becausedet[[]2 =0